Basis of a subspace

WEBVTT
Kind: captions
Language: en

00:00:00.590 --> 00:00:04.180
Let's say I have
the subspace v.

00:00:04.180 --> 00:00:07.730
And this is a subspace and we
learned all about subspaces in

00:00:07.730 --> 00:00:09.510
the last video.

00:00:09.510 --> 00:00:13.240
And it's equal to the span
of some set of vectors.

00:00:13.240 --> 00:00:15.910
And I showed in that video that
the span of any set of

00:00:15.910 --> 00:00:17.710
vectors is a valid subspace.

00:00:20.310 --> 00:00:24.910
It's going to be the span of v1,
v2, all the way, so it's

00:00:24.910 --> 00:00:26.030
going to be n vectors.

00:00:26.030 --> 00:00:30.100
So each of these are vectors.

00:00:30.100 --> 00:00:33.920
Now let me also say that all of
these vectors are linearly

00:00:33.920 --> 00:00:35.970
independent.

00:00:35.970 --> 00:00:43.850
So v1, v2, all the way to vn,
this set of vectors are

00:00:43.850 --> 00:00:45.100
linearly independent.

00:00:50.370 --> 00:00:52.800
Now before I kind of give you
the punchline, let's review

00:00:52.800 --> 00:00:54.600
what exactly span meant.

00:00:54.600 --> 00:01:00.280
Span meant that this set, this
subspace, represents all of

00:01:00.280 --> 00:01:02.380
the possible linear
combinations of

00:01:02.380 --> 00:01:04.800
all of these vectors.

00:01:04.800 --> 00:01:08.060
So you know, I could have all of
the combinations for all of

00:01:08.060 --> 00:01:08.880
the different c's.

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So c1 times v1 plus c2 times v2,
all the way to cn times vn

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for all of the possible c's
and the real numbers.

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If you take all of the
possibilities of these and you

00:01:24.140 --> 00:01:28.610
put all of those vectors into
a set, that is the span and

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that's what we're defining
the subspace v as.

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Now, the definition of linear
independence meant that the

00:01:35.810 --> 00:01:44.690
only solution to c1, v1, plus
c2, v2 plus all the way to cn,

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vn, that the only solution to
this equally the 0 vector--

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maybe I should put a little
vector sign up there-- is when

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all of these terms
are equal to 0.

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c1 is equal to c2, is equal
to all of these.

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All of them are equal to 0.

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Or kind of a more common sense
way to think of it is that you

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can't represent any one of these
vectors as a combination

00:02:07.670 --> 00:02:08.990
of the other vectors.

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Now, if both of these conditions
are true that the

00:02:13.660 --> 00:02:17.430
span of this set of vectors is
equal to this subspace or

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creates this subspace or it
spans this subspace, and that

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all of these vectors are
linearly independent, then we

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can say that the set of
vectors-- maybe we call this,

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we could call this
set of vectors s.

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Where we say s is equal to v1,
v2, all the way to vn.

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It's equal to that
set of vectors.

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We can then say and this
is the punchline.

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We can then say that S, the
set S is a basis for v.

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And this is the definition
I wanted to make.

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If something is a basis for a
set, that means that those

00:03:01.790 --> 00:03:05.230
vectors, if you take the span
of those vectors, you can

00:03:05.230 --> 00:03:09.390
construct-- you can get to any
of the vectors in that

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subspace and that those
vectors are linearly

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independent.

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So there's a couple of ways
to think about it.

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One is there's a lot of things
that might span for something.

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For example, if this spans for
v, then so would-- let me add

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another vector.

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Let me define another set.

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Let me define set T to be
all of set S: v1, v2,

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all the way to vn.

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But it also contains
this other vector.

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I'm going to call it the
v special vector.

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So it's going to be essentially,
the set S plus

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one more vector.

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Where this vector I'm just
saying is equal to v1 plus v2.

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So clearly, this is not a
linearly independent set.

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But if I had asked you what the
span of T is, the span of

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T is still going to be
this subspace, v.

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But I have this extra vector
in here that made it

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non-linearly independent.

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This set is not linearly
independent.

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So T is linearly dependent.

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So in this case, T is
not a basis for v.

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And I had showed you this
example because the way my

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head thinks about basis is,
the basis is really the

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minimum set of vectors that I
need, the minimum set-- and

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I'll write this down.

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This isn't a formal definition,
but I view a

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basis-- let me switch colors--
as really the-- let me get a

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good color here.

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As a basis is the minimum-- I'll
put it in quotes because

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I haven't defined that.

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The minimum set of vectors that
spans the space that it's

00:05:07.470 --> 00:05:10.810
a basis of, spans
the subspace.

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So in this case, this is the
minimum set of vectors.

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And I'm not going to prove it
just yet, but you can see

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that, look.

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This set of vectors right
here, it does span the

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subspace, but it's clearly not
the minimum set of vectors.

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Because the span of this thing,
I could still remove

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this last vector here.

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I could still remove that guy
and still-- and then the span

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of what's left over is still
going to span my subspace v.

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So this guy right here
was redundant.

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In a basis, you have
a no redundancy.

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Each one of these guys is needed
to be able to construct

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any of the vectors in
the subspace v.

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Let me do some examples.

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So let's just take some
vectors here.

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Let's say I had to find
my set of vectors, and

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I'll deal in r2.

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So let's say I have
the vector 2, 3.

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Let's say I have the
other vector 7, 0.

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So first of all, let's just
think about the span, the span

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of this set of vectors.

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This is a set of vectors.

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So what's the span of S?

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What's all of the linear
combinations of this?

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Well, let's see if
it's all of r2.

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So if it's all of r2 that means
the linear combination

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of this could be-- we could
always construct anything in

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r2 with the linear combination
of this.

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So if we have c1 times 2,
3 plus c2 times 7, 0.

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If it is true that this spans
all of r2, then we should be

00:07:04.240 --> 00:07:07.280
able to construct-- we should
always be able find a c1 and a

00:07:07.280 --> 00:07:11.140
c2 to construct any
point in r2.

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And let's see if we
can show that.

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So we get 2c1 plus 7c2
is equal to x1.

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And then we get 3c1 plus 0c2.

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Plus 0 is equal to x2.

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And if we take this second
equation and divide both sides

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by 3 we get c1 is equal
to x2 over 3.

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And then if we substitute that
back into this first equation,

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we get 2/3-- I'm just
substituting c1 in there.

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So 2/3 x2.

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2 times x2 over 3 is 2/3 x2.
plus 7c2 is equal to x1.

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And then, what can we do?

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We can subtract the 2/3
x2 from both sides.

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Let me do it right here.

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So we get 7c2 is equal
to x1 minus 2/3 x2.

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Divide both sides of this
by 7 and you get c2.

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Let me do it in yellow.

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You get c2 is equal to x1 over
7 minus 2 over 21 x2.

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So if you've given me any x1 and
any x2 where either x1 or

00:08:29.670 --> 00:08:34.980
x2 are a member of the real
numbers, we're talking about--

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well, everything we're going to
be dealing with right now

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is real numbers.

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You give me any two
real numbers.

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I take my x2 divided by 3 and
I'll give you your c1.

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And I'll take the x1 divided
by 7 and subtract

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2/21 times your x2.

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And I'll get you your c2.

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This will never break.

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There's no division
by any of these.

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You don't have to worry about
these equaling a 0.

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These two formulas
will always work.

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So you give me any x1 and any
x2, I can always find

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you a c1 or a c2.

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Which is essentially finding a
linear combination that will

00:09:04.730 --> 00:09:06.070
equal your vector.

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So the span of S is r2.

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Now the second question is, is
are these two vectors linearly

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independent?

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And linear independence means
that the only solution to the

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equation c-- let me
switch colors.

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The only solution to the
equation c1 times the first

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vector plus c2 times the second
vector equaling the 0

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vector, that the only solution
to this is when both

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of these equal 0.

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So let's see if that's true.

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We've already solved for it, so
if x1-- in this case, x1 is

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equal to 0 and x2
is equal to 0.

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This is just a special
case where I'm making

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them equal to 0 vector.

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If I want to get the 0 vector,
c1 is equal to 0/3.

00:10:01.340 --> 00:10:04.140
So c1 must be equal to 0.

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And c2 is equal to 0/7
minus 2/21 times 0.

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So c2 must also be equal to 0.

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So the only solution to this
was settings both of these

00:10:15.540 --> 00:10:20.040
guys equal to 0.

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So S is also a linearly
independent set.

00:10:29.700 --> 00:10:33.240
So it spans r2, it's linearly
independent.

00:10:33.240 --> 00:10:37.610
So we can say definitively, that
S-- that the set S, the

00:10:37.610 --> 00:10:46.280
set of vectors S is
a basis for r2.

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Now, is this the only
basis for r2?

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Well I could draw a trivially
simple vector, set of vectors.

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I could do this one.

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Let me call it T.

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If I define T to be the set 1, 0
and 0, 1, does this span r2?

00:11:10.290 --> 00:11:13.970
Let's say I want to generate
the-- I want to

00:11:13.970 --> 00:11:16.420
get to x1 and x2.

00:11:16.420 --> 00:11:19.820
How can I construct that out
of these two vectors?

00:11:19.820 --> 00:11:27.850
Well, if I always just do x1
times 0, 1 plus x2 times 0, 1,

00:11:27.850 --> 00:11:31.970
that'll always give me x1, x2.

00:11:31.970 --> 00:11:34.380
So this definitely
does span r2.

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Is it linearly independent?

00:11:41.670 --> 00:11:43.260
I could show it to you.

00:11:43.260 --> 00:11:45.880
If you wanted to make this
equal to the 0 vector.

00:11:48.630 --> 00:11:51.900
If this is a 0 and this is 0,
then this has to be a 0 and

00:11:51.900 --> 00:11:53.330
this has to be a 0.

00:11:53.330 --> 00:11:54.360
And that's kind of obvious.

00:11:54.360 --> 00:11:56.810
There's no way that you could
get one of the other vectors

00:11:56.810 --> 00:11:58.700
by some multiple of
the other one.

00:11:58.700 --> 00:12:00.810
There's no way you could get a
1 here by multiplying this by

00:12:00.810 --> 00:12:02.640
anything and vice versa.

00:12:02.640 --> 00:12:04.225
So it's also linearly
independent.

00:12:09.960 --> 00:12:12.260
And the whole reason why I
showed you this is because I

00:12:12.260 --> 00:12:16.300
wanted to show you that look,
this set T spans r2.

00:12:16.300 --> 00:12:19.620
It's also linearly independent,
so T is also a

00:12:19.620 --> 00:12:27.210
basis for r2.

00:12:27.210 --> 00:12:30.600
And I wanted to show you this
to show that if I look at a

00:12:30.600 --> 00:12:34.980
vector subspace and r2 is a
valid subspace of itself.

00:12:34.980 --> 00:12:36.640
You can verify that.

00:12:36.640 --> 00:12:40.880
But if I have a subspace, it
doesn't have just one basis.

00:12:40.880 --> 00:12:42.290
It could have multiple bases.

00:12:42.290 --> 00:12:44.450
In fact, it normally
has infinite bases.

00:12:44.450 --> 00:12:48.880
So in this case, S is a valid
basis and T is also a valid

00:12:48.880 --> 00:12:49.580
basis for r2.

00:12:49.580 --> 00:12:52.410
And actually, just so you know
what T is, the situation here,

00:12:52.410 --> 00:12:55.020
this is called a
standard basis.

00:12:55.020 --> 00:13:00.560
This is the standard basis.

00:13:00.560 --> 00:13:02.850
And this is what you're used
to dealing with in just

00:13:02.850 --> 00:13:04.550
regular calculus or
physics class.

00:13:04.550 --> 00:13:06.450
And if you remember from physics
class, this is the

00:13:06.450 --> 00:13:11.830
unit vector i and then this
is the unit vector j.

00:13:11.830 --> 00:13:16.180
And it's the standard basis for
two-dimensional Cartesian

00:13:16.180 --> 00:13:18.320
coordinates.

00:13:18.320 --> 00:13:20.990
What's useful about a basis is
that you can always-- and it's

00:13:20.990 --> 00:13:23.310
not just true of the standard
basis, is that you can

00:13:23.310 --> 00:13:27.940
represent any vector
in your subspace.

00:13:27.940 --> 00:13:31.480
You can represent any vector
in your subspace by some

00:13:31.480 --> 00:13:36.050
unique combination of the
vectors in your basis.

00:13:36.050 --> 00:13:38.220
So let me show you that.

00:13:38.220 --> 00:13:47.220
So let's say that the set v1,
v2, all the way to vn, let's

00:13:47.220 --> 00:13:57.240
say that this is a basis for--
I don't know-- just some

00:13:57.240 --> 00:13:58.480
subspace U.

00:13:58.480 --> 00:13:59.730
So this is a subspace.

00:14:02.970 --> 00:14:05.720
So that means that these guys
are linearly independent.

00:14:05.720 --> 00:14:08.640
And also means that the span of
these guys, or all of the

00:14:08.640 --> 00:14:12.880
linear combinations of these
vectors, will get you all of

00:14:12.880 --> 00:14:15.260
the vectors, all of the possible
components, all of

00:14:15.260 --> 00:14:19.950
the difference members of U.

00:14:19.950 --> 00:14:24.370
Now what I want to show you is
each member of U can only be

00:14:24.370 --> 00:14:28.300
uniquely defined by a unique set
of-- a unique combination

00:14:28.300 --> 00:14:28.910
of these guys.

00:14:28.910 --> 00:14:32.520
Let me be clear about that.

00:14:32.520 --> 00:14:37.440
Let's say my vector a is a
member of our subspace U.

00:14:37.440 --> 00:14:41.370
That means that a can be
represented by some linear

00:14:41.370 --> 00:14:42.730
combination of these guys.

00:14:42.730 --> 00:14:45.230
These guys span U.

00:14:45.230 --> 00:14:47.950
So that means that we can
represent our vector a as

00:14:47.950 --> 00:14:52.010
being c1 times v1 plus
c2 times v2.

00:14:52.010 --> 00:14:53.430
These are vectors.

00:14:53.430 --> 00:14:57.930
All the way to cn times vn.

00:14:57.930 --> 00:15:00.860
Now I want to show you that this
is a unique combination.

00:15:00.860 --> 00:15:06.700
And to show that I'm going to
prove by contradiction.

00:15:06.700 --> 00:15:08.270
Let's say that there's
another combination.

00:15:08.270 --> 00:15:11.080
Let's say I could also represent
a by some other

00:15:11.080 --> 00:15:17.410
combination, d1 times v2 plus
d2 times v2 plus all the way

00:15:17.410 --> 00:15:20.940
to dn times vn.

00:15:20.940 --> 00:15:24.130
Now, what happens if I
subtract a from a?

00:15:24.130 --> 00:15:25.410
I'm going to get the 0 vector.

00:15:25.410 --> 00:15:28.160
Let me subtract these
two things.

00:15:28.160 --> 00:15:30.880
If I subtract a from
a, a minus a is

00:15:30.880 --> 00:15:32.970
clearly the 0 vector.

00:15:32.970 --> 00:15:36.250
It's clearly the 0 vector and
if I subtract this side from

00:15:36.250 --> 00:15:38.250
that side, what do we get?

00:15:38.250 --> 00:15:39.720
I'll do it in a different
color.

00:15:39.720 --> 00:15:49.390
We get c1 minus d1 times v1 plus
c2 minus d2 times v2, all

00:15:49.390 --> 00:15:53.400
the way to-- I'm at the point on
my board where it starts to

00:15:53.400 --> 00:15:54.260
malfunction.

00:15:54.260 --> 00:16:00.510
All the way to c-- you can't
see it. cn minus vn.

00:16:00.510 --> 00:16:01.760
It's showing up somehow.

00:16:03.890 --> 00:16:07.070
cn minus-- no, it's
messing me up.

00:16:07.070 --> 00:16:09.670
Let me rewrite it on the left
right here where it's less

00:16:09.670 --> 00:16:11.050
likely to mess up.

00:16:11.050 --> 00:16:16.580
The 0 vector, I'll write
it like that.

00:16:16.580 --> 00:16:23.920
Is equal to c1 minus d1 times
v1 plus all the way up to cn

00:16:23.920 --> 00:16:27.740
minus dn times vn.

00:16:27.740 --> 00:16:30.570
I just subtracted the
vector by itself.

00:16:30.570 --> 00:16:34.510
Now, I told you that
these are a basis.

00:16:34.510 --> 00:16:38.520
There's two things that-- when
you say a basis, it says that

00:16:38.520 --> 00:16:42.270
the span of these guys
makes the subspace.

00:16:42.270 --> 00:16:44.970
Or the span of these guys
is the subspace.

00:16:44.970 --> 00:16:47.040
And it also tells you that
these guys are linearly

00:16:47.040 --> 00:16:48.360
independent.

00:16:48.360 --> 00:16:51.530
So if they're linearly
independent, the only solution

00:16:51.530 --> 00:16:55.120
to this equation-- this is just
a constant times v1 plus

00:16:55.120 --> 00:16:56.660
another constant times
v2, all the way to a

00:16:56.660 --> 00:16:58.250
constant times vn.

00:16:58.250 --> 00:17:01.730
The only solution to this
equation is if each of these

00:17:01.730 --> 00:17:04.490
constants equal 0.

00:17:04.490 --> 00:17:06.990
So all of those constants
have to be equal to 0.

00:17:06.990 --> 00:17:10.179
Over here before it messed up,
this has to equal 0, this has

00:17:10.179 --> 00:17:10.910
to equal 0.

00:17:10.910 --> 00:17:13.440
That was a definition of
linear independence.

00:17:13.440 --> 00:17:16.220
And we know that this is a
linearly independent set.

00:17:16.220 --> 00:17:19.660
So if all of those constants are
equal to 0, then we know

00:17:19.660 --> 00:17:25.170
that c1-- if this is equal to 0,
then c1 is equal to d1, c2

00:17:25.170 --> 00:17:30.690
is equal to d2, all the way
to cn is equal to dn.

00:17:30.690 --> 00:17:34.660
So by the fact that it's
linearly independent, all of

00:17:34.660 --> 00:17:37.250
these-- each of these constants
have to be equal to

00:17:37.250 --> 00:17:37.540
each other.

00:17:37.540 --> 00:17:38.410
And that's our contradiction.

00:17:38.410 --> 00:17:41.280
I assume they're different, but
the linear independence

00:17:41.280 --> 00:17:43.370
forced them to be the same.

00:17:43.370 --> 00:17:47.380
So if you have a basis for some
subspace, any member of

00:17:47.380 --> 00:17:50.060
that subspace can be uniquely
determined by a unique

00:17:50.060 --> 00:17:52.870
combination of those vectors.

00:17:52.870 --> 00:17:57.370
And just to hit the point home,
I told you that this was

00:17:57.370 --> 00:17:59.270
a basis for r2.

00:17:59.270 --> 00:18:02.130
And my next question is,
and I just want to kind

00:18:02.130 --> 00:18:02.980
of backtrack a bit.

00:18:02.980 --> 00:18:05.120
If I just added another vector
here, if I just added the

00:18:05.120 --> 00:18:10.310
vector 1, 0, is S now
a basis for r2?

00:18:10.310 --> 00:18:14.170
Well no, it clearly will
continue to span r2, but this

00:18:14.170 --> 00:18:15.210
guy is redundant.

00:18:15.210 --> 00:18:18.330
This guy is in r2.

00:18:18.330 --> 00:18:22.760
And I already told you that
these two guys alone span r2.

00:18:22.760 --> 00:18:25.630
That anything in r2 can be
represented by a linear

00:18:25.630 --> 00:18:27.800
combination of these two guys.

00:18:27.800 --> 00:18:31.160
This guy is clearly in r2, so
he can be represented by a

00:18:31.160 --> 00:18:33.000
linear combination of
these two guys.

00:18:33.000 --> 00:18:36.660
So therefore, this is not a
linearly independent set.

00:18:36.660 --> 00:18:38.085
This is linearly dependent.

00:18:41.830 --> 00:18:44.250
And because it's literally
dependent, I have redundant

00:18:44.250 --> 00:18:45.250
information here.

00:18:45.250 --> 00:18:47.700
And then this would no
longer be a basis.

00:18:47.700 --> 00:18:49.770
So in order for these to be a
basis I kind of have to create

00:18:49.770 --> 00:18:54.620
the minimum set of vectors that
can span, or the most

00:18:54.620 --> 00:18:58.500
efficient set of vectors that
can span, in this case, r2.


Linear subspaces

WEBVTT
Kind: captions
Language: en

00:00:00.480 --> 00:00:04.820
We now have the tools, I think,
to understand the idea

00:00:04.820 --> 00:00:07.680
of a linear subspace of Rn.

00:00:07.680 --> 00:00:10.900
Let me write that down.

00:00:10.900 --> 00:00:13.130
I'll just always call
it a subspace of Rn.

00:00:13.130 --> 00:00:14.630
Everything we're doing
is linear.

00:00:14.630 --> 00:00:20.090
Subspace of Rn.

00:00:20.090 --> 00:00:21.710
I'm going to make a
definition here.

00:00:21.710 --> 00:00:24.450
I'm going to say that
a set of vectors V.

00:00:24.450 --> 00:00:32.410
So V is some subset of vectors,
some subset of Rn.

00:00:35.020 --> 00:00:37.430
So we already said Rn, when we
think about it, it's really

00:00:37.430 --> 00:00:42.010
just really an infinitely large
set of vectors, where

00:00:42.010 --> 00:00:46.390
each of those vectors
have n components.

00:00:46.390 --> 00:00:48.620
I'm going to not formally define
it, but this is just a

00:00:48.620 --> 00:00:49.460
set of vectors.

00:00:49.460 --> 00:00:52.240
I mean sometimes we visualize it
as multi-dimensional space

00:00:52.240 --> 00:00:55.290
and all that, but if we wanted
to be just as abstract about

00:00:55.290 --> 00:00:57.770
it as possible, it's
just all the set.

00:00:57.770 --> 00:01:02.400
It's the set of all of the --
you know we could call x1, x2,

00:01:02.400 --> 00:01:07.920
all the way to xn-- where each
of these, where each of the

00:01:07.920 --> 00:01:14.250
xi's are a member of the real
numbers for all of the i's.

00:01:14.250 --> 00:01:14.470
Right?

00:01:14.470 --> 00:01:16.410
That was our definition of Rn.

00:01:16.410 --> 00:01:18.880
It's just a huge
set of vectors.

00:01:18.880 --> 00:01:20.880
An infinitely large
set of vectors.

00:01:20.880 --> 00:01:25.040
V, I'm calling that, I'm going
to call that a subset of Rn,

00:01:25.040 --> 00:01:28.590
and which means it's just some
-- you know, it could be all

00:01:28.590 --> 00:01:30.550
of these vectors, and I'll talk
about that in a second.

00:01:30.550 --> 00:01:33.140
Or it could be some subset
of these vectors.

00:01:33.140 --> 00:01:36.040
Maybe it's all of them but
one particular vector.

00:01:36.040 --> 00:01:41.090
In order for this V to be a
subspace-- so I'm already

00:01:41.090 --> 00:01:44.280
saying it's a subset of Rn.

00:01:44.280 --> 00:01:45.240
Maybe this'll help you.

00:01:45.240 --> 00:01:48.300
If I draw all of Rn here
as this big blob.

00:01:48.300 --> 00:01:52.760
So these are all of the vectors
that are in Rn.

00:01:52.760 --> 00:01:54.390
V is some subset of it.

00:01:54.390 --> 00:01:55.220
It could be all of Rn.

00:01:55.220 --> 00:01:56.385
I'll show that a second.

00:01:56.385 --> 00:01:59.950
But let's just say
that this is V.

00:01:59.950 --> 00:02:01.470
V is a subset of vectors.

00:02:04.970 --> 00:02:07.500
Now in order for V to be a
subspace, and this is a

00:02:07.500 --> 00:02:16.140
definition, if V is a subspace,
or linear subspace

00:02:16.140 --> 00:02:23.210
of Rn, this means, this
is my definition,

00:02:23.210 --> 00:02:25.270
this means three things.

00:02:25.270 --> 00:02:28.605
This means that V contains
the 0 vector.

00:02:31.970 --> 00:02:36.530
I'll do it really, that's
the 0 vector.

00:02:36.530 --> 00:02:40.620
This is equal to 0 all the
way and you have n 0's.

00:02:40.620 --> 00:02:42.960
So V contains the 0 vector,
and this is

00:02:42.960 --> 00:02:45.660
a big V right there.

00:02:45.660 --> 00:02:48.820
If we have some vector x in V.

00:02:48.820 --> 00:02:57.470
So let me write this, if my
vector x is in V, if x is one

00:02:57.470 --> 00:03:00.870
of these vectors that's included
in my V, then when I

00:03:00.870 --> 00:03:04.270
multiply x times any member
of the reals.

00:03:04.270 --> 00:03:10.470
So if x is in V, then if V is a
subspace of Rn, then x times

00:03:10.470 --> 00:03:16.290
any scalar is also in V.

00:03:16.290 --> 00:03:18.360
This has to be the case.

00:03:18.360 --> 00:03:20.410
For those of you who are
familiar with the term, this

00:03:20.410 --> 00:03:21.900
term is called closure.

00:03:21.900 --> 00:03:25.700
If I have any element of a set,
this is closure under

00:03:25.700 --> 00:03:26.520
multiplication.

00:03:26.520 --> 00:03:29.180
Let me write that down--
in a new color.

00:03:29.180 --> 00:03:36.283
This is closure under scalar
multiplication.

00:03:40.720 --> 00:03:44.720
And that's just a fancy way of
saying, look, if I take some

00:03:44.720 --> 00:03:52.860
member of my set and I multiply
it by some scalar,

00:03:52.860 --> 00:03:55.840
I'm still going to
be in my set.

00:03:55.840 --> 00:03:58.500
If I multiplied it by some
scalar and I end up outside of

00:03:58.500 --> 00:04:02.940
my set, if I ended up with some
other vector that's not

00:04:02.940 --> 00:04:06.080
included in my subset, then this
wouldn't be a subspace.

00:04:06.080 --> 00:04:10.020
In order for it to be a
subspace, if I multiply any

00:04:10.020 --> 00:04:15.940
vector in my subset by a real
scalar, I'm defining this

00:04:15.940 --> 00:04:19.570
subspace over real numbers, if
I multiply it by any real

00:04:19.570 --> 00:04:24.130
number, I should also get
another member of this subset.

00:04:24.130 --> 00:04:25.880
So this is one of the
requirements.

00:04:25.880 --> 00:04:31.170
And then the other requirement
is if I take two vectors,

00:04:31.170 --> 00:04:33.410
let's say I have vector a,
it's in here, and I have

00:04:33.410 --> 00:04:36.360
vector b in here.

00:04:36.360 --> 00:04:38.320
So this is my other requirement

00:04:38.320 --> 00:04:39.710
for v being a subspace.

00:04:39.710 --> 00:04:44.430
If a is in a-- sorry-- if vector
a is in my set V, and

00:04:44.430 --> 00:04:51.100
vector b is in my set V, then if
V is a subspace of Rn, that

00:04:51.100 --> 00:04:58.470
tells me that a and b must
be in V as well.

00:04:58.470 --> 00:05:02.120
So this is closure
under addition.

00:05:02.120 --> 00:05:03.840
Let me write that down.

00:05:03.840 --> 00:05:08.910
Closure under addition.

00:05:08.910 --> 00:05:12.430
Once again, just a very fancy
way of saying, look, if you

00:05:12.430 --> 00:05:17.040
give me two elements that's in
my subset, and if I add them

00:05:17.040 --> 00:05:19.740
to each other -- these could be
any two arbitrary elements

00:05:19.740 --> 00:05:21.990
in my subset -- and I add them
to each other, I'm going to

00:05:21.990 --> 00:05:23.980
get another element
in my subset.

00:05:23.980 --> 00:05:26.320
That's what closure under
addition means.

00:05:26.320 --> 00:05:29.330
That when you add two vectors in
your set, you still end up

00:05:29.330 --> 00:05:30.660
with another vector
in your set.

00:05:30.660 --> 00:05:32.540
You don't somehow end up
with a vector that's

00:05:32.540 --> 00:05:35.380
outside of your set.

00:05:35.380 --> 00:05:39.910
If I have a subset of Rn, so
some subset of vectors of Rn,

00:05:39.910 --> 00:05:45.060
that contains the 0 vector,
and it's closed under

00:05:45.060 --> 00:05:48.930
multiplication and addition,
then I have a subspace.

00:05:48.930 --> 00:05:51.870
So subspace implies all of these
things, and all of these

00:05:51.870 --> 00:05:54.540
things imply a subspace.

00:05:54.540 --> 00:05:57.600
This is the definition
of a subspace.

00:05:57.600 --> 00:06:00.870
This might seem all abstract to
you right now, so let's do

00:06:00.870 --> 00:06:01.860
a couple of examples.

00:06:01.860 --> 00:06:04.650
And I don't know if these
examples will make it any more

00:06:04.650 --> 00:06:06.860
concrete, but I think if we do
it enough, you'll kind of get

00:06:06.860 --> 00:06:09.810
the intuitive sense of
what a space implies.

00:06:12.550 --> 00:06:13.740
Let me just do some examples.

00:06:13.740 --> 00:06:15.370
Because I want to
stay relatively

00:06:15.370 --> 00:06:17.470
mathematically formal.

00:06:17.470 --> 00:06:21.210
Let's just say I have the almost
trivially basic set.

00:06:21.210 --> 00:06:25.760
Let's say my set of vectors, I
only have one vector in it and

00:06:25.760 --> 00:06:27.470
I have the 0 vector.

00:06:27.470 --> 00:06:29.670
So I'll just do a really
bold 0 there.

00:06:29.670 --> 00:06:33.335
Or I could write it like this,
the only vector in my set is

00:06:33.335 --> 00:06:35.410
the 0 vector.

00:06:35.410 --> 00:06:35.830
Now Let's say?

00:06:35.830 --> 00:06:37.830
We're talking about R3.

00:06:37.830 --> 00:06:41.860
So let's say my 0 vector
in R3 looks like that.

00:06:41.860 --> 00:06:53.710
What I want to know is, is my
set V a subspace of R3?

00:06:53.710 --> 00:06:56.670
Well, in order for it to be a
subspace, three conditions.

00:06:56.670 --> 00:06:59.030
It has to contain
the 0 vector.

00:06:59.030 --> 00:07:01.110
Well the only thing it does
contain is the 0 vector.

00:07:01.110 --> 00:07:03.430
So it definitely contains
the 0 vector.

00:07:03.430 --> 00:07:06.830
So 0 vector, check.

00:07:06.830 --> 00:07:09.980
Now, is it closed under
multiplication?

00:07:09.980 --> 00:07:12.650
So that means, if I take any
member of the set, there's

00:07:12.650 --> 00:07:15.810
only one of them, and I multiply
it by any scalar, I

00:07:15.810 --> 00:07:17.440
should get another member
of the set.

00:07:17.440 --> 00:07:18.820
Or I should get maybe itself.

00:07:18.820 --> 00:07:21.640
So let's see, there's only
one member of the set.

00:07:21.640 --> 00:07:24.670
So the one member of the
set is the 0 vector.

00:07:24.670 --> 00:07:26.810
If I multiply it times
any scalar, what

00:07:26.810 --> 00:07:27.370
am I going to get?

00:07:27.370 --> 00:07:31.120
I'm going to get c times 0 which
is 0, c times 0, which

00:07:31.120 --> 00:07:33.610
is 0, and c times 0.

00:07:33.610 --> 00:07:35.280
I'm going to get its
only member.

00:07:35.280 --> 00:07:36.250
But it is closed.

00:07:36.250 --> 00:07:42.140
So it is closed under
multiplication.

00:07:42.140 --> 00:07:45.720
You can multiply this one vector
times any scalar, and

00:07:45.720 --> 00:07:47.750
you're just going to get
this vector again.

00:07:47.750 --> 00:07:52.840
So you're going to end up being
in your 0 vector set.

00:07:52.840 --> 00:07:54.010
That's a check.

00:07:54.010 --> 00:07:56.000
Is it closed under addition?

00:07:56.000 --> 00:08:00.660
Well, clearly if I add any
member of this set to itself,

00:08:00.660 --> 00:08:02.260
I mean, there's only one
member, to another

00:08:02.260 --> 00:08:03.070
member of the set.

00:08:03.070 --> 00:08:04.650
There's only one option here.

00:08:04.650 --> 00:08:07.990
If I just add that to
that, what do I get?

00:08:07.990 --> 00:08:09.430
I just get that.

00:08:09.430 --> 00:08:10.750
I just get it again.

00:08:10.750 --> 00:08:18.550
So it definitely is closed
under addition.

00:08:18.550 --> 00:08:19.270
Check.

00:08:19.270 --> 00:08:23.980
So it does turn out that this
trivially basic subset of r3,

00:08:23.980 --> 00:08:29.520
that just contains the 0 vector,
it is a subspace.

00:08:29.520 --> 00:08:33.860
Maybe a trivially simple
subspace, but it satisfies our

00:08:33.860 --> 00:08:35.140
constraints of a subspace.

00:08:35.140 --> 00:08:37.750
You can't do anything with the
vectors in it, they'll somehow

00:08:37.750 --> 00:08:39.230
get you out of that subspace.

00:08:39.230 --> 00:08:40.640
Or at least if you're
dealing with scalar

00:08:40.640 --> 00:08:42.549
multiplication or addition.

00:08:42.549 --> 00:08:45.450
Let me do one that maybe the
idea will be a little clearer

00:08:45.450 --> 00:08:47.340
if I show you an example
of something

00:08:47.340 --> 00:08:49.700
that is not a subspace.

00:08:49.700 --> 00:08:52.660
Let me get my coordinate
axes is over here.

00:08:52.660 --> 00:08:58.880
Let's say I were to find some
subspace, some subset.

00:08:58.880 --> 00:09:00.170
I don't know whether
it's a subspace.

00:09:00.170 --> 00:09:03.830
Let me call it my set S.

00:09:03.830 --> 00:09:19.010
And it equals all the vectors
x1, x2 that are a member of R2

00:09:19.010 --> 00:09:22.330
such that, I'm going to make a
little constraint here, such

00:09:22.330 --> 00:09:30.670
that x1 is greater than
or equal to 0.

00:09:30.670 --> 00:09:35.160
It contains all of the vectors
in R2 that are at least is 0

00:09:35.160 --> 00:09:36.960
or greater for the first term.

00:09:36.960 --> 00:09:39.850
So if we were to graph that
here, what do you get?

00:09:39.850 --> 00:09:40.780
We can get anything.

00:09:40.780 --> 00:09:43.510
We can move up or down
in any direction.

00:09:43.510 --> 00:09:45.110
Right?

00:09:45.110 --> 00:09:47.530
We can go up and down in any
direction, but we're

00:09:47.530 --> 00:09:48.330
constraining ourselves.

00:09:48.330 --> 00:09:50.520
These are all going to
be 0 or greater.

00:09:50.520 --> 00:09:53.530
So all of these first
coordinates are going to be 0

00:09:53.530 --> 00:09:55.530
or greater.

00:09:55.530 --> 00:09:57.950
And this one, we can go up
and down arbitrarily.

00:09:57.950 --> 00:10:04.390
So we're essentially, this
subset of R2, R2 is my entire

00:10:04.390 --> 00:10:06.020
Cartesian plane.

00:10:06.020 --> 00:10:10.690
But this subset of R2 will
include the vertical axis,

00:10:10.690 --> 00:10:12.125
often referred to
as the y-axis.

00:10:12.125 --> 00:10:14.950
It will include the vertical
axis, and essentially the

00:10:14.950 --> 00:10:16.175
first and fourth quadrants.

00:10:16.175 --> 00:10:18.210
If you remember your
quadrant labelling.

00:10:18.210 --> 00:10:19.920
So that's the first
quadrant and

00:10:19.920 --> 00:10:21.890
that's the fourth quadrant.

00:10:21.890 --> 00:10:33.810
So my question to you is,
is S a subspace of R2.

00:10:33.810 --> 00:10:37.080
So the first question, does
it contain the 0 vector?

00:10:37.080 --> 00:10:43.870
So in the case of R2, does
it contain 0, 0?

00:10:43.870 --> 00:10:44.395
Well, sure.

00:10:44.395 --> 00:10:48.790
It includes 0, 0 right there.

00:10:48.790 --> 00:10:52.610
We said x is greater than or
equal to 0, so this could be 0

00:10:52.610 --> 00:10:54.190
and obviously, there's no
constraint on this, so

00:10:54.190 --> 00:10:56.290
definitely the 0, 0 vector
is definitely

00:10:56.290 --> 00:10:58.870
contained in our set S.

00:10:58.870 --> 00:11:02.410
So that is a check.

00:11:02.410 --> 00:11:03.840
Let's try another one.

00:11:03.840 --> 00:11:09.520
If I add any two vectors in this
set, is that also going

00:11:09.520 --> 00:11:10.820
to show up in my set?

00:11:10.820 --> 00:11:12.270
Let me just do a couple
of examples.

00:11:12.270 --> 00:11:14.230
Maybe this isn't a proof.

00:11:14.230 --> 00:11:17.860
If I add that vector to
that vector, what

00:11:17.860 --> 00:11:18.690
am I going to get?

00:11:18.690 --> 00:11:21.730
If I put this up here, I'm
going to get that vector.

00:11:21.730 --> 00:11:26.660
If I add that vector to
that vector, what

00:11:26.660 --> 00:11:27.330
am I going to get?

00:11:27.330 --> 00:11:29.530
I could put this one heads to
tails, I would get a vector

00:11:29.530 --> 00:11:34.140
that looks like that.

00:11:34.140 --> 00:11:39.200
And if I did it formally, if I
add -- let's say that I have

00:11:39.200 --> 00:11:41.550
two vectors that are a
member of our set.

00:11:41.550 --> 00:11:45.503
Let's say the first one is
a, b and I add it to c,

00:11:45.503 --> 00:11:48.570
d, what do I get?

00:11:48.570 --> 00:11:55.400
I get a plus c over -- this
was a d -- over b plus d.

00:11:55.400 --> 00:11:58.100
So this thing is going
to be greater than 0.

00:11:58.100 --> 00:12:00.120
This thing is also going
to be greater than 0.

00:12:00.120 --> 00:12:02.550
That was my requirement
for being in the set.

00:12:02.550 --> 00:12:05.060
So if both of these are greater
than 0 and we add them

00:12:05.060 --> 00:12:07.190
to each other, this thing
is also going to be

00:12:07.190 --> 00:12:08.200
greater than 0.

00:12:08.200 --> 00:12:10.190
And we don't care what these,
these can be anything, I

00:12:10.190 --> 00:12:13.320
didn't put any constraints
on the second

00:12:13.320 --> 00:12:14.830
component of my vector.

00:12:14.830 --> 00:12:17.410
So it does seem like it is
closed under addition.

00:12:22.880 --> 00:12:26.790
Now what about scalar
multiplication?

00:12:26.790 --> 00:12:30.620
Let's take a particular
case here.

00:12:30.620 --> 00:12:34.170
Let's take my a, b again.

00:12:34.170 --> 00:12:37.070
I have my vector a, b.

00:12:37.070 --> 00:12:40.340
Now I can pick any
real scalar.

00:12:40.340 --> 00:12:41.400
So any real scalar.

00:12:41.400 --> 00:12:43.830
What if I just multiply
it by minus 1?

00:12:43.830 --> 00:12:45.440
So minus 1.

00:12:45.440 --> 00:12:51.140
So if I multiply it by minus
1, I get minus a, minus b.

00:12:51.140 --> 00:12:54.880
If I were to draw it visually,
if this is-- let's say a, b

00:12:54.880 --> 00:12:56.235
was the vector 2, 4.

00:12:56.235 --> 00:12:59.500
So it's like this.

00:12:59.500 --> 00:13:01.910
When I multiply it by minus
1, what do I get?

00:13:01.910 --> 00:13:04.650
I get minus a, minus b.

00:13:04.650 --> 00:13:06.580
I get this vector.

00:13:06.580 --> 00:13:11.300
Which you can be visually
clearly see falls out of, if

00:13:11.300 --> 00:13:14.540
we view these as kind of
position vectors, it falls out

00:13:14.540 --> 00:13:16.040
of our subspace.

00:13:16.040 --> 00:13:18.500
Or if you just view it not even
visually, if you just do

00:13:18.500 --> 00:13:23.070
it mathematically, clearly if
this is positive then this is

00:13:23.070 --> 00:13:25.040
going to-- and let's say if we
assume this is positive, and

00:13:25.040 --> 00:13:26.250
definitely not 0.

00:13:26.250 --> 00:13:27.610
So it's definitely a
positive number.

00:13:27.610 --> 00:13:30.300
So this is definitely going
to be a negative number.

00:13:30.300 --> 00:13:34.000
So when we multiply it by
negative 1, for really any

00:13:34.000 --> 00:13:36.635
element of this that doesn't
have a 0 there, you're going

00:13:36.635 --> 00:13:39.460
to end up with something that
falls out of it, right?

00:13:39.460 --> 00:13:43.460
This is not a member of this
set, because to be a member of

00:13:43.460 --> 00:13:46.270
the set, your first component
had to be greater than 0.

00:13:46.270 --> 00:13:48.530
This first component
is less than 0.

00:13:48.530 --> 00:13:54.040
So this subset that I drew out
here, the subset of R2, is not

00:13:54.040 --> 00:13:54.920
a subspace, Because?

00:13:54.920 --> 00:13:57.650
It's not closed under
multiplication or scalar

00:13:57.650 --> 00:13:58.900
multiplication.

00:14:08.330 --> 00:14:17.910
This is not a sub space of R2.

00:14:17.910 --> 00:14:20.560
Now I'll ask you one interesting
question.

00:14:20.560 --> 00:14:25.570
What if I ask you just the span
of some set of vectors?

00:14:25.570 --> 00:14:34.980
Let's say I want to know the
span of, I don't know, let's

00:14:34.980 --> 00:14:40.390
sat I have vector
v1, v2, and v3.

00:14:40.390 --> 00:14:42.660
I'm not even going to tell you
how many elements each of

00:14:42.660 --> 00:14:52.590
these vectors have. Is this
a valid subspace of Rn?

00:14:52.590 --> 00:14:55.150
Where n is the number of
elements that each of these

00:14:55.150 --> 00:14:58.500
have.

00:14:58.500 --> 00:15:02.320
Let's pick one of
the elements.

00:15:02.320 --> 00:15:09.610
Let me define, let me just call
u to be the set-- the set

00:15:09.610 --> 00:15:12.030
of all linear combinations
of this is the span.

00:15:12.030 --> 00:15:13.740
So let me just define
u to be the span.

00:15:13.740 --> 00:15:17.830
So I want to know, is
u a valid subspace?

00:15:17.830 --> 00:15:18.890
So let's think about
it this way.

00:15:18.890 --> 00:15:22.630
Let me just pick out a
random element of u.

00:15:25.660 --> 00:15:29.090
Actually, does this contain
the 0 vector?

00:15:29.090 --> 00:15:29.590
Well, sure.

00:15:29.590 --> 00:15:32.950
If we just multiply all of these
times 0, if we just say

00:15:32.950 --> 00:15:37.530
0 times v1, plus 0 times v2,
these are all the vectors, I

00:15:37.530 --> 00:15:41.480
didn't write them bold, plus
0 times v3, we get

00:15:41.480 --> 00:15:43.390
the 0 vector, right?

00:15:43.390 --> 00:15:44.790
We did everything
just zeroed out.

00:15:44.790 --> 00:15:46.720
So it definitely contains
the 0 vector.

00:15:46.720 --> 00:15:49.685
This is a linear combination
of those three vectors, so

00:15:49.685 --> 00:15:52.120
it's included in the span.

00:15:52.120 --> 00:15:58.740
Now let me just pick some
arbitrary member of this span.

00:15:58.740 --> 00:16:02.650
So in order to be a member of
this set, it just means that

00:16:02.650 --> 00:16:05.550
you can be represented-- let me
just call it the vector x--

00:16:05.550 --> 00:16:07.500
it means that you can be
represented as a linear

00:16:07.500 --> 00:16:09.040
combination of these vectors.

00:16:09.040 --> 00:16:16.420
So some combination, c1 times
v1 plus c2 times v2

00:16:16.420 --> 00:16:19.145
plus c3 times v3.

00:16:19.145 --> 00:16:19.940
Right?

00:16:19.940 --> 00:16:21.900
I'm just representing this
vector x, it's a member of

00:16:21.900 --> 00:16:23.880
this, so it can be represented
as a linear combination of

00:16:23.880 --> 00:16:26.030
those three vectors.

00:16:26.030 --> 00:16:29.570
Is this set closed under
multiplication?

00:16:29.570 --> 00:16:32.850
Well let's just multiply this
times some arbitrary constant.

00:16:32.850 --> 00:16:34.980
What is c times x?

00:16:34.980 --> 00:16:37.240
Let me scroll down
a little bit.

00:16:37.240 --> 00:16:39.600
What does c times x equal?

00:16:39.600 --> 00:16:42.370
Let me do a different
constant actually.

00:16:42.370 --> 00:16:45.400
Let me multiply it times some
arbitrary constant a.

00:16:45.400 --> 00:16:47.810
What is a times x?

00:16:47.810 --> 00:16:52.200
What's a times c1 times v1--
I'm just multiplying both

00:16:52.200 --> 00:16:56.300
sides equation times a--
a times c2 times v2

00:16:56.300 --> 00:17:00.710
plus a times c3, v3.

00:17:00.710 --> 00:17:03.280
Right?

00:17:03.280 --> 00:17:05.109
If this was an arbitrary
constant, you could just write

00:17:05.109 --> 00:17:06.520
this as another arbitrary
constant.

00:17:06.520 --> 00:17:08.050
This is another arbitrary
constant.

00:17:08.050 --> 00:17:09.869
This is another arbitrary
constant.

00:17:09.869 --> 00:17:10.290
I want to be clear.

00:17:10.290 --> 00:17:12.520
All I did is I just multiplied
both sides of the equation

00:17:12.520 --> 00:17:13.640
times a scalar.

00:17:13.640 --> 00:17:17.410
But clearly, this expression
right here, I mean I could

00:17:17.410 --> 00:17:23.040
write this, I could rewrite this
as c4 times v1 plus c5

00:17:23.040 --> 00:17:26.150
times v2, where this
is c5, this is c4.

00:17:26.150 --> 00:17:28.890
Plus c6 times v3.

00:17:28.890 --> 00:17:31.410
This is clearly another
linear combination

00:17:31.410 --> 00:17:33.320
of these three vectors.

00:17:33.320 --> 00:17:36.300
So, the span is the set of all
of the linear combinations of

00:17:36.300 --> 00:17:37.370
these three vectors.

00:17:37.370 --> 00:17:40.100
So clearly this is one of the
linear combinations, so it's

00:17:40.100 --> 00:17:42.080
also included in the span.

00:17:42.080 --> 00:17:44.980
So this is also in u.

00:17:44.980 --> 00:17:47.620
It's also in the span of
those three vectors.

00:17:47.620 --> 00:17:51.410
So it is closed under
multiplication.

00:17:55.130 --> 00:17:57.160
Now we just have to show that
it's closed under addition,

00:17:57.160 --> 00:18:00.540
and then we know that the span
of -- and I did three, here

00:18:00.540 --> 00:18:03.600
but you can extend an arbitrary
n number of vectors

00:18:03.600 --> 00:18:08.620
if the span of any set of
vectors is a valid subspace.

00:18:08.620 --> 00:18:10.930
Let me prove that.

00:18:10.930 --> 00:18:12.410
We already defined one x here.

00:18:12.410 --> 00:18:16.140
Let me define another vector
that's in u, or that's in the

00:18:16.140 --> 00:18:17.640
span of these vectors.

00:18:17.640 --> 00:18:21.780
And it equals, I don't know,
let's say it equals d1 times

00:18:21.780 --> 00:18:29.120
v1 plus d2 times v2
plus d3 times v3.

00:18:29.120 --> 00:18:35.900
Now what is x plus y?

00:18:35.900 --> 00:18:39.090
If I add these two vectors,
what does it equal to?

00:18:39.090 --> 00:18:41.550
Well, I could just add.

00:18:41.550 --> 00:18:45.570
x plus y means all of this stuff
plus all of this stuff.

00:18:45.570 --> 00:18:46.620
So what does that equal?

00:18:46.620 --> 00:18:52.020
It means if you just add these
together you get c1 plus d1

00:18:52.020 --> 00:19:04.020
times v1 plus c2 plus d2 times
v2 plus c3 plus d3 times v3.

00:19:04.020 --> 00:19:04.930
Right?

00:19:04.930 --> 00:19:08.260
You had a v3 here, you had a
v3 there, you just add up

00:19:08.260 --> 00:19:09.480
their coefficients.

00:19:09.480 --> 00:19:11.930
Clearly this is just another
linear combination.

00:19:11.930 --> 00:19:13.600
These are just constants
again.

00:19:13.600 --> 00:19:15.960
That's an arbitrary constant,
that's an arbitrary constant,

00:19:15.960 --> 00:19:17.420
that's an arbitrary constant.

00:19:17.420 --> 00:19:22.660
So this thing is just a linear
combination of v1, v2, and v3.

00:19:22.660 --> 00:19:27.880
So it must be, by definition, in
the span of v1, v2, and v3.

00:19:27.880 --> 00:19:36.340
So we are definitely closed
under addition.

00:19:36.340 --> 00:19:38.400
Now, you might say, hey, Sal,
you're saying that the span of

00:19:38.400 --> 00:19:43.500
any vector is a valid subspace,
but let me show you

00:19:43.500 --> 00:19:47.440
an example that clearly, if I
just took the span of one

00:19:47.440 --> 00:19:52.900
vector, let me just define u
to be equal to the span of

00:19:52.900 --> 00:19:56.250
just the vector, let me just
do a really simple one.

00:19:56.250 --> 00:19:58.500
Just the vector 1, 1.

00:19:58.500 --> 00:20:01.550
Clearly this can't be
a valid subspace.

00:20:01.550 --> 00:20:03.540
Let's think about
this visually.

00:20:03.540 --> 00:20:06.190
What does vector
1, 1 look like?

00:20:06.190 --> 00:20:08.622
Vector 1, 1 looks like this.

00:20:08.622 --> 00:20:09.680
Right?

00:20:09.680 --> 00:20:13.260
And the span of vector 1, 1--
this is in its standard

00:20:13.260 --> 00:20:16.300
position -- the span of vector
1, 1 is all of the linear

00:20:16.300 --> 00:20:18.040
combinations of this vector.

00:20:18.040 --> 00:20:19.880
Well, there's nothing else to
add it to, so it's really just

00:20:19.880 --> 00:20:21.710
going to be all of the
scaled up and scaled

00:20:21.710 --> 00:20:23.050
down versions of this.

00:20:23.050 --> 00:20:25.020
So if you scale it up
you get things that

00:20:25.020 --> 00:20:26.900
look more like that.

00:20:26.900 --> 00:20:28.970
If you scale it down, you get
things that look more like

00:20:28.970 --> 00:20:31.250
that if you go into the
negative domain.

00:20:31.250 --> 00:20:35.060
So just by multiplying this
vector times different values,

00:20:35.060 --> 00:20:37.370
and if you're going to put
them all into a standard

00:20:37.370 --> 00:20:40.240
position, you'd essentially
get a line

00:20:40.240 --> 00:20:41.880
that looks like that.

00:20:41.880 --> 00:20:45.070
You say, gee, that doesn't look
like a whole subspace.

00:20:45.070 --> 00:20:46.180
But a couple of things.

00:20:46.180 --> 00:20:49.800
Clearly, it contains
the 0 vector.

00:20:52.670 --> 00:20:56.060
We can just scale it by 0.

00:20:56.060 --> 00:20:58.180
The span is just all of the
different scales of this.

00:20:58.180 --> 00:20:59.820
And if there are other vectors,
you would add it to

00:20:59.820 --> 00:21:00.480
those as well.

00:21:00.480 --> 00:21:02.750
But this is clearly going
to be the 0 vector.

00:21:02.750 --> 00:21:05.580
So it contains the 0 vector.

00:21:05.580 --> 00:21:07.980
Is it closed under
multiplication?

00:21:07.980 --> 00:21:12.120
Well, the span is the set of all
the vectors, where, if you

00:21:12.120 --> 00:21:15.230
take all of the real numbers
for c and you multiply it

00:21:15.230 --> 00:21:17.920
times 1, 1, that is the span.

00:21:17.920 --> 00:21:20.130
Clearly, you multiply this times
anything it's going to

00:21:20.130 --> 00:21:24.590
equal another thing that's
definitely in your span.

00:21:24.590 --> 00:21:26.540
The last thing, is it closed
under addition?

00:21:29.270 --> 00:21:33.190
So any two vectors in the span
could, let's say that I have

00:21:33.190 --> 00:21:35.690
one vector a that's
in my span.

00:21:35.690 --> 00:21:43.660
I can represent it as c1 some
scalar times my vector there.

00:21:43.660 --> 00:21:47.305
And then I have another vector
b, and I could represent it

00:21:47.305 --> 00:21:52.690
with c2 times my one vector
in my set right there.

00:21:52.690 --> 00:21:54.020
And so what is this going
to be equal to?

00:21:54.020 --> 00:21:57.390
This is going to be equal to,
this is essentially going to

00:21:57.390 --> 00:22:01.550
be equal to c-- well, get a
little more space-- this is

00:22:01.550 --> 00:22:07.030
going to be equal to c1 plus
c2 times my vector.

00:22:07.030 --> 00:22:09.150
This is almost trivially
obvious.

00:22:09.150 --> 00:22:11.080
But clearly this
is in the span.

00:22:11.080 --> 00:22:12.820
It's just a scaled up
version of this.

00:22:12.820 --> 00:22:15.260
This is in the span, it's in a
scaled up version of this.

00:22:15.260 --> 00:22:18.370
And this is also going to be
in the span of this vector,

00:22:18.370 --> 00:22:20.190
because this is just
another scalar.

00:22:20.190 --> 00:22:22.380
We could call that c3.

00:22:22.380 --> 00:22:28.110
If you just do it visually, if I
take this vector right there

00:22:28.110 --> 00:22:33.700
and I were to add it to this
vector, if you put them head

00:22:33.700 --> 00:22:38.080
to tails, you would end
up with this vector.

00:22:38.080 --> 00:22:39.200
Right there in green.

00:22:39.200 --> 00:22:40.200
I don't know if you
can see it.

00:22:40.200 --> 00:22:42.410
I'll do it in red right there.

00:22:42.410 --> 00:22:44.720
You end up with that vector.

00:22:44.720 --> 00:22:47.160
And you could do that any vector
plus any other vector

00:22:47.160 --> 00:22:50.020
on this line is going to equal
another vector on this line.

00:22:50.020 --> 00:22:52.585
Any vector on this line
multiplied by some scalar is

00:22:52.585 --> 00:22:55.130
just going to be another
vector on this line.

00:22:55.130 --> 00:22:58.280
So you're closed under
multiplication.

00:22:58.280 --> 00:22:59.830
Your closed under addition.

00:22:59.830 --> 00:23:03.880
And you include the 0 vector.

00:23:03.880 --> 00:23:09.640
So even this trivially simple
span is a valid subspace.

00:23:09.640 --> 00:23:12.460
And that just backs up the
idea that we showed here.

00:23:12.460 --> 00:23:14.290
That, in general, I could
have just made

00:23:14.290 --> 00:23:16.740
this a set of n vectors.

00:23:16.740 --> 00:23:19.060
I picked three vectors right
here, but it could've been n

00:23:19.060 --> 00:23:21.490
vectors and I could have used
the same argument that the

00:23:21.490 --> 00:23:26.980
span of n vectors is a
valid subspace of Rn.

00:23:26.980 --> 00:23:29.050
And I showed it right there.


Span and linear independence example

WEBVTT
Kind: captions
Language: en

00:00:00.430 --> 00:00:03.440
I want to bring everything we've
learned about linear

00:00:03.440 --> 00:00:07.450
independence and dependence,
and the span of a set of

00:00:07.450 --> 00:00:10.850
vectors together in one
particularly hairy problem,

00:00:10.850 --> 00:00:12.850
because if you understand what
this problem is all about, I

00:00:12.850 --> 00:00:15.640
think you understand what we're
doing, which is key to

00:00:15.640 --> 00:00:19.120
your understanding of linear
algebra, these two concepts.

00:00:19.120 --> 00:00:21.230
So the first question I'm going
to ask about the set of

00:00:21.230 --> 00:00:25.050
vectors s, and they're all
three-dimensional vectors,

00:00:25.050 --> 00:00:34.020
they have three components, Is
the span of s equal to R3?

00:00:34.020 --> 00:00:35.070
It seems like it might be.

00:00:35.070 --> 00:00:37.160
If each of these add new
information, it seems like

00:00:37.160 --> 00:00:40.680
maybe I could describe any
vector in R3 by these three

00:00:40.680 --> 00:00:43.500
vectors, by some combination
of these three vectors.

00:00:43.500 --> 00:00:46.540
And the second question I'm
going to ask is are they

00:00:46.540 --> 00:00:51.910
linearly independent?

00:00:51.910 --> 00:00:54.700
And maybe I'll be able to answer
them at the same time.

00:00:54.700 --> 00:00:57.240
So let's answer the first one.

00:00:57.240 --> 00:00:58.600
Do they span R3?

00:00:58.600 --> 00:01:01.760
To span R3, that means some
linear combination of these

00:01:01.760 --> 00:01:06.540
three vectors should be able to
construct any vector in R3.

00:01:06.540 --> 00:01:10.340
So let me give you a linear
combination of these vectors.

00:01:10.340 --> 00:01:18.450
I could have c1 times the first
vector, 1, minus 1, 2

00:01:18.450 --> 00:01:22.260
plus some other arbitrary
constant c2, some scalar,

00:01:22.260 --> 00:01:28.890
times the second vector, 2, 1,
2 plus some third scaling

00:01:28.890 --> 00:01:34.680
vector times the third
vector minus 1, 0, 2.

00:01:34.680 --> 00:01:39.070
I should be able to, using some
arbitrary constants, take

00:01:39.070 --> 00:01:42.010
a combination of these vectors
that sum up to

00:01:42.010 --> 00:01:43.990
any vector in R3.

00:01:43.990 --> 00:01:47.620
And I'm going to represent any
vector in R3 by the vector a,

00:01:47.620 --> 00:01:53.090
b, and c, where a, b, and
c are any real numbers.

00:01:53.090 --> 00:01:55.850
So if you give me any a, b, and
c, and I can give you a

00:01:55.850 --> 00:01:59.140
formula for telling you what
your c3's, your c2's and your

00:01:59.140 --> 00:02:01.640
c1's are, then than essentially
means that it

00:02:01.640 --> 00:02:04.200
spans R3, because if you give me
a vector, I can always tell

00:02:04.200 --> 00:02:06.850
you how to construct that
vector with these three.

00:02:06.850 --> 00:02:09.650
So Let's see if I can do that.

00:02:09.650 --> 00:02:13.790
Just from our definition of
scalar multiplication of a

00:02:13.790 --> 00:02:16.810
vector, we know that c1 times
this vector, I could rewrite

00:02:16.810 --> 00:02:17.380
it if I want.

00:02:17.380 --> 00:02:18.660
I normally skip this
step, but I really

00:02:18.660 --> 00:02:19.370
want to make it clear.

00:02:19.370 --> 00:02:22.770
So c1 times, I could just
rewrite as 1 times c-- it's

00:02:22.770 --> 00:02:24.950
each of the terms times c1.

00:02:24.950 --> 00:02:27.840
Similarly, c2 times this is the
same thing as each of the

00:02:27.840 --> 00:02:30.790
terms times c2.

00:02:30.790 --> 00:02:33.040
And c3 times this is the
same thing as each of

00:02:33.040 --> 00:02:35.535
the terms times c3.

00:02:35.535 --> 00:02:37.760
I want to show you that
everything we do it just

00:02:37.760 --> 00:02:42.220
formally comes from our
definition of multiplication

00:02:42.220 --> 00:02:44.980
of a vector times a scalar,
which is what we just did, or

00:02:44.980 --> 00:02:46.550
vector addition, which is
what we're about to do.

00:02:46.550 --> 00:02:50.500
So vector addition tells us that
this term plus this term

00:02:50.500 --> 00:02:53.010
plus this term needs
to equal that term.

00:02:53.010 --> 00:02:55.100
So let me write that down.

00:02:55.100 --> 00:03:05.470
We get c1 plus 2c2 minus
c3 will be equal to a.

00:03:05.470 --> 00:03:07.950
Likewise, we can do the same
thing with the next row.

00:03:07.950 --> 00:03:14.160
Minus c1 plus c2 plus 0c3
must be equal to b.

00:03:14.160 --> 00:03:21.600
So we get minus c1 plus c2 plus
0c3-- so we don't even

00:03:21.600 --> 00:03:25.240
have to write that-- is going
to be equal to b.

00:03:25.240 --> 00:03:27.940
And then finally, let's
just do that last row.

00:03:27.940 --> 00:03:43.390
2c1 plus 3c2 plus 2c3 is
going to be equal to c.

00:03:43.390 --> 00:03:47.430
Now, let's see if we can solve
for our different constants.

00:03:47.430 --> 00:03:48.810
I'm going to do it
by elimination.

00:03:48.810 --> 00:03:50.750
I think you might be familiar
with this process.

00:03:50.750 --> 00:03:53.160
I think I've done it in some of
the earlier linear algebra

00:03:53.160 --> 00:03:55.480
videos before I started doing
a formal presentation of it.

00:03:55.480 --> 00:03:58.540
And I'm going to review it again
in a few videos from

00:03:58.540 --> 00:04:00.190
now, but I think you
understand how to

00:04:00.190 --> 00:04:01.260
solve it this way.

00:04:01.260 --> 00:04:03.150
What I'm going to do is I'm
going to first eliminate these

00:04:03.150 --> 00:04:05.390
two terms and then I'm going
to eliminate this term, and

00:04:05.390 --> 00:04:08.150
then I can solve for my
various constants.

00:04:08.150 --> 00:04:10.750
If I want to eliminate this term
right here, what I could

00:04:10.750 --> 00:04:13.050
do is I could add this equation
to that equation.

00:04:13.050 --> 00:04:16.209
Or even better, I can replace
this equation with the sum of

00:04:16.209 --> 00:04:17.750
these two equations.

00:04:17.750 --> 00:04:19.269
Let me do that.

00:04:19.269 --> 00:04:25.120
I'm just going to add these two
equations to each other

00:04:25.120 --> 00:04:27.350
and replace this one
with that sum.

00:04:27.350 --> 00:04:30.140
So minus c1 plus c1, that
just gives you 0.

00:04:30.140 --> 00:04:31.380
I can ignore it.

00:04:31.380 --> 00:04:33.630
Then c2 plus 2c2, that's 3c2.

00:04:36.450 --> 00:04:42.910
And then 0 plus minus c3
is equal to minus c3.

00:04:42.910 --> 00:04:47.400
Minus c3 is equal to-- and I'm
replacing this with the sum of

00:04:47.400 --> 00:04:48.900
these two, so b plus a.

00:04:48.900 --> 00:04:51.110
It equals b plus a.

00:04:51.110 --> 00:04:53.070
Let me write down that first
equation on the top.

00:04:53.070 --> 00:04:55.810
So the first equation, I'm
not doing anything to it.

00:04:55.810 --> 00:05:03.730
So I get c1 plus 2c2 minus
c3 is equal to a.

00:05:03.730 --> 00:05:08.120
Now, in this last equation, I
want to eliminate this term.

00:05:08.120 --> 00:05:13.260
Let's take this equation and
subtract from it 2 times this

00:05:13.260 --> 00:05:14.510
top equation.

00:05:18.700 --> 00:05:21.810
You can also view it as let's
add this to minus 2 times this

00:05:21.810 --> 00:05:23.090
top equation.

00:05:23.090 --> 00:05:24.890
Since we're almost done using
this when we actually even

00:05:24.890 --> 00:05:26.900
wrote it, let's just multiply
this times minus 2.

00:05:26.900 --> 00:05:42.200
So this becomes a minus 2c1
minus 4c2 plus 2c3 is

00:05:42.200 --> 00:05:45.300
equal to minus 2a.

00:05:45.300 --> 00:05:46.890
If you just multiply each of
these terms-- I want to be

00:05:46.890 --> 00:05:47.540
very careful.

00:05:47.540 --> 00:05:48.600
I don't want to make
a careless mistake.

00:05:48.600 --> 00:05:54.700
Minus 2 times c1 minus 4 plus
2 and then minus 2.

00:05:54.700 --> 00:05:56.830
And now we can add these
two together.

00:05:56.830 --> 00:05:58.520
And what do we get?

00:05:58.520 --> 00:06:01.380
2c1 minus 2c1, that's a 0.

00:06:01.380 --> 00:06:02.540
I don't have to write it.

00:06:02.540 --> 00:06:06.300
3c2 minus 4c2, that's
a minus c2.

00:06:09.030 --> 00:06:16.890
And then you have your 2c3 plus
another 2c3, so that is

00:06:16.890 --> 00:06:26.710
equal to plus 4c3 is equal
to c minus 2a.

00:06:26.710 --> 00:06:31.820
All I did is I replaced this
with this minus 2 times that,

00:06:31.820 --> 00:06:33.070
and I got this.

00:06:35.740 --> 00:06:39.580
Now I'm going to keep my top
equation constant again.

00:06:39.580 --> 00:06:41.160
I'm not going to do anything
to it, so I'm just going to

00:06:41.160 --> 00:06:42.230
move it to the right.

00:06:42.230 --> 00:06:48.440
So I get c1 plus 2c2 minus
c3 is equal to a.

00:06:48.440 --> 00:06:51.320
I'm also going to keep my second
equation the same, so I

00:06:51.320 --> 00:06:59.040
get 3c2 minus c3 is
equal to b plus a.

00:06:59.040 --> 00:07:00.900
Let me scroll over a good bit.

00:07:00.900 --> 00:07:03.620
And then this last equation
I want to eliminate.

00:07:03.620 --> 00:07:07.520
My goal is to eliminate
this term right here.

00:07:07.520 --> 00:07:10.960
What I want to do is I want to
multiply this bottom equation

00:07:10.960 --> 00:07:16.070
times 3 and add it to this
middle equation to eliminate

00:07:16.070 --> 00:07:18.000
this term right here.

00:07:18.000 --> 00:07:21.840
So if I multiply this bottom
equation times 3-- let me just

00:07:21.840 --> 00:07:25.190
do-- well, actually, I don't
want to make things messier,

00:07:25.190 --> 00:07:30.010
so this becomes a minus 3 plus
a 3, so those cancel out.

00:07:30.010 --> 00:07:34.940
This becomes a 12 minus a 1.

00:07:34.940 --> 00:07:43.040
So this becomes 12c3 minus
c3, which is 11c3.

00:07:43.040 --> 00:07:49.690
And then this becomes a-- oh,
sorry, I was already done.

00:07:49.690 --> 00:07:52.570
When I do 3 times this plus
that, those canceled out.

00:07:52.570 --> 00:07:56.540
And then when I multiplied 3
times this, I get 12c3 minus a

00:07:56.540 --> 00:08:00.810
c3, so that's 11c3.

00:08:00.810 --> 00:08:11.770
And I multiplied this times 3
plus this, so I get 3c minus

00:08:11.770 --> 00:08:17.280
6a-- I'm just multiplying
this times 3-- plus

00:08:17.280 --> 00:08:23.490
this, plus b plus a.

00:08:23.490 --> 00:08:25.280
So what can I rewrite this by?

00:08:25.280 --> 00:08:26.600
Actually, I want to make
something very clear.

00:08:26.600 --> 00:08:29.810
This c is different than these
c1's, c2's and c3's

00:08:29.810 --> 00:08:30.710
that I had up here.

00:08:30.710 --> 00:08:31.940
I think you realize that.

00:08:31.940 --> 00:08:35.010
But I just realized that I used
the letters c twice, and

00:08:35.010 --> 00:08:36.490
I just didn't want any
confusion here.

00:08:36.490 --> 00:08:38.850
So this c that doesn't have any
subscript is a different

00:08:38.850 --> 00:08:42.059
constant then all of these
things over here.

00:08:42.059 --> 00:08:44.930
Let's see if we can
simplify this.

00:08:44.930 --> 00:08:48.550
We have an a and a minus 6a,
so let's just add them.

00:08:48.550 --> 00:08:53.500
So let's get rid of that a and
this becomes minus 5a.

00:08:53.500 --> 00:08:55.630
If we divide both sides
of this equation by

00:08:55.630 --> 00:08:59.520
11, what do we get?

00:08:59.520 --> 00:09:13.310
We get c3 is equal to 1/11
times 3c minus 5a.

00:09:13.310 --> 00:09:15.570
So you give me any a or
c and I'll already

00:09:15.570 --> 00:09:16.810
tell you what c3 is.

00:09:16.810 --> 00:09:18.190
What is c2?

00:09:18.190 --> 00:09:21.590
c2 is equal to-- let
me simplify this

00:09:21.590 --> 00:09:22.960
equation right here.

00:09:22.960 --> 00:09:24.710
Let me do it right there.

00:09:24.710 --> 00:09:27.250
So if I just add c3 to both
sides of the equation, I get

00:09:27.250 --> 00:09:34.370
3c2 is equal to b
plus a plus c3.

00:09:34.370 --> 00:09:40.160
And if I divide both sides of
this by 3, I get c2 is equal

00:09:40.160 --> 00:09:47.930
to 1/3 times b plus a plus c3.

00:09:47.930 --> 00:09:50.000
I'll just leave it like
that for now.

00:09:50.000 --> 00:09:53.190
Then what is c1 equal to?

00:09:53.190 --> 00:09:58.000
I could just rewrite this top
equation as if I subtract 2c2

00:09:58.000 --> 00:10:06.430
and add c3 to both sides,
I get c1 is equal to a

00:10:06.430 --> 00:10:10.990
minus 2c2 plus c3.

00:10:10.990 --> 00:10:12.240
What have I just shown you?

00:10:14.660 --> 00:10:18.580
You can give me any vector in
R3 that you want to find.

00:10:18.580 --> 00:10:21.040
So you can give me any real
number for a, any real number

00:10:21.040 --> 00:10:23.760
for b, any real number for c.

00:10:23.760 --> 00:10:26.660
And if you give me those
numbers, I'm claiming now that

00:10:26.660 --> 00:10:30.150
I can always tell you some
combination of these three

00:10:30.150 --> 00:10:31.980
vectors that will
add up to those.

00:10:31.980 --> 00:10:34.160
And I've actually already solved
for what I have to

00:10:34.160 --> 00:10:37.260
multiply each of those
vectors by to add up

00:10:37.260 --> 00:10:38.620
to this third vector.

00:10:38.620 --> 00:10:41.770
So you give me your a's, b's
and c's, I just have to

00:10:41.770 --> 00:10:45.240
substitute into the a's and
the c's right here.

00:10:45.240 --> 00:10:47.220
Oh, sorry.

00:10:47.220 --> 00:10:48.780
I forgot this b over here.

00:10:48.780 --> 00:10:50.230
There's also a b.

00:10:50.230 --> 00:10:52.130
It was suspicious that I didn't
have to deal with a b.

00:10:52.130 --> 00:10:55.040
So there was a b right there.

00:10:55.040 --> 00:11:00.090
So this is 3c minus 5a plus b.

00:11:00.090 --> 00:11:01.860
Let me write that.

00:11:01.860 --> 00:11:04.300
There's a b right there
in a parentheses.

00:11:04.300 --> 00:11:06.720
But I think you get
the general idea.

00:11:06.720 --> 00:11:09.100
You give me your a's,
b's and c's, any

00:11:09.100 --> 00:11:11.290
real numbers can apply.

00:11:11.290 --> 00:11:13.290
There's no division over here,
so I don't have to worry about

00:11:13.290 --> 00:11:14.810
dividing by zero.

00:11:14.810 --> 00:11:16.600
So this is just a linear
combination of any real

00:11:16.600 --> 00:11:18.950
numbers, so I can clearly
get another real number.

00:11:18.950 --> 00:11:21.050
So you give me your a's,
b's and c's, I'm going

00:11:21.050 --> 00:11:23.800
to give you a c3.

00:11:23.800 --> 00:11:25.170
Now, you gave me a's,
b's and c's.

00:11:25.170 --> 00:11:26.070
I got a c3.

00:11:26.070 --> 00:11:28.290
This is just going to be
another real number.

00:11:28.290 --> 00:11:30.750
I'm just going to take that with
your former a's and b's

00:11:30.750 --> 00:11:34.030
and I'm going to be able
to give you a c2.

00:11:34.030 --> 00:11:37.530
We were already able to solve
for a c2 and a c3, and then I

00:11:37.530 --> 00:11:40.325
just use your a as well,
and then I'm going

00:11:40.325 --> 00:11:41.575
to give you a c1.

00:11:43.690 --> 00:11:47.490
Hopefully, you're seeing that no
matter what a, b, and c you

00:11:47.490 --> 00:11:50.600
give me, I can give you
a c1, c2, or c3.

00:11:50.600 --> 00:11:53.200
There's no reason that any a's,
b's or c's should break

00:11:53.200 --> 00:11:54.100
down these formulas.

00:11:54.100 --> 00:11:56.410
We're not doing any division, so
it's not like a zero would

00:11:56.410 --> 00:11:58.070
break it down.

00:11:58.070 --> 00:12:02.380
I can say definitively that the
set of vectors, of these

00:12:02.380 --> 00:12:06.930
three vectors, does
indeed span R3.

00:12:06.930 --> 00:12:07.980
Let me ask you another
question.

00:12:07.980 --> 00:12:09.120
I already asked it.

00:12:09.120 --> 00:12:12.530
Are these vectors linearly
independent?

00:12:16.340 --> 00:12:19.570
We said in order for them to be
linearly independent, the

00:12:19.570 --> 00:12:28.900
only solution to c1 times my
first vector, 1, minus 1, 2,

00:12:28.900 --> 00:12:36.920
plus c2 times my second vector,
2, 1, 3, plus c3 times

00:12:36.920 --> 00:12:41.670
my third vector,
minus 1, 0, 2.

00:12:41.670 --> 00:12:44.510
If something is linearly
independent that means that

00:12:44.510 --> 00:12:51.970
the only solution to this
equation-- so I want to find

00:12:51.970 --> 00:12:55.320
some set of combinations of
these vectors that add up to

00:12:55.320 --> 00:12:59.830
the zero vector, and I did that
in the previous video.

00:12:59.830 --> 00:13:03.900
If they are linearly dependent,
there must be some

00:13:03.900 --> 00:13:05.890
non-zero solution.

00:13:05.890 --> 00:13:08.260
One of these constants, at least
one of these constants,

00:13:08.260 --> 00:13:11.440
would be non-zero for
this solution.

00:13:11.440 --> 00:13:13.670
You can always make them zero,
no matter what, but if they

00:13:13.670 --> 00:13:16.120
are linearly dependent,
then one of

00:13:16.120 --> 00:13:17.870
these could be non-zero.

00:13:17.870 --> 00:13:23.080
If they're linearly independent
then all of these

00:13:23.080 --> 00:13:25.740
have to be-- the only solution
to this equation

00:13:25.740 --> 00:13:27.570
would be c1, c2, c3.

00:13:27.570 --> 00:13:32.870
All have to be equal to
0. c1, c2, c3 all have

00:13:32.870 --> 00:13:35.070
to be equal to 0.

00:13:35.070 --> 00:13:38.460
Linear independence implies
this, this implies linear

00:13:38.460 --> 00:13:40.256
independence.

00:13:40.256 --> 00:13:44.770
Now, this is the exact same
thing we did here, but in this

00:13:44.770 --> 00:13:48.330
case, I'm just picking my a's,
b's and c's to be zero.

00:13:48.330 --> 00:13:55.170
This is a, this is b and
this is c, right?

00:13:55.170 --> 00:13:59.030
I can pick any vector in R3
for my a's, b's and c's.

00:13:59.030 --> 00:14:01.030
I'm now picking the
zero vector.

00:14:01.030 --> 00:14:03.740
So let's see what our c1's,
c2's and c3's are.

00:14:03.740 --> 00:14:07.820
So my a equals b is equal
to c is equal to 0.

00:14:07.820 --> 00:14:09.940
I'm setting it equal
to the zero vector.

00:14:09.940 --> 00:14:13.980
What linear combination of these
three vectors equal the

00:14:13.980 --> 00:14:15.490
zero vector?

00:14:15.490 --> 00:14:20.150
Well, if a, b, and c are all
equal to 0, that term is 0,

00:14:20.150 --> 00:14:21.910
that is 0, that is 0.

00:14:21.910 --> 00:14:26.350
You have 1/11 times
0 minus 0 plus 0.

00:14:26.350 --> 00:14:27.380
That's just 0.

00:14:27.380 --> 00:14:30.140
So c3 is equal to 0.

00:14:30.140 --> 00:14:33.320
Now, if c3 is equal to 0, we
already know that a is equal

00:14:33.320 --> 00:14:35.210
to 0 and b is equal to 0.

00:14:35.210 --> 00:14:39.090
C2 is 1/3 times 0,
so it equals 0.

00:14:39.090 --> 00:14:39.850
Now what's c1?

00:14:39.850 --> 00:14:42.400
Well, it's c3, which is 0.

00:14:42.400 --> 00:14:45.240
c2 is 0, so 2 times 0 is 0.

00:14:45.240 --> 00:14:47.460
So c1 is just going
to be equal to a.

00:14:47.460 --> 00:14:49.960
I just said a is equal to 0.

00:14:49.960 --> 00:14:58.210
So the only solution to this
equation right here, the only

00:14:58.210 --> 00:15:01.570
linear combination of these
three vectors that result in

00:15:01.570 --> 00:15:05.690
the zero vector are when you
weight all of them by zero.

00:15:05.690 --> 00:15:11.600
So I just showed you that c1, c2
and c3 all have to be zero.

00:15:11.600 --> 00:15:14.820
And because they're all zero,
we know that this is a

00:15:14.820 --> 00:15:22.240
linearly independent
set of vectors.

00:15:22.240 --> 00:15:25.610
Or that none of these vectors
can be represented as a

00:15:25.610 --> 00:15:27.680
combination of the other two.

00:15:32.150 --> 00:15:32.750
This is interesting.

00:15:32.750 --> 00:15:36.170
I have exactly three vectors
that span R3 and they're

00:15:36.170 --> 00:15:37.820
linearly independent.

00:15:37.820 --> 00:15:40.810
And linearly independent, in my
brain that means, look, I

00:15:40.810 --> 00:15:43.640
don't have any redundant
vectors, anything that could

00:15:43.640 --> 00:15:46.340
have just been built with the
other vectors, and I have

00:15:46.340 --> 00:15:49.190
exactly three vectors,
and it's spanning R3.

00:15:49.190 --> 00:15:51.970
So in general, and I haven't
proven this to you, but I

00:15:51.970 --> 00:15:55.070
could, is that if you have
exactly three vectors and they

00:15:55.070 --> 00:15:58.590
do span R3, they have to be
linearly independent.

00:15:58.590 --> 00:16:01.680
If they weren't linearly
independent, then one of these

00:16:01.680 --> 00:16:03.930
would be redundant.

00:16:03.930 --> 00:16:05.490
Let's say that that guy
was a redundant one.

00:16:05.490 --> 00:16:07.150
I always pick the third one, but
let's say this guy would

00:16:07.150 --> 00:16:10.580
be redundant, which means that
the span of this would be

00:16:10.580 --> 00:16:13.790
equal to the span of
these two, right?

00:16:13.790 --> 00:16:16.200
Because if this guy is
redundant, he could just be

00:16:16.200 --> 00:16:18.730
part of the span of
these two guys.

00:16:18.730 --> 00:16:22.760
And the span of two of vectors
could never span R3.

00:16:22.760 --> 00:16:24.850
Or the other way you could go,
if you have three linear

00:16:24.850 --> 00:16:29.950
independent-- three tuples, and
they're all independent,

00:16:29.950 --> 00:16:31.960
then you can also say
that that spans R3.

00:16:31.960 --> 00:16:33.820
I haven't proven that to you,
but hopefully, you get the

00:16:33.820 --> 00:16:37.450
sense that each of these
is contributing new

00:16:37.450 --> 00:16:38.960
directionality, right?

00:16:38.960 --> 00:16:41.220
One is going like that.

00:16:41.220 --> 00:16:43.510
They're not completely
orthogonal to each other, but

00:16:43.510 --> 00:16:46.040
they're giving just enough
directionality that you can

00:16:46.040 --> 00:16:49.190
add a new dimension to
what's going on.

00:16:49.190 --> 00:16:51.360
Hopefully, that helped you a
bit, and I'll see you in the

00:16:51.360 --> 00:16:52.840
next video.


More on linear independence

WEBVTT
Kind: captions
Language: en

00:00:00.610 --> 00:00:03.380
I think by now we have a
reasonable sense of what

00:00:03.380 --> 00:00:07.990
linear dependence means.

00:00:07.990 --> 00:00:11.240
So let's just do a slightly
more formal definition of

00:00:11.240 --> 00:00:13.010
linear dependence.

00:00:13.010 --> 00:00:16.390
So we're going to say that a set
of vectors-- Let me just

00:00:16.390 --> 00:00:18.800
define my set of vectors.

00:00:18.800 --> 00:00:24.230
Let me call my set s
of vectors, v1, v2,

00:00:24.230 --> 00:00:27.760
all the way to vn.

00:00:27.760 --> 00:00:29.845
I'm going to say that they
are linearly dependent.

00:00:36.970 --> 00:00:38.320
If and only if.

00:00:38.320 --> 00:00:42.950
So sometimes it's written if, if
with a lot of f's in there.

00:00:42.950 --> 00:00:45.410
So sometimes it's written
if and only if.

00:00:45.410 --> 00:00:49.570
Sometimes it's shown like an
arrow in two directions.

00:00:49.570 --> 00:00:54.290
If and only if I can satisfy
this equation, I can find a

00:00:54.290 --> 00:00:59.250
set of constants c1 times v1.

00:00:59.250 --> 00:01:03.750
I can take a linear combination
of my vectors all

00:01:03.750 --> 00:01:10.260
the way to cn vn, that satisfy
the equation that I can create

00:01:10.260 --> 00:01:13.070
this into the 0 vector.

00:01:13.070 --> 00:01:16.760
Sometimes it's just written as
a bold 0, and sometimes you

00:01:16.760 --> 00:01:18.300
could just write it -- I
mean we don't know the

00:01:18.300 --> 00:01:20.100
dimensionality of this vector.

00:01:20.100 --> 00:01:22.610
It would be a bunch of 0's.

00:01:22.610 --> 00:01:29.040
We don't know how many actual
elements are in each of these

00:01:29.040 --> 00:01:32.280
vectors, but you get the idea.

00:01:32.280 --> 00:01:35.770
My set of vectors is linearly
dependent-- remember I'm

00:01:35.770 --> 00:01:39.030
saying dependent, not
independent --is linearly

00:01:39.030 --> 00:01:43.010
dependant, if and only if I can
satisfy this equation for

00:01:43.010 --> 00:01:51.335
some ci's where not all of
them are equal to 0.

00:01:54.050 --> 00:01:58.650
This is key, not all are 0.

00:01:58.650 --> 00:02:00.990
Or you could say it
the other way.

00:02:00.990 --> 00:02:06.465
You could say at least
one is non-zero.

00:02:09.590 --> 00:02:12.010
So how does this gel with what
we were talking about in the

00:02:12.010 --> 00:02:15.300
previous video where I said
look, a set is linearly

00:02:15.300 --> 00:02:19.730
dependent if one of the of
vectors can be represented by

00:02:19.730 --> 00:02:24.690
the combination of the
other vectors?

00:02:24.690 --> 00:02:25.800
Let me write that down.

00:02:25.800 --> 00:02:35.320
In the last few I said, look,
one vector can be -- Let me

00:02:35.320 --> 00:02:37.220
write it this way.

00:02:37.220 --> 00:02:39.330
One vector being represented
by the some of the other

00:02:39.330 --> 00:02:41.050
vectors, I can just write
it like this.

00:02:41.050 --> 00:02:43.160
I can write it a little
bit more math-y.

00:02:43.160 --> 00:02:46.380
In the last video, I said that
linear dependence means that--

00:02:46.380 --> 00:02:50.280
let me just pick an arbitrary
vector, v1.

00:02:50.280 --> 00:02:53.480
Let's say that v1, you know
this is arbitrary, v1 one

00:02:53.480 --> 00:02:55.490
could be represented
by some combination

00:02:55.490 --> 00:02:57.570
of the other vectors.

00:02:57.570 --> 00:03:04.070
Let me call them a1 times v --
let me be careful -- a2 times

00:03:04.070 --> 00:03:11.170
v2 plus a3 times v3 plus all
the way up to an times vn.

00:03:11.170 --> 00:03:13.980
This is what we said in
the previous video.

00:03:13.980 --> 00:03:16.860
If this is linear dependence,
any one of these guys can be

00:03:16.860 --> 00:03:19.930
represented as some combination
of the other ones.

00:03:19.930 --> 00:03:21.980
So how does this imply that?

00:03:21.980 --> 00:03:24.790
In order show this if and only
if, I have to show that this

00:03:24.790 --> 00:03:29.410
implies that and I have to show
that that implies this.

00:03:29.410 --> 00:03:31.870
So this is almost a trivially
easy proof.

00:03:31.870 --> 00:03:36.340
Because if I subtract v1 from
both sides of this equation I

00:03:36.340 --> 00:03:49.320
get 0 is equal to minus 1 v1
plus a2 v2 plus a3 v3 all the

00:03:49.320 --> 00:03:52.730
way to an vn.

00:03:52.730 --> 00:03:55.930
And clearly I've just
said, well,

00:03:55.930 --> 00:03:57.670
this is linearly dependent.

00:03:57.670 --> 00:04:00.160
That means that I can represent
this vector as a sum

00:04:00.160 --> 00:04:04.510
of the other vectors, which
means that minus 1 times v1

00:04:04.510 --> 00:04:07.770
plus some combination of the
other vectors is equal to 0,

00:04:07.770 --> 00:04:12.850
which means that I've been able
to satisfy this equation,

00:04:12.850 --> 00:04:15.960
and at least one of my constants
is non-zero.

00:04:19.740 --> 00:04:23.160
So I've shown you that, if I
can represent one of the

00:04:23.160 --> 00:04:26.440
vectors by a sum of the other
ones, then this condition is

00:04:26.440 --> 00:04:28.160
definitely going to be true.

00:04:28.160 --> 00:04:29.380
Now let me go the other way.

00:04:29.380 --> 00:04:32.810
Let me show you if I have this
situation that I can

00:04:32.810 --> 00:04:35.110
definitely represent one
of the vectors as

00:04:35.110 --> 00:04:36.890
the sum of the others.

00:04:36.890 --> 00:04:39.730
So let's say that
this is true.

00:04:42.270 --> 00:04:46.190
And one of these constants,
remember it's not just this,

00:04:46.190 --> 00:04:48.370
it's at least one,
is non-zero.

00:04:48.370 --> 00:04:50.860
So let me just assume, just for
the sake of simplicity-- I

00:04:50.860 --> 00:04:53.790
mean these are all arbitrary.

00:04:53.790 --> 00:04:55.270
I'll do it in a new color.

00:04:55.270 --> 00:04:57.610
Let me do it in the magenta.

00:04:57.610 --> 00:05:03.090
Let me assume that c1
is not equal to 0.

00:05:03.090 --> 00:05:06.370
If c1 is not equal to 0, then
I can divide both sides of

00:05:06.370 --> 00:05:08.180
this equation by c1.

00:05:08.180 --> 00:05:09.240
And what do I get?

00:05:09.240 --> 00:05:19.860
I get v1 plus c2 over c1 v2 plus
all the way up to cn over

00:05:19.860 --> 00:05:23.420
c1 is equal to 0.

00:05:23.420 --> 00:05:28.350
And then I can multiply both
sides of this, or I could add

00:05:28.350 --> 00:05:31.510
negative v1 to both sides of
this equation or subtract v1

00:05:31.510 --> 00:05:32.290
from both sides.

00:05:32.290 --> 00:05:41.290
And I get c2 over c1 v2 plus all
the way up to cn over c1

00:05:41.290 --> 00:05:46.890
vn-- there's a vn here --is
equal to minus v1.

00:05:46.890 --> 00:05:49.400
Now if I just multiply both
sides of this by negative 1, I

00:05:49.400 --> 00:05:52.200
get a minus, and all these
become minuses and this

00:05:52.200 --> 00:05:53.480
becomes a plus.

00:05:53.480 --> 00:05:56.090
And I just showed you that
if at least one of these

00:05:56.090 --> 00:06:02.160
constants is non-zero, that I
can represent my vector v1 as

00:06:02.160 --> 00:06:04.920
some combination of
the other vectors.

00:06:04.920 --> 00:06:07.920
So we're able to go
this way too.

00:06:07.920 --> 00:06:10.200
If this condition is true, then
I can represent one of

00:06:10.200 --> 00:06:12.040
the vectors as a combination
of the others.

00:06:12.040 --> 00:06:14.590
If I can represent one of those
vectors as a combination

00:06:14.590 --> 00:06:17.360
of the others, then this
condition is true.

00:06:17.360 --> 00:06:20.710
Hopefully that kind of proves
that these two definitions are

00:06:20.710 --> 00:06:21.240
equivalent.

00:06:21.240 --> 00:06:22.930
Maybe it's a little
bit of overkill.

00:06:22.930 --> 00:06:26.260
Let's apply that definition
now, to actually test. You

00:06:26.260 --> 00:06:28.690
might say, hey Sal, why'd you go
through all of this effort?

00:06:28.690 --> 00:06:30.180
I went through all of this
effort because this is

00:06:30.180 --> 00:06:36.770
actually a really useful way
to test whether things are

00:06:36.770 --> 00:06:39.460
linearly independent
or dependent.

00:06:39.460 --> 00:06:40.250
Let's try it out.

00:06:40.250 --> 00:06:43.170
Let's use our newly
found tool.

00:06:43.170 --> 00:06:47.520
Let's say I have the
set of vectors--

00:06:47.520 --> 00:06:48.460
Let me do it up here.

00:06:48.460 --> 00:06:51.980
I want to be efficient
with my space usage.

00:06:51.980 --> 00:06:59.450
So let's say I have the set
of vectors 2,1 and 3,2.

00:06:59.450 --> 00:07:03.470
And my question to you is, are
these linearly independent or

00:07:03.470 --> 00:07:06.270
are they linearly dependent?

00:07:06.270 --> 00:07:11.440
In order for them to be linearly
dependent, that means

00:07:11.440 --> 00:07:18.600
that if some constant times 2,1
plus some other constant

00:07:18.600 --> 00:07:23.090
times this second vector,
3,2 where this

00:07:23.090 --> 00:07:24.250
should be equal to 0.

00:07:24.250 --> 00:07:27.450
Where these both aren't
necessarily 0.

00:07:27.450 --> 00:07:30.570
Before I go up for this problem,
let's remember what

00:07:30.570 --> 00:07:32.210
we're going to find out.

00:07:32.210 --> 00:07:38.790
If either of these are non-zero,
if c1 or c2 are

00:07:38.790 --> 00:07:44.630
non-zero, then this implies
that we are dealing with a

00:07:44.630 --> 00:07:48.380
dependent, linearly
dependent set.

00:07:51.660 --> 00:07:58.430
If c1 and c2 are both 0, if the
only way to satisfy this

00:07:58.430 --> 00:08:02.060
equation -- I mean you can
always satisfy it by sitting

00:08:02.060 --> 00:08:02.680
everything equal to 0.

00:08:02.680 --> 00:08:06.520
But if the only way to satisfy
it is by making both of these

00:08:06.520 --> 00:08:08.830
guys 0, then we're
dealing with a

00:08:08.830 --> 00:08:10.570
linearly independent set.

00:08:13.860 --> 00:08:15.110
Let's try to do some math.

00:08:15.110 --> 00:08:19.280
And this'll just take us back
to our Algebra 1 days.

00:08:19.280 --> 00:08:25.880
In order for this to be true,
that means 2 times c1 plus 3

00:08:25.880 --> 00:08:29.870
times c2 is equal to -- when I
say this is equal to 0, it's

00:08:29.870 --> 00:08:30.980
really the 0 vector.

00:08:30.980 --> 00:08:34.669
I can rewrite this as 0,0.

00:08:34.669 --> 00:08:37.120
So 2 times c1 plus 3
times c2 would be

00:08:37.120 --> 00:08:39.159
equal to that 0 there.

00:08:39.159 --> 00:08:46.480
And then we'd have 1 times
c1 plus 2 times c2 is

00:08:46.480 --> 00:08:47.730
equal to that 0.

00:08:50.600 --> 00:08:53.870
And now this is just a system,
two equations, two unknowns.

00:08:53.870 --> 00:08:55.420
A couple of things
we could do.

00:08:55.420 --> 00:09:02.340
Let's just multiply this
top equation by 1/2.

00:09:02.340 --> 00:09:12.170
If you multiply it by 1/2 you
get c1 plus 3/2 plus 3/2 c2 is

00:09:12.170 --> 00:09:15.200
equal to 0.

00:09:15.200 --> 00:09:18.460
And then if we subtract the
green equation from the red

00:09:18.460 --> 00:09:20.410
equation this becomes 0.

00:09:20.410 --> 00:09:25.290
2 minus 1 and 1/2-- 3/2 is 1 and
1/2 --of this is just 1/2

00:09:25.290 --> 00:09:27.960
c2 is equal to 0.

00:09:27.960 --> 00:09:29.390
And this is easy to solve.

00:09:29.390 --> 00:09:32.090
c2 is equal to 0.

00:09:32.090 --> 00:09:33.770
So what's c1?

00:09:33.770 --> 00:09:36.350
Well, just substitute this back
in. c2 is equal to 0.

00:09:36.350 --> 00:09:38.270
So this is equal to 0.

00:09:38.270 --> 00:09:40.720
So c1 plus 0 is equal to 0.

00:09:40.720 --> 00:09:42.940
So c1 is also equal to 0.

00:09:42.940 --> 00:09:44.190
We could have substituted
it back into that

00:09:44.190 --> 00:09:45.440
top equation as well.

00:09:45.440 --> 00:09:52.320
So the only solution to this
equation involves both c1 and

00:09:52.320 --> 00:09:54.610
c2 being equal to 0.

00:09:54.610 --> 00:09:56.680
So they both have to be 0.

00:09:56.680 --> 00:10:04.110
So this is a linearly
independent set of vectors.

00:10:04.110 --> 00:10:06.790
Which means that neither
of them are redundant

00:10:06.790 --> 00:10:07.480
of the other one.

00:10:07.480 --> 00:10:10.600
You can't represent one as a
combination of the other.

00:10:10.600 --> 00:10:13.550
And since we have two vectors
here, and they're linearly

00:10:13.550 --> 00:10:16.790
independent, we can actually
know that this will span r2.

00:10:22.320 --> 00:10:26.910
The span of my r vectors
is equal to r2.

00:10:26.910 --> 00:10:29.290
If one of these vectors was
just some multiple of the

00:10:29.290 --> 00:10:32.610
other, than the span would have
been some line within r2,

00:10:32.610 --> 00:10:33.260
not all of.

00:10:33.260 --> 00:10:35.400
But now I can represent any
vector in r2 as some

00:10:35.400 --> 00:10:37.910
combination of those.

00:10:37.910 --> 00:10:41.300
Let's do another example.

00:10:41.300 --> 00:10:46.170
Let me scroll to the right,
because sometimes this thing,

00:10:46.170 --> 00:10:49.060
when I go too far down, I
haven't figured out why, when

00:10:49.060 --> 00:10:51.920
I go too far down it
starts messing up.

00:10:51.920 --> 00:10:56.480
So my next example is
the set of vectors.

00:10:56.480 --> 00:11:00.110
So I have the vector 2,1.

00:11:00.110 --> 00:11:04.520
I have the vector 3,2.

00:11:04.520 --> 00:11:08.490
And I have the vector 1,2.

00:11:08.490 --> 00:11:11.690
And I want to know are these
linearly dependent or linearly

00:11:11.690 --> 00:11:12.440
independent.

00:11:12.440 --> 00:11:13.800
So I go to through
the same drill.

00:11:13.800 --> 00:11:17.890
I use that little theorem that
I proved at the beginning of

00:11:17.890 --> 00:11:19.090
this video.

00:11:19.090 --> 00:11:24.710
In order for them to be linearly
dependent there must

00:11:24.710 --> 00:11:28.840
be some set of weights that
I can multiply these guys.

00:11:28.840 --> 00:11:37.100
So c1 times this vector plus c2
times this vector plus c3

00:11:37.100 --> 00:11:43.060
times that vector, that will
equal the 0 vector.

00:11:43.060 --> 00:11:45.690
And if one of these is non-zero
then we're dealing

00:11:45.690 --> 00:11:49.390
with a linearly dependent
set of vectors.

00:11:49.390 --> 00:11:52.790
And if all of them are 0,
then it's independent.

00:11:52.790 --> 00:11:54.445
Let's just do our
linear algebra.

00:12:00.430 --> 00:12:10.520
So this means that 2 times c1
plus 3 times c2 plus c3 is

00:12:10.520 --> 00:12:13.280
equal to that 0 up there.

00:12:13.280 --> 00:12:15.280
And then if we do the bottom
rows-- Remember when you

00:12:15.280 --> 00:12:17.270
multiply a scalar times a vector
you multiply it by each

00:12:17.270 --> 00:12:19.880
of these terms. So c1 times 1.

00:12:19.880 --> 00:12:29.200
1c1 plus 2c2 plus 2c3
is equal to 0.

00:12:29.200 --> 00:12:31.350
There's a couple of giveaways
on this problem.

00:12:31.350 --> 00:12:34.570
If you have three
two-dimensional vectors, one

00:12:34.570 --> 00:12:37.070
of them is going to
be redundant.

00:12:37.070 --> 00:12:41.650
Because, in the very best case,
even if you assume that

00:12:41.650 --> 00:12:44.540
that vector and that vector are
linearly independent, then

00:12:44.540 --> 00:12:45.740
these would span r2.

00:12:45.740 --> 00:12:49.270
Which means that any point,
any vector, in your

00:12:49.270 --> 00:12:51.250
two-dimensional space can
be represented by some

00:12:51.250 --> 00:12:53.140
combination of those two.

00:12:53.140 --> 00:12:56.280
In which case, this is going to
be one of them because this

00:12:56.280 --> 00:12:57.935
is just a vector in
two-dimensional space.

00:12:57.935 --> 00:12:59.940
So it would be linearly
dependent.

00:12:59.940 --> 00:13:01.890
And then, if you say, well,
these aren't linearly

00:13:01.890 --> 00:13:07.030
independent then, they're just
multiples of each other.

00:13:07.030 --> 00:13:09.900
In which case, this would
definitely be a linearly

00:13:09.900 --> 00:13:11.450
dependent set.

00:13:11.450 --> 00:13:15.290
When you see three vectors that
are each only vectors in

00:13:15.290 --> 00:13:17.910
r2, that are each
two-dimensional vectors, it's

00:13:17.910 --> 00:13:20.580
a complete giveaway that this
is linearly dependent.

00:13:20.580 --> 00:13:25.460
But I'm going to show it to you
using our dependent, using

00:13:25.460 --> 00:13:27.560
our little theorem here.

00:13:27.560 --> 00:13:30.940
So I'm going to show you that I
can get non-zero c3's, c2's,

00:13:30.940 --> 00:13:34.270
and c1's such that I
can get a 0 here.

00:13:34.270 --> 00:13:37.000
If all of these had to be 0-- I
mean you can always set them

00:13:37.000 --> 00:13:37.440
equal to 0.

00:13:37.440 --> 00:13:39.580
But if they had to be equal to
0, then it would be linearly

00:13:39.580 --> 00:13:41.190
independent.

00:13:41.190 --> 00:13:43.220
Let me just show you.

00:13:43.220 --> 00:13:45.290
I can just pick some
random c3.

00:13:45.290 --> 00:13:49.420
Let me pick c3 to be equal
to negative 1.

00:13:49.420 --> 00:13:51.610
So what would these two
equations reduce to?

00:13:51.610 --> 00:13:53.920
I mean you have just three
unknowns and two equations, it

00:13:53.920 --> 00:13:57.050
means you don't have enough
constraints on your system.

00:13:57.050 --> 00:13:59.640
So if I just set c3-- I just
pick that out of a hat.

00:13:59.640 --> 00:14:01.290
I could have picked
c3 to be anything.

00:14:01.290 --> 00:14:05.460
But if I set c3 to be equal to
negative 1, what do these

00:14:05.460 --> 00:14:06.230
equations become?

00:14:06.230 --> 00:14:12.790
You get 2c1 plus 3c2 minus
1 is equal to 0.

00:14:12.790 --> 00:14:19.560
And you get c1 plus 2c2
minus 2 is equal to 0.

00:14:19.560 --> 00:14:19.830
Right?

00:14:19.830 --> 00:14:22.780
2 times minus 1.

00:14:22.780 --> 00:14:24.010
What can I do here?

00:14:24.010 --> 00:14:28.950
If I multiply this second
equation by 2, what do I get?

00:14:28.950 --> 00:14:37.360
I get 2 plus 4c2 minus
4 is equal to 0.

00:14:37.360 --> 00:14:41.580
And now let's subtract this
equation from that equation.

00:14:41.580 --> 00:14:43.600
So the c1's cancel out.

00:14:43.600 --> 00:14:47.960
3c2 minus 4c2 is minus c2.

00:14:47.960 --> 00:14:53.840
And then minus 1 minus minus 4,
so that's minus 1 plus 4.

00:14:53.840 --> 00:14:58.620
That's plus 3 is equal to 0.

00:14:58.620 --> 00:15:01.860
And so we get our -- Let me make
sure I got that right.

00:15:06.640 --> 00:15:10.190
We have a minus 1
minus a minus 4.

00:15:10.190 --> 00:15:11.150
So plus 4.

00:15:11.150 --> 00:15:13.170
So we have a plus 3.

00:15:13.170 --> 00:15:15.410
So that is a minus 2.

00:15:20.150 --> 00:15:28.225
So minus c2 is equal to minus
3 or c2 is equal to 3.

00:15:28.225 --> 00:15:32.480
And if c2 is equal to 3 and c3
is equal to minus 1-- Let's

00:15:32.480 --> 00:15:38.510
just substitute here, so we
get c1 plus 2 times c2, so

00:15:38.510 --> 00:15:41.760
plus 6, plus 2 times c3.

00:15:41.760 --> 00:15:45.980
So minus 2 is equal to 0.

00:15:45.980 --> 00:15:48.830
c1 plus 4 is equal to 0.

00:15:48.830 --> 00:15:51.310
c1 is equal to minus 4.

00:15:54.250 --> 00:15:56.720
I'm giving you a combination
of c's that

00:15:56.720 --> 00:15:58.210
will give us a 0 vector.

00:15:58.210 --> 00:16:03.920
If I multiply minus 4 times our
first vector, 2,1, that's

00:16:03.920 --> 00:16:13.150
c1, plus 3 times our second
vector, 3,2 minus 1 times our

00:16:13.150 --> 00:16:16.720
third vector, 1,2 this
should be equal to 0.

00:16:16.720 --> 00:16:18.510
Let's verify it just for fun.

00:16:18.510 --> 00:16:24.660
Minus 4 times 2 is minus
8 plus 9 minus 1.

00:16:24.660 --> 00:16:26.590
Yeah, that's minus 9 plus 9.

00:16:26.590 --> 00:16:27.490
That's 0.

00:16:27.490 --> 00:16:36.010
Minus 4 times plus 6 minus
2 that's also 0.

00:16:36.010 --> 00:16:39.370
So we've just shown a linear
combination of these vectors,

00:16:39.370 --> 00:16:41.820
where actually none of
the constants are 0.

00:16:41.820 --> 00:16:43.840
But all we had to show was
that at least one of the

00:16:43.840 --> 00:16:46.680
constants had to be non-zero,
and we actually showed all

00:16:46.680 --> 00:16:47.690
three of them were.

00:16:47.690 --> 00:16:49.850
But at least one of these
had to be non-zero.

00:16:49.850 --> 00:16:53.100
And I was able to satisfy this
equation, I was able to make

00:16:53.100 --> 00:16:54.560
them into the zero vector.

00:16:54.560 --> 00:17:02.460
So this shows, this proves,
that this is a linearly

00:17:02.460 --> 00:17:07.520
dependent set of vectors.

00:17:07.520 --> 00:17:09.450
Which means one of the
vectors is redundant.

00:17:09.450 --> 00:17:11.520
And you can never just say,
oh, this is the redundant

00:17:11.520 --> 00:17:13.410
vector, because I can represent
this as combination

00:17:13.410 --> 00:17:14.170
of those two.

00:17:14.170 --> 00:17:16.589
You could just as easily pick
this guy as the redundant

00:17:16.589 --> 00:17:19.960
vector, and say, hey, I can
represent this guy as the sum

00:17:19.960 --> 00:17:20.609
of those two.

00:17:20.609 --> 00:17:23.380
There's not one bad apple
in the bunch.

00:17:23.380 --> 00:17:26.640
Any of them can be represented
by the combination of some

00:17:26.640 --> 00:17:29.580
other, by all of the
rest of them.

00:17:29.580 --> 00:17:33.180
So hopefully you have a better
intuition of linear dependence

00:17:33.180 --> 00:17:33.780
and independence.

00:17:33.780 --> 00:17:34.580
Maybe I'll continue.

00:17:34.580 --> 00:17:37.400
I'll do a few more examples
in the next video.