Proof: Any subspace basis has same number of elements

WEBVTT
Kind: captions
Language: en

00:00:00.490 --> 00:00:06.170
Let's say I've got some set
A of the vectors a1, a2,

00:00:06.170 --> 00:00:08.109
all the way to an.

00:00:08.109 --> 00:00:12.620
And I know for a fact that
it's a basis for

00:00:12.620 --> 00:00:14.990
the subspace V.

00:00:14.990 --> 00:00:18.400
What I want to show you in this
video is that if this guy

00:00:18.400 --> 00:00:25.210
has n elements right here, that
any set that spans V has

00:00:25.210 --> 00:00:29.210
to have at least n elements,
[typing]

00:00:43.070 --> 00:00:46.160
or n members, or cardinality
of n.

00:00:46.160 --> 00:00:48.020
There's all different ways
of saying you had n

00:00:48.020 --> 00:00:49.270
vectors in this set.

00:00:52.700 --> 00:00:56.920
I'm saying that every set that
spans V must have at least n

00:00:56.920 --> 00:01:01.340
elements, if the sum basis
set has n elements for V.

00:01:01.340 --> 00:01:04.959
Let's see if we can kind of run
with the set that has less

00:01:04.959 --> 00:01:07.820
than n elements and see if we
reach any contradictions.

00:01:07.820 --> 00:01:13.155
So let's say that I have sum set
B here and it's equal to

00:01:13.155 --> 00:01:18.330
the vectors b1, b2,
all the way to bm.

00:01:18.330 --> 00:01:22.420
And m is less than n, so I have
sum set of vectors here

00:01:22.420 --> 00:01:27.060
that have fewer elements
than my set A.

00:01:27.060 --> 00:01:30.180
So you come to me one day and
say, look I found you this set

00:01:30.180 --> 00:01:31.390
of vectors right here.

00:01:31.390 --> 00:01:34.140
And not only does it have
fewer elements than A,

00:01:34.140 --> 00:01:37.500
but it spans V.

00:01:37.500 --> 00:01:39.460
I look at you very suspiciously
because I always

00:01:39.460 --> 00:01:41.440
thought that this green
statement was true.

00:01:41.440 --> 00:01:43.860
So we start a little bit of
a thought experiment.

00:01:43.860 --> 00:01:46.970
And I say OK, you claim
that your set spans

00:01:46.970 --> 00:01:48.900
V, so let's do something.

00:01:48.900 --> 00:01:50.500
Let me define a new set.

00:01:50.500 --> 00:01:55.080
Let me call this new set B1
prime, and you'll see why I'm

00:01:55.080 --> 00:01:57.430
doing this kind of
strange notation.

00:01:57.430 --> 00:02:03.230
What's essentially going to be
is a set B plus my vector a1.

00:02:03.230 --> 00:02:07.070
So it's a1, and then I have
all of my elements of B.

00:02:07.070 --> 00:02:12.350
So be b1, b2, all
the way to bm.

00:02:12.350 --> 00:02:15.390
Now I think you and I could both
agree that this set is

00:02:15.390 --> 00:02:18.530
linearly dependent.

00:02:18.530 --> 00:02:19.240
How do I know that?

00:02:19.240 --> 00:02:22.240
Linear dependence means that at
least one of the elements

00:02:22.240 --> 00:02:24.100
of the set can be represented
as a linear

00:02:24.100 --> 00:02:25.720
combination of the others.

00:02:25.720 --> 00:02:30.610
Well, we know that a1 is one of
the basis vectors for V for

00:02:30.610 --> 00:02:31.800
this definition of a basis.

00:02:31.800 --> 00:02:36.510
But all the basis vectors
are members of V.

00:02:41.520 --> 00:02:44.970
If this set is a basis for V,
then this means that this set

00:02:44.970 --> 00:02:48.360
spans V, or that every member
of V can be represented as a

00:02:48.360 --> 00:02:49.990
linear combination
of these guys.

00:02:49.990 --> 00:02:51.980
Or in other ways every
linear combination of

00:02:51.980 --> 00:02:53.710
these guys is in V.

00:02:53.710 --> 00:02:55.600
And one of the linear
combinations of these guys is

00:02:55.600 --> 00:03:01.280
you just set the coefficient
on a1 to be 1, and the

00:03:01.280 --> 00:03:03.240
coefficients on everyone
else to be 0.

00:03:03.240 --> 00:03:05.610
So obviously a1 is
also in the set.

00:03:05.610 --> 00:03:12.110
So if a1 is in V, and all
of these guys span V, by

00:03:12.110 --> 00:03:15.680
definition if these guys span
V some linear combination of

00:03:15.680 --> 00:03:19.390
these guys can be used to
construct any member of V.

00:03:19.390 --> 00:03:21.910
So you can take some linear
combination of these guys to

00:03:21.910 --> 00:03:23.010
construct a1.

00:03:23.010 --> 00:03:29.590
So you could say a1 is equal to
d1, where the d's are the

00:03:29.590 --> 00:03:37.950
constants, d1 b1 plus d2 b2, all
the way to dm bm, and at

00:03:37.950 --> 00:03:40.080
least one of these have
to be non-zero.

00:03:40.080 --> 00:03:42.010
We know that a is a
non-zero vector.

00:03:42.010 --> 00:03:44.800
If it was a 0 vector, this
couldn't be a basis because it

00:03:44.800 --> 00:03:47.570
wouldn't be linearly
independent, because you can

00:03:47.570 --> 00:03:51.210
always represent a 0 vector
as really just 0

00:03:51.210 --> 00:03:52.270
times any other vector.

00:03:52.270 --> 00:03:53.830
So this won't be a 0 vector.

00:03:53.830 --> 00:03:56.390
So at least one of these
are non-zero.

00:03:56.390 --> 00:04:01.130
So let's just say, for the sake
of argument, that dj --

00:04:01.130 --> 00:04:04.550
so the coefficient on
bj is non-zero.

00:04:04.550 --> 00:04:08.610
So dj does not equal 0.

00:04:08.610 --> 00:04:11.910
So we could actually solve
for that term.

00:04:11.910 --> 00:04:16.959
So over here someplace you have
the term dj bj, plus a

00:04:16.959 --> 00:04:18.010
bunch of other stuff.

00:04:18.010 --> 00:04:21.110
We can solve for this term if we
subtract it from both sides

00:04:21.110 --> 00:04:25.070
of the equation, and then divide
both sides by minus dj,

00:04:25.070 --> 00:04:29.360
and put this minus a1 on the
other side, what do we get?

00:04:29.360 --> 00:04:30.960
I know that was a lot of
operations, but that's just

00:04:30.960 --> 00:04:31.750
straight up algebra.

00:04:31.750 --> 00:04:34.470
I think you can say that we
could rewrite this right here.

00:04:34.470 --> 00:04:35.720
We can solve for bj.

00:04:41.360 --> 00:04:46.470
That should be equal to minus
1 over its coefficient.

00:04:46.470 --> 00:04:51.620
And, if we subtract the a1 from
both sides and then plus

00:04:51.620 --> 00:04:57.640
all of these guys, plus d1 b1
plus all the way -- you're

00:04:57.640 --> 00:04:59.300
going to have a little
bit of a gap here.

00:04:59.300 --> 00:05:00.080
I'll just draw it like that.

00:05:00.080 --> 00:05:02.370
It's very unconventional
notation where this guy sat.

00:05:02.370 --> 00:05:06.090
All the way to dm bm.

00:05:06.090 --> 00:05:10.160
I'm doing this all of this to
show you that, by definition,

00:05:10.160 --> 00:05:12.240
you can write a1 as a
linear combination

00:05:12.240 --> 00:05:13.050
of these other guys.

00:05:13.050 --> 00:05:14.400
But you can just rearrange
things.

00:05:14.400 --> 00:05:18.160
You can rearrange it so that you
can write one of the other

00:05:18.160 --> 00:05:20.180
guys as a linear combination
of the rest of the

00:05:20.180 --> 00:05:21.910
other guys and a1.

00:05:24.490 --> 00:05:27.330
This guy's now redundant.

00:05:27.330 --> 00:05:30.780
I don't need this guy any longer
to continue to span V.

00:05:30.780 --> 00:05:32.860
Clearly this set
still spans V.

00:05:32.860 --> 00:05:35.270
I added an extra vector here.

00:05:35.270 --> 00:05:41.390
But I can remove this guy right
here from my set b1

00:05:41.390 --> 00:05:43.980
prime and still span V.

00:05:43.980 --> 00:05:45.080
And how do I know that?

00:05:45.080 --> 00:05:47.410
Because I can achieve him.

00:05:47.410 --> 00:05:49.220
By removing him, I don't
lose anything.

00:05:49.220 --> 00:05:51.550
Because if I needed this vector
to create some other

00:05:51.550 --> 00:05:54.310
vector, I can construct him with
a linear combination of

00:05:54.310 --> 00:05:59.180
the rest of the b's,
plus my a1.

00:05:59.180 --> 00:06:02.680
So let's get rid of him,
and call that set v1.

00:06:02.680 --> 00:06:05.270
And actually, just for the sake
of notation, let me just

00:06:05.270 --> 00:06:07.600
change his name.

00:06:07.600 --> 00:06:08.840
This is little unconventional.

00:06:08.840 --> 00:06:11.350
You won't see it done like
this in any textbook.

00:06:11.350 --> 00:06:13.470
But I think it's a little bit
easier, instead of having to

00:06:13.470 --> 00:06:16.420
keep talking about these little
guys that are embedded

00:06:16.420 --> 00:06:18.250
someplace in the middle
of the stream.

00:06:18.250 --> 00:06:24.380
I mean these names, b1, b2, bn,
they're arbitrary names.

00:06:24.380 --> 00:06:25.690
Let me swap the labels.

00:06:25.690 --> 00:06:29.860
Let me just say that bj is equal
to b1, and that b1 is

00:06:29.860 --> 00:06:30.660
equal to bj.

00:06:30.660 --> 00:06:34.230
I'm just swapping their names.

00:06:34.230 --> 00:06:36.590
I took that guy, I
renamed him b1.

00:06:36.590 --> 00:06:40.120
I renamed b1, bj, so that
I could swap them.

00:06:40.120 --> 00:06:42.610
So I'm essentially just going
to remove b1 one from the

00:06:42.610 --> 00:06:44.210
vector just to make my
notation easier.

00:06:44.210 --> 00:06:46.490
You could just keep saying I'll
remove this bj from the

00:06:46.490 --> 00:06:48.980
middle, but it becomes
very confusing then.

00:06:48.980 --> 00:06:53.150
So let me call my new set after
removing the bj that

00:06:53.150 --> 00:06:58.880
I've renamed as b1, let me just
call that straight up B1.

00:06:58.880 --> 00:07:06.470
So my straight set B1 is equal
to a1, and then remember I

00:07:06.470 --> 00:07:09.100
removed the bj and I renamed
that as b1, and then I

00:07:09.100 --> 00:07:10.380
renamed b1 as bj.

00:07:10.380 --> 00:07:12.840
So now my set looks like
this -- let me go

00:07:12.840 --> 00:07:14.900
to the other color.

00:07:14.900 --> 00:07:18.130
b2 -- and for all we know
bj might have been

00:07:18.130 --> 00:07:20.830
b1, we don't know.

00:07:20.830 --> 00:07:23.930
There are probably multiple of
these that are non-zero, so we

00:07:23.930 --> 00:07:26.480
could pick any of those
to be our bj.

00:07:26.480 --> 00:07:30.150
But anyway, we took our
bj renamed it b1,

00:07:30.150 --> 00:07:31.590
and removed the b1.

00:07:31.590 --> 00:07:36.130
So now out set looks like this:
b3 all the way to bm.

00:07:39.320 --> 00:07:44.100
This set still spans V.

00:07:44.100 --> 00:07:47.230
And we know that because the
guy we removed can be

00:07:47.230 --> 00:07:50.270
constructed with any linear
combination of these guys.

00:07:50.270 --> 00:07:53.130
So we haven't lost our ability
to construct all of the

00:07:53.130 --> 00:07:54.480
vectors in V.

00:07:54.480 --> 00:07:55.730
Now let me create
another vector.

00:07:59.420 --> 00:08:01.160
We'll do a new color.

00:08:01.160 --> 00:08:05.430
Let's say I have the
vector B2 prime.

00:08:05.430 --> 00:08:07.110
And what I'm going to do here
is now I'm going to take

00:08:07.110 --> 00:08:10.700
another element from
our basis of V.

00:08:10.700 --> 00:08:13.560
I'll take the second element;
I'll take a2.

00:08:13.560 --> 00:08:15.820
I'll take a2 and throw
it on this guy.

00:08:15.820 --> 00:08:23.970
So now we have the set B2 prime
is equal to -- I'm going

00:08:23.970 --> 00:08:25.220
to add a2 to this guy.

00:08:25.220 --> 00:08:31.090
So you have a1, a2, and then you
have all the rest of these

00:08:31.090 --> 00:08:36.360
guys, b2, b3 all the way bm.

00:08:36.360 --> 00:08:39.169
Of course this still spans V,
I just added something here.

00:08:39.169 --> 00:08:41.460
But this is definitely
linearly dependent.

00:08:41.460 --> 00:08:43.240
Remember I didn't say in the
beginning whether this was

00:08:43.240 --> 00:08:44.710
linear dependent or not.

00:08:44.710 --> 00:08:45.700
It may or may not be.

00:08:45.700 --> 00:08:47.650
But when you add this other
vector that's in V, you

00:08:47.650 --> 00:08:49.490
definitely know that you're
linear dependent, because

00:08:49.490 --> 00:08:51.730
these guys can construct
that guy.

00:08:51.730 --> 00:08:56.380
Similarly we know that this
vector B1 spans V.

00:08:56.380 --> 00:09:01.550
So when we add this new element
here, we know that it

00:09:01.550 --> 00:09:03.950
can be written as a linear
combination of the other one.

00:09:03.950 --> 00:09:06.940
So we know that this right here
is linearly dependent.

00:09:06.940 --> 00:09:12.770
And we could say that a2 is
equal to some constant times

00:09:12.770 --> 00:09:28.260
c1 times a1 plus c2 b2 plus c3
b3, all the way to cm bm.

00:09:28.260 --> 00:09:32.240
Now what I'm going to claim is
that at least one of these

00:09:32.240 --> 00:09:33.610
coefficients is non-zero.

00:09:33.610 --> 00:09:37.160
So at least one of the ci's
does not equal to 0.

00:09:37.160 --> 00:09:40.800
And I'll make the further claim
that there's at least

00:09:40.800 --> 00:09:43.170
one that's outside
of this one.

00:09:46.440 --> 00:09:50.670
That at least one of the
coefficients on these B terms

00:09:50.670 --> 00:09:52.230
has to be non-zero.

00:09:52.230 --> 00:09:55.350
And the way you can think about
is, what if all of these

00:09:55.350 --> 00:09:56.750
guys were 0?

00:09:56.750 --> 00:10:00.070
If all of these guys were 0,
then that means that a2 is a

00:10:00.070 --> 00:10:01.490
linear combination of a1.

00:10:01.490 --> 00:10:03.640
All of these guys would cancel
out, and you'd have a2 is

00:10:03.640 --> 00:10:06.440
equal to some non-zero
constant times a1.

00:10:06.440 --> 00:10:09.440
We know that's not the case
because these two guys come

00:10:09.440 --> 00:10:13.630
from the same linearly
independent set.

00:10:13.630 --> 00:10:17.150
They both come from that
spanning basis.

00:10:17.150 --> 00:10:20.180
The fact that they are a basis
-- the word spanning basis, I

00:10:20.180 --> 00:10:21.600
shouldn't say it like that,
because it's redundant.

00:10:21.600 --> 00:10:25.710
A basis is a spanning set that
is linearly independent.

00:10:25.710 --> 00:10:28.140
If they're linearly independent
we know that a2

00:10:28.140 --> 00:10:31.240
cannot be represented as some
linear combination of the rest

00:10:31.240 --> 00:10:32.310
of these guys.

00:10:32.310 --> 00:10:37.710
So we know that one of the
coefficients on the B terms

00:10:37.710 --> 00:10:39.780
has to be non-zero.

00:10:39.780 --> 00:10:46.170
And let's just say that you're
going to have a cj bj, this is

00:10:46.170 --> 00:10:47.930
a different one than
we had before.

00:10:47.930 --> 00:10:51.760
And we know that this guy, at
least one of them has to be

00:10:51.760 --> 00:10:54.760
non-zero, because if all of
these guys were non-zero, then

00:10:54.760 --> 00:10:56.910
you wouldn't be able to say
that this vector and that

00:10:56.910 --> 00:10:58.780
vector are linearly independent,
because they

00:10:58.780 --> 00:11:00.900
would be scalar multiples
of each other.

00:11:00.900 --> 00:11:03.050
So we're going to do
the same exercise.

00:11:03.050 --> 00:11:12.600
This guy right here, cj bj,
obviously this coefficient is

00:11:12.600 --> 00:11:14.920
non-zero so we can
solve for our bj.

00:11:14.920 --> 00:11:20.500
Once again we can say that bj is
equal minus 1 over cj times

00:11:20.500 --> 00:11:28.070
minus a2 plus c1 a1, all
the way to cm bm.

00:11:28.070 --> 00:11:31.860
So we have some bj here that can
be represented as a linear

00:11:31.860 --> 00:11:35.420
combination of the rest of the
people, including our new a2.

00:11:35.420 --> 00:11:38.250
And so, just like we did before,
let's remove him.

00:11:38.250 --> 00:11:39.640
Let's take him out of the set.

00:11:39.640 --> 00:11:41.290
And before I take him
out of the set, I'm

00:11:41.290 --> 00:11:42.230
going to rename him.

00:11:42.230 --> 00:11:46.750
Just solely for the purpose of
notational simplicity, I'm

00:11:46.750 --> 00:11:51.770
just going to rename our bj, b2,
and our b2 is equal to bj.

00:11:51.770 --> 00:11:53.600
So I'm just rearranging
the names.

00:11:53.600 --> 00:11:56.380
And I'm going to
remove our b2.

00:11:56.380 --> 00:11:58.700
Or I'm going to remove
what I now call b2.

00:11:58.700 --> 00:12:02.260
It was whatever was out here
that could be represented as a

00:12:02.260 --> 00:12:04.340
linear combination of
everything else,

00:12:04.340 --> 00:12:05.590
including our new a2.

00:12:08.290 --> 00:12:14.000
When I remove one of those
terms right here, and I

00:12:14.000 --> 00:12:15.250
renamed it B2.

00:12:18.210 --> 00:12:25.670
And now it's equal to a1
a2, and now I have the

00:12:25.670 --> 00:12:28.030
leftovers of my b's.

00:12:28.030 --> 00:12:33.290
So I have b3, b4, all
the way to bm.

00:12:33.290 --> 00:12:36.780
Notice I still have exactly
m elements, and

00:12:36.780 --> 00:12:40.410
this still spans V.

00:12:40.410 --> 00:12:43.870
It spans V because the element
that I took out of it can be

00:12:43.870 --> 00:12:46.650
represented as a linear
combination of these guys.

00:12:46.650 --> 00:12:48.780
So if I ever want to construct
anything that need that, I can

00:12:48.780 --> 00:12:50.750
construct that vector with some

00:12:50.750 --> 00:12:51.610
combination of these guys.

00:12:51.610 --> 00:12:52.780
So it wasn't necessary.

00:12:52.780 --> 00:12:54.460
It still spans V.

00:12:54.460 --> 00:12:57.300
So this process I'm doing, I
can just keep repeating it.

00:12:57.300 --> 00:13:00.700
I can add an a3.

00:13:00.700 --> 00:13:04.600
I can define B3 prime.

00:13:04.600 --> 00:13:09.960
I can just add a3 to this
set right here, a2 a3.

00:13:09.960 --> 00:13:17.350
And then I have my b3, b4,
all the way to bm.

00:13:17.350 --> 00:13:21.050
And I'll say this is linearly
dependent because

00:13:21.050 --> 00:13:22.830
this guy spans V.

00:13:22.830 --> 00:13:25.130
So everything but this
guy spans V.

00:13:25.130 --> 00:13:27.750
So obviously you can construct
this guy with a linear

00:13:27.750 --> 00:13:29.460
combination of the
rest of them.

00:13:29.460 --> 00:13:37.610
So you could say a3 is equal to
sum a1 plus sum a2 plus c3

00:13:37.610 --> 00:13:41.670
b3, all the way to cm bm.

00:13:41.670 --> 00:13:44.310
And we know that at least one
of the coefficients on the B

00:13:44.310 --> 00:13:48.190
terms has to be non-zero.

00:13:48.190 --> 00:13:51.040
Because if all of those were
0, then you would be saying

00:13:51.040 --> 00:13:54.060
that a3 could be a linear
combination of the A terms.

00:13:54.060 --> 00:13:57.640
And we know that a3 can't be
a representative linear

00:13:57.640 --> 00:14:00.570
combination of the A terms,
because all these a terms come

00:14:00.570 --> 00:14:02.900
from a linearly independent
set.

00:14:02.900 --> 00:14:06.430
So you do the same operation.

00:14:06.430 --> 00:14:12.430
Let's say that cj is non-zero.

00:14:12.430 --> 00:14:14.550
Then you could solve
for that bj.

00:14:14.550 --> 00:14:17.650
And then I do that little
renaming thing I do, where I

00:14:17.650 --> 00:14:21.220
rename the bj b3, and rename b3
bj, and then I remove b3.

00:14:24.100 --> 00:14:31.840
And I get the set B3 is
equal to a1, a2, a3.

00:14:31.840 --> 00:14:35.750
And I have b4 all
the way to bm.

00:14:35.750 --> 00:14:39.200
And this still spans V.

00:14:39.200 --> 00:14:40.640
And I keep doing it.

00:14:40.640 --> 00:14:42.500
So what's going to happen
eventually?

00:14:42.500 --> 00:14:46.210
What's going to happen if I keep
doing this process over

00:14:46.210 --> 00:14:48.600
and over and over again?

00:14:48.600 --> 00:14:53.460
Eventually I'll essentially
replace all of the bm's.

00:14:53.460 --> 00:14:56.730
Or I'll replace all of the n
terms, so eventually I'll have

00:14:56.730 --> 00:14:58.580
a set that looks like this.

00:14:58.580 --> 00:15:02.470
I'll have a set that looks Bm,
where I will have replaced

00:15:02.470 --> 00:15:04.150
each of these guys with an a.

00:15:07.060 --> 00:15:12.480
So I have a1, a2, a3,
all the way to am.

00:15:12.480 --> 00:15:15.330
You can always do this by
definition if you started with

00:15:15.330 --> 00:15:17.850
that initial set B that
is a spanning set.

00:15:17.850 --> 00:15:20.345
And once you do this process
you'll get the same result

00:15:20.345 --> 00:15:24.090
that this also spans V.

00:15:24.090 --> 00:15:25.440
Now let me let me write this.

00:15:25.440 --> 00:15:28.890
This is the result we got by
starting off with a spanning

00:15:28.890 --> 00:15:34.490
set B that has m elements,
where we said that m

00:15:34.490 --> 00:15:36.190
is less than n.

00:15:36.190 --> 00:15:38.530
So we always have enough a
elements to do this, because

00:15:38.530 --> 00:15:40.260
we have more a elements
than there were b

00:15:40.260 --> 00:15:41.450
elements to begin with.

00:15:41.450 --> 00:15:44.370
And we get result that
this spans V.

00:15:44.370 --> 00:15:51.840
But we already said that the
set A which is equal to a1,

00:15:51.840 --> 00:16:00.430
a2, all the way to am, and then
am plus 1, I don't know

00:16:00.430 --> 00:16:02.580
how many more terms there are
between m and n, but then you

00:16:02.580 --> 00:16:03.610
go all the way to an.

00:16:03.610 --> 00:16:07.000
Remember we said that
n is greater than m.

00:16:07.000 --> 00:16:10.610
Or when we defined B, we said
that m is less than n.

00:16:10.610 --> 00:16:12.990
Same thing, that this
was a smaller set.

00:16:12.990 --> 00:16:16.380
Now we're saying that this spans
V, but at the same time

00:16:16.380 --> 00:16:18.990
we said this was a basis.

00:16:18.990 --> 00:16:20.880
This is just our starting
fact, that this

00:16:20.880 --> 00:16:24.410
is a basis for V.

00:16:24.410 --> 00:16:28.900
Basis means two things: it means
it spans V and it means

00:16:28.900 --> 00:16:30.910
it's linearly independent.

00:16:30.910 --> 00:16:35.430
Now we just got this result by
assuming that we had some set

00:16:35.430 --> 00:16:38.100
B that's smaller than this
set here that spans V.

00:16:38.100 --> 00:16:41.420
We were able to construct this
by saying that a1 through am

00:16:41.420 --> 00:16:44.350
also spans V.

00:16:44.350 --> 00:16:48.250
The result we got is
that this spans V.

00:16:48.250 --> 00:16:53.440
But if this subset of A spans
V, then A becomes linearly

00:16:53.440 --> 00:16:53.900
independent.

00:16:53.900 --> 00:17:00.260
Because if this subset spans V,
that means that an can be

00:17:00.260 --> 00:17:05.240
represented as some linear
combination of these guys.

00:17:05.240 --> 00:17:08.560
So that implies that you're
linearly dependent, which is a

00:17:08.560 --> 00:17:12.300
contradiction with our original
statement that set A

00:17:12.300 --> 00:17:14.839
is a basis for V, because
that means it's linearly

00:17:14.839 --> 00:17:15.869
independent.

00:17:15.869 --> 00:17:18.810
If you're able to do this, then
this means that there was

00:17:18.810 --> 00:17:22.390
some smaller spanning set, you
get the result that A has to

00:17:22.390 --> 00:17:24.589
be linearly dependant, even
though we said it was linearly

00:17:24.589 --> 00:17:25.700
independent.

00:17:25.700 --> 00:17:28.860
So we now know, we get our
contradiction, we say that

00:17:28.860 --> 00:17:30.580
there cannot be [typing]

00:17:43.480 --> 00:17:59.640
a spanning set B that has
fewer elements than A.

00:17:59.640 --> 00:18:05.010
And this is a pretty neat
outcome, because now, if I

00:18:05.010 --> 00:18:12.160
come up to you and say I
found some set X spans

00:18:12.160 --> 00:18:16.960
the subspace V again.

00:18:16.960 --> 00:18:22.550
Then you know that X
has five elements.

00:18:22.550 --> 00:18:26.870
You now know that no set that
spans the subspace V can have

00:18:26.870 --> 00:18:28.560
fewer than five elements.

00:18:28.560 --> 00:18:36.230
Even better, if I told you that
X is a basis for V, and I

00:18:36.230 --> 00:18:42.370
told you it has five elements,
and Y is a basis for V.

00:18:42.370 --> 00:18:43.620
[typing]

00:18:58.130 --> 00:19:01.320
You know that Y also has to have
exactly five elements.

00:19:05.460 --> 00:19:07.310
How do I know that?

00:19:07.310 --> 00:19:13.450
Well, if Y is a basis, then that
means that it spans V.

00:19:13.450 --> 00:19:15.000
And we know it can't
have anything

00:19:15.000 --> 00:19:16.310
less than five elements.

00:19:16.310 --> 00:19:18.450
We just proved that.

00:19:18.450 --> 00:19:21.380
One way we know that Y has to
have greater than or equal to

00:19:21.380 --> 00:19:23.730
five elements.

00:19:23.730 --> 00:19:30.430
But on the other hand we know if
Y is a basis for V and X is

00:19:30.430 --> 00:19:32.590
a basis, X also spans V.

00:19:32.590 --> 00:19:35.360
So we know that X has to have
fewer elements than Y.

00:19:35.360 --> 00:19:37.395
So we know that [typing]

00:19:40.060 --> 00:19:49.010
Y's elements have to be greater
than X's elements,

00:19:49.010 --> 00:19:52.170
because any spanning set has to
have more elements, or at

00:19:52.170 --> 00:19:58.170
least as many elements, as a
basis set, X's elements.

00:19:58.170 --> 00:20:03.570
And then since X is a spanning
set, X's elements have to be

00:20:03.570 --> 00:20:05.310
greater than or equal
to Y's elements,

00:20:05.310 --> 00:20:06.560
because why is a basis.

00:20:11.650 --> 00:20:15.530
But if this guy's elements are
less than this guy's elements,

00:20:15.530 --> 00:20:19.250
but it's also greater than or
equal to, we know that the

00:20:19.250 --> 00:20:23.190
number of elements that X has--
X's elements, or the

00:20:23.190 --> 00:20:26.180
cardinality or the number of
elements in it --is equal to

00:20:26.180 --> 00:20:27.430
Y's elements.

00:20:30.200 --> 00:20:39.190
And so now that we know that any
basis for a vector space--

00:20:39.190 --> 00:20:40.610
Let me just go back
to our set A.

00:20:40.610 --> 00:20:45.080
A is equal to a1 a2,
all the way to an.

00:20:45.080 --> 00:20:51.040
We can now say that any basis
for some vector, for some

00:20:51.040 --> 00:20:54.700
subspace V, they all have the
same number of elements.

00:20:54.700 --> 00:20:57.770
And so we can define a
new term called the

00:20:57.770 --> 00:21:03.510
dimension of V.

00:21:03.510 --> 00:21:07.270
Sometimes it's written just as
dimension of V, is equal to

00:21:07.270 --> 00:21:10.530
the number of elements,
sometimes called the

00:21:10.530 --> 00:21:17.670
cardinality, of any
basis of V.

00:21:17.670 --> 00:21:19.880
And I went through great pains
in this video to show that any

00:21:19.880 --> 00:21:22.740
basis of V all has the same
number of elements, so this is

00:21:22.740 --> 00:21:23.870
well-defined.

00:21:23.870 --> 00:21:26.120
You can't have one basis that
has five elements and

00:21:26.120 --> 00:21:26.900
one that has six.

00:21:26.900 --> 00:21:28.920
By definition they would either
both have to have five,

00:21:28.920 --> 00:21:30.520
or they both would
have to have six.

00:21:30.520 --> 00:21:32.550
And so we can define
our dimensionality.


Visualizing a column space as a plane in R3

WEBVTT
Kind: captions
Language: en

00:00:00.590 --> 00:00:03.690
In the last video, I started
with this matrix right here,

00:00:03.690 --> 00:00:06.980
and right from the get go, we
said the span of this matrix

00:00:06.980 --> 00:00:11.240
is just the span of the column vectors
of it, and I just wrote it right there.

00:00:11.240 --> 00:00:13.740
But we weren't clear whether
this was linearly independent,

00:00:13.740 --> 00:00:17.050
and if it's not linearly independent,
it won't be a sufficient basis.

00:00:17.050 --> 00:00:20.360
And then we go off and
we figure out the null space of A.

00:00:20.360 --> 00:00:23.140
We find out that the null space
of A contains more than

00:00:23.140 --> 00:00:24.860
just the zero vector.

00:00:24.860 --> 00:00:27.870
It's just the span of these two
vectors here, which means

00:00:27.870 --> 00:00:31.150
that these columns are not
linearly independent.

00:00:31.150 --> 00:00:33.180
We saw that several
videos ago.

00:00:33.180 --> 00:00:35.140
And we used that information
that they're not linearly

00:00:35.140 --> 00:00:37.370
independent to try to make them
linearly independent by

00:00:37.370 --> 00:00:38.830
getting rid of the
redundant ones.

00:00:38.830 --> 00:00:41.610
So we were able to get rid of
this guy and this guy, because

00:00:41.610 --> 00:00:45.140
these were essentially the
columns associated with the

00:00:45.140 --> 00:00:46.990
free variables.

00:00:46.990 --> 00:00:49.500
And we were able to do it using
this little technique

00:00:49.500 --> 00:00:53.370
down here, where we set one of
these equal to 0, the other

00:00:53.370 --> 00:00:54.830
one equal to negative
1 and then solved

00:00:54.830 --> 00:00:56.140
for the pivot variables.

00:00:56.140 --> 00:00:57.890
And then we set the other one
equal to 0 and the other one

00:00:57.890 --> 00:01:00.700
equal to negative 1 and solved
for the pivot variables.

00:01:00.700 --> 00:01:02.470
And you could imagine
that this is a

00:01:02.470 --> 00:01:05.390
generalizable process.

00:01:05.390 --> 00:01:07.890
If you have a bunch of free
variables, you can set all of

00:01:07.890 --> 00:01:12.210
them but one to equal 0, and
then that one that you didn't

00:01:12.210 --> 00:01:14.090
set to 0, you set it to
equal negative 1.

00:01:14.090 --> 00:01:17.960
And you can express it as a sum
of the pivot variables,

00:01:17.960 --> 00:01:19.280
where the pivot variables are a

00:01:19.280 --> 00:01:22.660
function of the free variables.

00:01:22.660 --> 00:01:25.700
In general, this would be
a quick way of doing it.

00:01:25.700 --> 00:01:29.910
We had to do it more
slowly over here.

00:01:29.910 --> 00:01:34.590
If I have some matrix A, and I
want to figure out the basis

00:01:34.590 --> 00:01:38.220
for the column space, the column
space is just the span

00:01:38.220 --> 00:01:41.960
of these, but if I wanted a
linearly independent basis, I

00:01:41.960 --> 00:01:43.920
need to figure out some set
of these that are linearly

00:01:43.920 --> 00:01:44.880
independent.

00:01:44.880 --> 00:01:50.010
What I can do is put this guy
into reduced row echelon form.

00:01:50.010 --> 00:01:52.500
When I put them in reduced row
echelon form, and I did that

00:01:52.500 --> 00:02:00.700
over here, this is the reduced
row echelon form of A, I can

00:02:00.700 --> 00:02:02.280
look at the variables
that are associated

00:02:02.280 --> 00:02:04.380
with the pivot entries.

00:02:04.380 --> 00:02:07.360
So this is x1.

00:02:07.360 --> 00:02:09.320
Let me scroll up a little bit.

00:02:09.320 --> 00:02:10.479
This is associated
with x1, right?

00:02:10.479 --> 00:02:14.160
When you multiply this times x1,
you get this column times

00:02:14.160 --> 00:02:18.580
x1, this column times x2, this
column times x3, this column

00:02:18.580 --> 00:02:19.950
times x4 like that.

00:02:19.950 --> 00:02:22.570
When you look at just regular A,
when you look at just your

00:02:22.570 --> 00:02:23.990
matrix A, it's the same thing.

00:02:23.990 --> 00:02:27.100
If you were to write Ax equal
to 0, this column would be

00:02:27.100 --> 00:02:29.380
associated with x1, this column
would be associated

00:02:29.380 --> 00:02:33.350
with x2, x3 and x4 like that.

00:02:33.350 --> 00:02:36.290
What you can do is you put it
in reduced row echelon form.

00:02:36.290 --> 00:02:39.390
You say which columns have
my pivot entries or are

00:02:39.390 --> 00:02:40.765
associated with pivot
variables?

00:02:40.765 --> 00:02:43.830
You say, OK, x1 and x2 are
associated with pivot

00:02:43.830 --> 00:02:46.730
variables, or they are the pivot
variables, and they're

00:02:46.730 --> 00:02:48.930
associated with these first
two columns, and so those

00:02:48.930 --> 00:02:55.640
first two columns would be a
basis for the column space.

00:02:55.640 --> 00:02:57.810
How did I get this?

00:02:57.810 --> 00:02:59.660
Am I just making up
things on the fly?

00:02:59.660 --> 00:03:01.630
Well, no!

00:03:01.630 --> 00:03:06.650
It all comes from the reality
that you can always construct

00:03:06.650 --> 00:03:10.100
a situation where the vectors
associated with the free

00:03:10.100 --> 00:03:15.830
variables you can write them as
linear combinations of the

00:03:15.830 --> 00:03:18.640
vectors associated with the
pivot variables, and we did a

00:03:18.640 --> 00:03:20.630
special case of that
last time.

00:03:20.630 --> 00:03:23.540
But a very quick and dirty way
of doing it, and I don't know

00:03:23.540 --> 00:03:27.260
if it's actually dirty, is just
take your matrix, put it

00:03:27.260 --> 00:03:31.050
in it reduced row echelon form,
and you say, look, this

00:03:31.050 --> 00:03:33.920
column and this column
are associated

00:03:33.920 --> 00:03:36.310
with my free variables.

00:03:36.310 --> 00:03:40.200
Therefore, this column and this
column must be associated

00:03:40.200 --> 00:03:41.610
with my free variables.

00:03:41.610 --> 00:03:44.910
The solution sets are the same
to Ax equal to 0 or the

00:03:44.910 --> 00:03:49.330
reduced row echelon form
of Ax is equal to 0.

00:03:49.330 --> 00:03:50.040
So they're the same.

00:03:50.040 --> 00:03:51.810
So if this column and this
column are associated with

00:03:51.810 --> 00:03:54.970
free variables, so are this
column and this column, which

00:03:54.970 --> 00:03:58.590
means that they can be expressed
just by judiciously

00:03:58.590 --> 00:04:01.930
selecting your values of your
free variables as linear

00:04:01.930 --> 00:04:05.730
combinations of the columns
associated with the pivot

00:04:05.730 --> 00:04:09.470
variables, with the pivot
entries, which are that column

00:04:09.470 --> 00:04:10.140
and that column.

00:04:10.140 --> 00:04:14.070
So this column and this column
would be a basis for

00:04:14.070 --> 00:04:15.500
A, and we see that.

00:04:15.500 --> 00:04:17.959
We see that all the
way down here.

00:04:17.959 --> 00:04:21.860
1, 2, 3 and 1, 1, 4, We did a
lot of work and we got all the

00:04:21.860 --> 00:04:25.170
way there, and we said this
is a basis for the

00:04:25.170 --> 00:04:29.860
column span of A.

00:04:29.860 --> 00:04:32.180
Now, doing all of that work,
let's see if we can actually

00:04:32.180 --> 00:04:36.630
visualize what the column
space of A looks like.

00:04:36.630 --> 00:04:39.840
I have a strange feeling I might
have said column span a

00:04:39.840 --> 00:04:41.800
couple of times, but the
column space, what

00:04:41.800 --> 00:04:43.310
does it look like?

00:04:43.310 --> 00:04:45.800
Well, there's a couple of
ways to think about

00:04:45.800 --> 00:04:47.480
what it looks like.

00:04:47.480 --> 00:04:51.990
One way is we can say, look,
this span this is 2-- that's a

00:04:51.990 --> 00:04:53.720
member of R3.

00:04:53.720 --> 00:04:55.800
That's a vector in R3 and
this is a vector in R3.

00:04:59.700 --> 00:05:07.070
Let me draw my x, z and-- well,
normally it's drawn--

00:05:07.070 --> 00:05:12.920
this is normally y, x, and
z-axes in R3, if I'm want to

00:05:12.920 --> 00:05:15.270
represent them as
three-dimensional space.

00:05:15.270 --> 00:05:20.050
Then the vector 1, 2, 3 might
look like this where it's one,

00:05:20.050 --> 00:05:23.980
one, two, one, two, three,
so we go out one down

00:05:23.980 --> 00:05:24.930
here, then up three.

00:05:24.930 --> 00:05:27.900
So the vector will look like
that in its standard form.

00:05:27.900 --> 00:05:29.340
That's that one right there.

00:05:29.340 --> 00:05:33.670
And the vector 1, 1, 4 would be
one, one and go up four, so

00:05:33.670 --> 00:05:36.710
it might look something
like this.

00:05:36.710 --> 00:05:38.600
It's hard to actually draw
them very well in three

00:05:38.600 --> 00:05:40.230
dimensions, but you
get the idea.

00:05:40.230 --> 00:05:42.690
But the column space is the span
of these, so all of the

00:05:42.690 --> 00:05:44.630
linear combinations
of these two guys.

00:05:44.630 --> 00:05:46.860
So all of the linear
combinations of these two guys

00:05:46.860 --> 00:05:49.380
are going to create
a plane that

00:05:49.380 --> 00:05:51.140
contains these two vectors.

00:05:51.140 --> 00:05:53.670
If you just sum these guys up in
multiple combinations, you

00:05:53.670 --> 00:05:55.000
can get any vector up there.

00:05:55.000 --> 00:05:56.280
If you just add them
up, you'll get that

00:05:56.280 --> 00:05:57.100
vector right there.

00:05:57.100 --> 00:05:59.480
If you add this guy plus 2 times
that, you'll get some

00:05:59.480 --> 00:06:00.550
vector up here.

00:06:00.550 --> 00:06:02.890
So if you view them as position
vectors, they'll

00:06:02.890 --> 00:06:06.490
essentially form
a plane in R3.

00:06:06.490 --> 00:06:09.220
But let's see if we can get the
equation for that plane.

00:06:09.220 --> 00:06:10.700
Well, how can we do that?

00:06:10.700 --> 00:06:12.990
Well, we know that we can figure
out the equation of a

00:06:12.990 --> 00:06:18.080
plane based on the fact that
a normal vector dotted with

00:06:18.080 --> 00:06:23.490
any-- let me write my normal
vector like this.

00:06:23.490 --> 00:06:29.440
The normal vector dotted with
any position vector specifying

00:06:29.440 --> 00:06:30.870
a position on the plane.

00:06:30.870 --> 00:06:37.800
So let me call that x minus any
point on the plane or any

00:06:37.800 --> 00:06:39.020
vector position on the plane.

00:06:39.020 --> 00:06:45.720
So I could do that minus
the vector 1, 2, 3 has

00:06:45.720 --> 00:06:47.000
to be equal to 0.

00:06:47.000 --> 00:06:49.140
And we can use this information
to figure out the

00:06:49.140 --> 00:06:51.270
equation for this plane.

00:06:51.270 --> 00:06:52.670
But what is a normal vector?

00:06:52.670 --> 00:06:56.660
How can we find a normal
vector to this plane?

00:06:56.660 --> 00:06:59.930
So this would be a vector.

00:06:59.930 --> 00:07:01.590
Let me see if I can draw
this in a way that

00:07:01.590 --> 00:07:03.350
doesn't confuse the issue.

00:07:03.350 --> 00:07:05.050
If the plane is like that,
the normal vector would

00:07:05.050 --> 00:07:06.890
come out like that.

00:07:06.890 --> 00:07:09.450
So how can I create
a normal vector?

00:07:09.450 --> 00:07:12.690
Well, we learned that you take
the cross product of any two

00:07:12.690 --> 00:07:15.960
vectors in R3, and the cross
product I've only defined so

00:07:15.960 --> 00:07:20.020
far in R3, and I will get a
vector that's normal to both

00:07:20.020 --> 00:07:20.680
of those vectors.

00:07:20.680 --> 00:07:23.830
So let's take the
cross product.

00:07:23.830 --> 00:07:26.030
This is a nice way of thinking
about it, because it's really

00:07:26.030 --> 00:07:27.960
integrating everything that
we've covered so far.

00:07:27.960 --> 00:07:33.700
So let me define my normal
vector to be equal to 1, 2, 3

00:07:33.700 --> 00:07:36.795
cross 1, 1, 4.

00:07:36.795 --> 00:07:38.045
And what does this equal?

00:07:41.600 --> 00:07:43.450
So my first term,
I ignore that.

00:07:43.450 --> 00:07:45.850
I get 2 times 4 minus
3 times 1.

00:07:45.850 --> 00:07:47.940
2 times 4 is 8.

00:07:47.940 --> 00:07:50.100
2 times 4 minus 3 times 1.

00:07:50.100 --> 00:07:51.960
8 minus 3.

00:07:51.960 --> 00:07:56.920
Then for my second row, I have
1 times 4, and my temptation

00:07:56.920 --> 00:07:59.150
is do 1 times 4 minus
3 times 1.

00:07:59.150 --> 00:08:00.260
But you reverse it.

00:08:00.260 --> 00:08:04.585
You do 3 times 1, so it's
3, minus 1 times 4.

00:08:07.860 --> 00:08:09.090
We've done that multiple
times.

00:08:09.090 --> 00:08:12.030
You might want to review the
cross product video if that

00:08:12.030 --> 00:08:12.740
seems strange.

00:08:12.740 --> 00:08:16.250
You ignore the middle row, and
you normally do 1 times 4

00:08:16.250 --> 00:08:19.940
minus 3 times 1, but the
middle row you switch.

00:08:19.940 --> 00:08:23.850
We're only defining it for R3,
so instead, we do 3 times 1

00:08:23.850 --> 00:08:25.150
minus 1 times 4.

00:08:25.150 --> 00:08:27.210
And then finally for the last
row, we ignore it, and we say

00:08:27.210 --> 00:08:32.400
1 times 1, which is 1, minus 2
times 1, which is minus 2.

00:08:32.400 --> 00:08:38.970
And this is equal to the vector
5, minus 1, minus 1,

00:08:38.970 --> 00:08:41.059
which by definition of the cross
product, and I've shown

00:08:41.059 --> 00:08:44.950
this to you multiple times,
is normal to both of these

00:08:44.950 --> 00:08:47.280
vectors, and so it'll be
normal to any linear

00:08:47.280 --> 00:08:49.370
combination of these
two vectors.

00:08:49.370 --> 00:08:56.280
So now that we have our normal
vector, we can define the

00:08:56.280 --> 00:08:58.630
traditional equation
for the plane.

00:08:58.630 --> 00:09:03.470
So we now know that our normal
vector 5, minus 1, minus 1,

00:09:03.470 --> 00:09:08.880
that I got by taking the cross
product of our basis vectors

00:09:08.880 --> 00:09:12.450
dot any vector in our plane.

00:09:12.450 --> 00:09:13.665
So let me just write
any vector.

00:09:13.665 --> 00:09:16.080
Let me just write it x, y, z.

00:09:16.080 --> 00:09:21.840
So x, y, z since that's how
I defined my axes up here.

00:09:21.840 --> 00:09:22.920
This was the x-axis.

00:09:22.920 --> 00:09:24.530
x, y and z.

00:09:24.530 --> 00:09:28.040
x, y, z minus-- I just
picked one of these.

00:09:28.040 --> 00:09:32.100
I could have picked either of
them-- minus 1, 2, 3 has got

00:09:32.100 --> 00:09:33.840
to be equal to 0.

00:09:33.840 --> 00:09:35.510
So what's this?

00:09:35.510 --> 00:09:38.100
This is going to be equal to--
let me write it a little

00:09:38.100 --> 00:09:45.920
smaller, a little neater-- 5,
minus 1, minus 1 dot-- what's

00:09:45.920 --> 00:09:46.920
this guy going to be?

00:09:46.920 --> 00:09:53.780
x minus 1, y minus 2, and
z minus 3 has got

00:09:53.780 --> 00:09:54.840
to be equal to 0.

00:09:54.840 --> 00:09:56.050
And what's the dot product?

00:09:56.050 --> 00:10:03.890
It's 5 times x minus 1 plus-- or
maybe I should say minus 1,

00:10:03.890 --> 00:10:10.340
so it's plus minus 1 times y
minus 2 plus minus 1 times z

00:10:10.340 --> 00:10:12.410
minus 3 is equal to 0.

00:10:12.410 --> 00:10:14.860
That's just the definition
of our dot product.

00:10:14.860 --> 00:10:27.360
If I simplify this, I get 5x
minus 5 minus y plus 2 minus z

00:10:27.360 --> 00:10:29.660
plus 3 is equal to 0.

00:10:29.660 --> 00:10:32.850
You have 2 plus 3 is 5 minus
5, so those all cancel out.

00:10:32.850 --> 00:10:34.070
Those will equal 0.

00:10:34.070 --> 00:10:44.600
And we get 5x minus y minus
z is equal to 0.

00:10:44.600 --> 00:10:48.270
This plane in R3 is the
column space of A.

00:10:48.270 --> 00:10:52.250
So we've now shown you that
it's truly a plane in A.

00:10:52.250 --> 00:10:55.620
And actually, it makes
sense that this plane

00:10:55.620 --> 00:10:59.890
intersects the origin.

00:10:59.890 --> 00:11:01.740
If you set x, y, and
z equal to 0, it

00:11:01.740 --> 00:11:03.220
satisfies this equation.

00:11:03.220 --> 00:11:06.480
And that makes sense, because we
said that a column space of

00:11:06.480 --> 00:11:10.850
a matrix has to be a valid
subspace, and a valid subspace

00:11:10.850 --> 00:11:13.470
has to contain the
zero vector.

00:11:13.470 --> 00:11:18.320
And in R3 that's the
coordinate 0, 0, 0.

00:11:18.320 --> 00:11:20.810
Now, what I want to do now is
see if we can get at the same

00:11:20.810 --> 00:11:27.400
answer going a completely
different way, or approaching

00:11:27.400 --> 00:11:30.790
it in a completely
different way.

00:11:30.790 --> 00:11:35.220
Let me get my original A,
which I've forgotten.

00:11:35.220 --> 00:11:37.410
I've marked it up a good
bit, but let me just

00:11:37.410 --> 00:11:39.460
copy and paste it.

00:11:39.460 --> 00:11:43.020
So that's my original
A right there.

00:11:45.930 --> 00:11:48.562
Let me copy it.

00:11:48.562 --> 00:11:50.700
Let me paste it.

00:11:50.700 --> 00:11:51.640
Nope.

00:11:51.640 --> 00:11:53.900
That's not what I
wanted to do.

00:11:53.900 --> 00:11:57.860
Let's see, so my original
A-- I copied and

00:11:57.860 --> 00:11:59.110
pasted the wrong thing.

00:12:01.680 --> 00:12:05.520
Let me do it a little-- don't
want to waste your time.

00:12:05.520 --> 00:12:08.570
Edit, copy, edit, paste.

00:12:08.570 --> 00:12:11.090
There you go and let me scroll
this down and get to a point

00:12:11.090 --> 00:12:13.436
that's relatively clean.

00:12:13.436 --> 00:12:15.130
Bring my A down.

00:12:15.130 --> 00:12:18.430
I've used a lot of space.

00:12:18.430 --> 00:12:19.250
So here you go.

00:12:19.250 --> 00:12:23.030
This is my original
A right there.

00:12:23.030 --> 00:12:25.030
And what I want to do is see
if I can get this result

00:12:25.030 --> 00:12:25.700
completely different.

00:12:25.700 --> 00:12:28.310
I got this result by figuring
out the basis of the column

00:12:28.310 --> 00:12:32.875
span, finding a normal vector by
taking the cross product of

00:12:32.875 --> 00:12:37.800
our two basis vectors, and then
using the dot product of

00:12:37.800 --> 00:12:40.770
the normal vector with the
difference-- this vector right

00:12:40.770 --> 00:12:44.730
here, where you take any vector
on our plane minus one

00:12:44.730 --> 00:12:46.500
of our basis vectors,
that's to find some

00:12:46.500 --> 00:12:47.380
vector in the plane.

00:12:47.380 --> 00:12:48.780
This is some vector
in the plane.

00:12:48.780 --> 00:12:53.200
So any vector in the plane
dotted with my normal vector

00:12:53.200 --> 00:12:57.230
is going to be equal to 0.

00:12:57.230 --> 00:12:59.260
And actually, I should probably
make a side note

00:12:59.260 --> 00:13:02.490
here, that the only reason I
was able to say that the

00:13:02.490 --> 00:13:04.990
normal vector is a cross
product of my two basis

00:13:04.990 --> 00:13:07.850
vectors, is because I knew that
these two basis vectors,

00:13:07.850 --> 00:13:10.980
not only do they specify some
point on the plane-- so let's

00:13:10.980 --> 00:13:13.970
say that this guy right here
is this blue vector.

00:13:13.970 --> 00:13:17.030
Not only does he specify some
point on the plane right

00:13:17.030 --> 00:13:23.810
there, but the vector lies
completely on the plane.

00:13:23.810 --> 00:13:24.860
How did I know that?

00:13:24.860 --> 00:13:29.950
Because I knew from the get go
that the 0, 0 vector is in my

00:13:29.950 --> 00:13:31.980
span, right?

00:13:31.980 --> 00:13:36.000
I knew that if I draw this guy
in just standard position, the

00:13:36.000 --> 00:13:42.265
point 0, 0, 0, is in my span,
and I know its end point is in

00:13:42.265 --> 00:13:45.870
the span, so this whole vector
has to be in my plane, and

00:13:45.870 --> 00:13:48.640
likewise, this whole vector
has to be in my plane.

00:13:48.640 --> 00:13:51.260
So if I take the cross product,
anything normal to

00:13:51.260 --> 00:13:53.340
these guys or any combination
of these guys is going be

00:13:53.340 --> 00:13:56.130
normal to the plane, and we got
this result right here.

00:13:56.130 --> 00:13:58.820
But let me take this right
here and use our other

00:13:58.820 --> 00:14:00.250
definition of column span.

00:14:00.250 --> 00:14:02.060
Our other definition, or it's
really an equivalent

00:14:02.060 --> 00:14:13.110
definition, is all of the valid
solutions to Ax where x

00:14:13.110 --> 00:14:18.460
is a member of Rn.

00:14:18.460 --> 00:14:22.570
Or another way we could think of
it is, we could view it as

00:14:22.570 --> 00:14:36.810
all of the valid b's where Ax
is equal to b, and x is a

00:14:36.810 --> 00:14:37.840
member of Rn.

00:14:37.840 --> 00:14:39.340
These are equivalent
statements.

00:14:39.340 --> 00:14:42.940
I'm just defining b here to be
Ax, so these are equivalent

00:14:42.940 --> 00:14:43.330
statements.

00:14:43.330 --> 00:14:46.350
But let's run with this
a little bit.

00:14:46.350 --> 00:14:49.840
So let's say that I define my
b, so b is going to be a

00:14:49.840 --> 00:14:53.070
vector in R3, right?

00:14:53.070 --> 00:14:54.580
We already have an intuition
like that.

00:14:58.030 --> 00:15:02.360
When I take Ax, I get a b
that's equal to x, y, z.

00:15:02.360 --> 00:15:05.600
And I want to figure out what x,
y's and z's can I get valid

00:15:05.600 --> 00:15:06.880
solutions for?

00:15:06.880 --> 00:15:16.780
So if I take my vector A and
then let me multiply it

00:15:16.780 --> 00:15:28.290
times-- well, actually, the
best way to do it, I think

00:15:28.290 --> 00:15:29.340
we're used to it right now.

00:15:29.340 --> 00:15:37.110
If I'm solving the equation
Ax is equal to b, I can

00:15:37.110 --> 00:15:41.290
essentially just create the
augmented matrix, where I have

00:15:41.290 --> 00:15:46.690
the matrix A and I can augment
it with b, and put this in

00:15:46.690 --> 00:15:49.420
reduced row echelon form, and
that'll essentially represent

00:15:49.420 --> 00:15:50.500
the solution set.

00:15:50.500 --> 00:15:51.420
So let me do that.

00:15:51.420 --> 00:15:55.155
So if I just augment this matrix
right here with b, so I

00:15:55.155 --> 00:15:56.790
write x, y, z.

00:15:56.790 --> 00:15:59.030
So this is A augmented with b.

00:15:59.030 --> 00:16:00.390
This is A, this is b.

00:16:00.390 --> 00:16:04.190
let me put this in reduced row
echelon form and find the

00:16:04.190 --> 00:16:05.440
solution set.

00:16:09.600 --> 00:16:13.440
And these are x, y, and z's
that define a valid b.

00:16:13.440 --> 00:16:15.400
So what do I get?

00:16:15.400 --> 00:16:17.710
The first thing I want to do,
and we've done this exercise

00:16:17.710 --> 00:16:22.160
before, let's keep my
first row the same.

00:16:22.160 --> 00:16:26.720
1, 1, 1, 1, and I get an x.

00:16:26.720 --> 00:16:32.560
And let's replace our second row
with the second row minus

00:16:32.560 --> 00:16:33.810
the first row.

00:16:39.800 --> 00:16:41.180
Actually let me do
it this way.

00:16:41.180 --> 00:16:43.620
Let me replace the second row
with 2 times the first row

00:16:43.620 --> 00:16:44.715
minus the second row.

00:16:44.715 --> 00:16:46.770
So 2 times the first row minus
the second row, we're going to

00:16:46.770 --> 00:16:49.550
get a 2x minus y up there.

00:16:49.550 --> 00:16:53.100
And then 2 times
1 minus 2 is 0.

00:16:53.100 --> 00:16:56.330
2 times 1 minus 1 is 1.

00:16:56.330 --> 00:16:59.990
2 times 1 minus 4 is minus 2.

00:16:59.990 --> 00:17:04.305
2 times 1 minus 3 is minus 1.

00:17:07.170 --> 00:17:08.260
Fair enough.

00:17:08.260 --> 00:17:17.930
And now let me replace my third
row with the third row

00:17:17.930 --> 00:17:20.380
minus 3 times the first row.

00:17:20.380 --> 00:17:25.690
So we're going to do the third
row minus-- no, let

00:17:25.690 --> 00:17:27.240
me do it this way.

00:17:27.240 --> 00:17:34.800
It's the third row minus
3 times the first row.

00:17:34.800 --> 00:17:37.670
So I'm just doing the b column
first, because I can remember

00:17:37.670 --> 00:17:38.580
what I did.

00:17:38.580 --> 00:17:40.120
The third row minus 3
times the first row.

00:17:40.120 --> 00:17:42.900
3 minus 3 times 1 is 0.

00:17:42.900 --> 00:17:45.090
4 minus 3 times 1 is 1.

00:17:45.090 --> 00:17:47.710
1 minus 3 times 1 is minus 2.

00:17:47.710 --> 00:17:57.750
And then 2 minus 3 times
1 is minus 1.

00:17:57.750 --> 00:18:01.450
Now, I could go all the way to
reduced row echelon form, but

00:18:01.450 --> 00:18:03.100
something interesting is
already happening.

00:18:03.100 --> 00:18:08.380
So let me from the get go try
to zero out this third row.

00:18:08.380 --> 00:18:11.710
And the best way to zero out
this third row is to just

00:18:11.710 --> 00:18:14.140
replace the third row.

00:18:14.140 --> 00:18:15.850
So the first row-- well,
I won't even

00:18:15.850 --> 00:18:16.350
write the first row.

00:18:16.350 --> 00:18:23.400
The second row is 0, 1, minus
2, minus 1, and 2x minus y.

00:18:23.400 --> 00:18:25.690
I'm not even going to worry
about the first row right now.

00:18:25.690 --> 00:18:28.110
But let's replace the third row,
just in our attempt to go

00:18:28.110 --> 00:18:30.100
into reduced row echelon form.

00:18:30.100 --> 00:18:33.370
Let's replace it with
the second row

00:18:33.370 --> 00:18:34.880
minus the third row.

00:18:34.880 --> 00:18:42.670
So you get 2x minus
y minus z plus 3x.

00:18:42.670 --> 00:18:44.560
I just took this minus this.

00:18:44.560 --> 00:18:47.300
So minus z plus 3x.

00:18:47.300 --> 00:18:49.420
So 0 minus 0 is 0.

00:18:49.420 --> 00:18:51.140
1 minus 1 is 0.

00:18:51.140 --> 00:18:56.040
Minus 2 minus minus 2 is
0, and that's also 0.

00:18:56.040 --> 00:19:02.110
So we're only going to have a
valid solution to Ax equals b

00:19:02.110 --> 00:19:08.270
if this guy right here
is equal to 0.

00:19:08.270 --> 00:19:09.830
What happens if he's
not equal to zero?

00:19:09.830 --> 00:19:13.110
Then we're going to have a bunch
of zeroes equaling some

00:19:13.110 --> 00:19:15.810
number, which tells us that
there's no solution.

00:19:15.810 --> 00:19:19.680
So if I pick a b where this guy
does not equal zero, then

00:19:19.680 --> 00:19:20.800
I'll have no solution.

00:19:20.800 --> 00:19:24.340
If this guy equals 5, if I pick
x, y, and z's such as

00:19:24.340 --> 00:19:27.820
that this expression is equal to
5, then Ax equal to b will

00:19:27.820 --> 00:19:30.090
have no solution, because it
will have 0 is equal to 5.

00:19:30.090 --> 00:19:32.840
So this has to equal 0.

00:19:32.840 --> 00:19:40.510
So 2x minus y minus z plus 3x
must be equal to 0 in order

00:19:40.510 --> 00:19:47.490
for b to be valid, in order for
b to be a member of the

00:19:47.490 --> 00:19:51.700
column space of A, in order for
it to be a valid vector

00:19:51.700 --> 00:19:56.510
that Ax can become, or the
product A times x can become

00:19:56.510 --> 00:19:58.180
for some x.

00:19:58.180 --> 00:19:59.670
So what does this equal to?

00:19:59.670 --> 00:20:09.570
If we add the 2x plus the 3x,
I get 5x minus y minus z is

00:20:09.570 --> 00:20:14.050
equal to 0, which is the exact
same outcome we got when we

00:20:14.050 --> 00:20:15.710
figured out the basis vectors.

00:20:15.710 --> 00:20:16.550
We said oh, you know what?

00:20:16.550 --> 00:20:19.260
The basis vectors, they have
to be in the column space

00:20:19.260 --> 00:20:21.490
themselves by definition.

00:20:21.490 --> 00:20:25.150
So let me find a normal vector
to them both by taking the

00:20:25.150 --> 00:20:26.230
cross product.

00:20:26.230 --> 00:20:30.720
I did that, and I said the cross
product times any valid

00:20:30.720 --> 00:20:34.910
vector in our space minus one of
the basis vectors has to be

00:20:34.910 --> 00:20:37.320
equal to zero, and then
I got this equation.

00:20:37.320 --> 00:20:38.750
Or we could have done
it the other way.

00:20:38.750 --> 00:20:42.440
We could've actually literally
solved this equation setting

00:20:42.440 --> 00:20:43.480
our b equal to this.

00:20:43.480 --> 00:20:47.100
We said what b's will give
us a valid solution?

00:20:47.100 --> 00:20:50.620
And our only valid solution will
be obtained when this guy

00:20:50.620 --> 00:20:53.700
has to be equal to zero, because
the rest of his row became zero.

00:20:53.700 --> 00:20:56.920
And when we set that equal
to zero, we got the exact same equation.

00:20:56.920 --> 00:21:00.000
So, hopefully, you found this
to be mildly satisfying,

00:21:00.000 --> 00:21:02.360
because we were able to tackle
the same problem

00:21:02.360 --> 00:21:04.895
from two different directions
and get the same result,

00:21:04.895 --> 00:21:07.480
and it kind of shows you
the beauty of linear algebra,

00:21:07.480 --> 00:21:09.970
how it all starts to fit together.


Null space and column space basis

WEBVTT
Kind: captions
Language: en

00:00:00.450 --> 00:00:02.890
What I want to do in this video
-- and it'll probably

00:00:02.890 --> 00:00:05.980
occur over several videos -- is
really integrate everything

00:00:05.980 --> 00:00:08.990
we know about matrices, and
null spaces, and column

00:00:08.990 --> 00:00:11.960
spaces, and linear
independence.

00:00:11.960 --> 00:00:15.100
So I have this matrix
here, this matrix A.

00:00:15.100 --> 00:00:17.510
And I guess a good place to
start is, let's figure out its

00:00:17.510 --> 00:00:19.380
column space and
its null space.

00:00:19.380 --> 00:00:22.510
The column space is actually
super easy to figure out.

00:00:22.510 --> 00:00:25.070
It's just the span of the
column vectors of A.

00:00:25.070 --> 00:00:27.940
So we can right from the get-go
write that the column

00:00:27.940 --> 00:00:34.240
space of our matrix A-- Let
me do it over here.

00:00:34.240 --> 00:00:38.700
I can write the column space of
my matrix A is equal to the

00:00:38.700 --> 00:00:46.050
span of the vectors 1, 2, 3.

00:00:46.050 --> 00:00:48.490
1, 1, 4.

00:00:52.550 --> 00:00:54.410
1, 4, 1.

00:00:54.410 --> 00:00:56.130
And 1, 3, 2.

00:00:56.130 --> 00:00:56.730
I'm done.

00:00:56.730 --> 00:00:58.410
That was pretty straightforward,
a lot easier

00:00:58.410 --> 00:01:00.290
than finding null spaces.

00:01:00.290 --> 00:01:03.220
Now this may or may not
be satisfying to you.

00:01:03.220 --> 00:01:04.660
And there's a lot of
open questions.

00:01:04.660 --> 00:01:07.940
Is this a basis for the
space, for example?

00:01:07.940 --> 00:01:10.640
Is this a linear independent
set of vectors?

00:01:10.640 --> 00:01:12.580
How can we visualize
this space?

00:01:12.580 --> 00:01:14.260
And I haven't answered
any of those yet.

00:01:14.260 --> 00:01:16.750
But if someone just says, hey
what's the column space of A?

00:01:16.750 --> 00:01:18.460
This is the column space of A.

00:01:18.460 --> 00:01:20.090
And then we can answer some
of those other questions.

00:01:23.390 --> 00:01:27.330
If this is a linearly
independent set of vectors,

00:01:27.330 --> 00:01:31.340
then these vectors would
be a basis for the

00:01:31.340 --> 00:01:32.170
column space of A.

00:01:32.170 --> 00:01:33.040
We don't know that yet.

00:01:33.040 --> 00:01:35.220
We don't know whether these
are linearly independent.

00:01:35.220 --> 00:01:37.700
But we can figure out if they're
linearly independent

00:01:37.700 --> 00:01:40.070
by looking at the
null space of A.

00:01:40.070 --> 00:01:42.530
Remember these are linearly
independent if the null space

00:01:42.530 --> 00:01:44.780
of A only contains
the 0 vector.

00:01:44.780 --> 00:01:47.110
So let's figure out what
the null space of A is.

00:01:47.110 --> 00:01:49.190
And remember, we can do a
little shortcut here.

00:01:49.190 --> 00:01:54.580
The null space of A is equal to
the null space of the row,

00:01:54.580 --> 00:01:58.140
the reduced row echelon
form of A.

00:01:58.140 --> 00:02:01.050
And I showed you that when we
first calculated the null

00:02:01.050 --> 00:02:04.200
space of a vector, because when
you performed these --

00:02:04.200 --> 00:02:06.580
essentially if you want to solve
for the null space of A,

00:02:06.580 --> 00:02:07.980
you create an augmented
matrix.

00:02:07.980 --> 00:02:11.490
And you put the augmented matrix
in reduced row echelon

00:02:11.490 --> 00:02:13.520
form, but the 0's
never change.

00:02:13.520 --> 00:02:15.300
So essentially you're just
taking A and putting it in

00:02:15.300 --> 00:02:16.630
reduced row echelon form.

00:02:16.630 --> 00:02:18.340
Let's do that.

00:02:18.340 --> 00:02:25.250
So I'll keep row one the
same, 1, 1, 1, 1.

00:02:25.250 --> 00:02:28.570
And then let me replace
row two with, row

00:02:28.570 --> 00:02:31.580
two minus row one.

00:02:34.390 --> 00:02:36.510
So what do I get?

00:02:36.510 --> 00:02:39.120
No, actually I want to
zero this out here.

00:02:39.120 --> 00:02:42.720
So row two minus,
2 times row one.

00:02:42.720 --> 00:02:44.130
Actually even better because
I eventually

00:02:44.130 --> 00:02:44.950
want to get a 1 here.

00:02:44.950 --> 00:02:48.700
So let me do 2 times row
one, minus row two.

00:02:48.700 --> 00:02:52.150
So let me say 2 times row
one, and I'm going

00:02:52.150 --> 00:02:53.170
to minus row two.

00:02:53.170 --> 00:02:57.670
So 2 times 1 minus 2 is
0, which is exactly

00:02:57.670 --> 00:02:58.800
what I wanted there.

00:02:58.800 --> 00:03:02.950
2 times 1 minus 1 is 1.

00:03:02.950 --> 00:03:04.680
That's nice to have
right there.

00:03:04.680 --> 00:03:10.300
2 times 1 minus 4 is minus 2.

00:03:10.300 --> 00:03:15.530
2 times 1 minus 3 is minus 1.

00:03:15.530 --> 00:03:20.240
All right, now let me see if I
can zero out this guy here.

00:03:20.240 --> 00:03:21.490
So what can I do?

00:03:26.960 --> 00:03:29.510
I could do any combination,
anything that essentially

00:03:29.510 --> 00:03:30.710
zeroes this guy out.

00:03:30.710 --> 00:03:33.640
But I want to minimize my number
of negative numbers.

00:03:33.640 --> 00:03:38.590
So let me take this third row,
minus 3 times this first row.

00:03:38.590 --> 00:03:43.060
So I'm going take minus 3 times
that first row and add

00:03:43.060 --> 00:03:45.250
it to this third row.

00:03:45.250 --> 00:03:49.336
So 3 minus 3 times 1 is 0.

00:03:52.250 --> 00:03:53.370
These are just going to
be a bunch of 3's.

00:03:53.370 --> 00:03:55.920
4 minus 3 times 1 is 1.

00:03:55.920 --> 00:03:58.990
1 minus 3 times 1 is minus 2.

00:03:58.990 --> 00:04:02.510
And 2 minus 3 times
1 is minus 1.

00:04:02.510 --> 00:04:05.290
Now if we want to get this into
reduced row echelon form

00:04:05.290 --> 00:04:08.970
we need to target that one
there and that one there.

00:04:08.970 --> 00:04:09.870
And what can we do?

00:04:09.870 --> 00:04:13.270
So let's keep my middle
row the same.

00:04:13.270 --> 00:04:16.000
My middle row is not
going to change.

00:04:16.000 --> 00:04:21.610
1, 1, minus 2, minus 1.

00:04:21.610 --> 00:04:24.550
And to get rid of this one up
here I can just replace my

00:04:24.550 --> 00:04:27.950
first row with my first row
minus my second row.

00:04:27.950 --> 00:04:29.460
Because then this
won't change.

00:04:29.460 --> 00:04:31.370
I'll have 1 minus 0 is 1.

00:04:31.370 --> 00:04:33.280
1 minus 1 is 0.

00:04:33.280 --> 00:04:34.140
That's what we wanted.

00:04:34.140 --> 00:04:39.210
1 minus minus 2 is 3.

00:04:39.210 --> 00:04:40.120
That's 1 plus 2.

00:04:40.120 --> 00:04:42.060
1 minus minus 1.

00:04:42.060 --> 00:04:43.000
That's 1 plus 1.

00:04:43.000 --> 00:04:44.700
That is 2.

00:04:44.700 --> 00:04:45.690
Fair enough?

00:04:45.690 --> 00:04:47.805
Now let me do my third row.

00:04:51.950 --> 00:04:55.000
Let me replace my third row with
my third row subtracted

00:04:55.000 --> 00:04:56.190
from my first row.

00:04:56.190 --> 00:04:57.240
They are obviously
the same thing.

00:04:57.240 --> 00:05:01.220
So if I subtract the third row
from the second row I'm just

00:05:01.220 --> 00:05:02.260
going to get a bunch of 0's.

00:05:02.260 --> 00:05:04.610
0 minus 0 is 0.

00:05:04.610 --> 00:05:07.060
1 minus 1 is 0.

00:05:07.060 --> 00:05:11.500
Minus 2 minus minus 2 is 0.

00:05:11.500 --> 00:05:13.070
And minus 1 minus minus 1.

00:05:13.070 --> 00:05:14.430
That's minus 1 plus 1.

00:05:14.430 --> 00:05:16.020
That's equal to 0.

00:05:16.020 --> 00:05:17.900
And just like that
we have it now in

00:05:17.900 --> 00:05:19.480
reduced row echelon form.

00:05:19.480 --> 00:05:24.690
So this right here is the
reduced row echelon form of A.

00:05:24.690 --> 00:05:26.030
That straightforward.

00:05:26.030 --> 00:05:29.450
Now the whole the reason why
we even went through this

00:05:29.450 --> 00:05:31.460
exercise is we wanted
to figure out the

00:05:31.460 --> 00:05:32.460
null space of A.

00:05:32.460 --> 00:05:34.740
And we already know that the
null space of A is equal to

00:05:34.740 --> 00:05:38.400
the null space of the reduced
row echelon form of A.

00:05:38.400 --> 00:05:41.360
So if this is the reduce row
echelon form of A, let's

00:05:41.360 --> 00:05:42.930
figure out its null space.

00:05:42.930 --> 00:05:48.030
So the null space is the set
of all of vectors in R4,

00:05:48.030 --> 00:05:50.170
because we have 4
columns here.

00:05:50.170 --> 00:05:54.170
1, 2, 3, 4.

00:05:54.170 --> 00:05:57.380
The null space is the set of
all of vectors that satisfy

00:05:57.380 --> 00:05:59.780
this equation, where
we're going to have

00:05:59.780 --> 00:06:01.770
three 0's right here.

00:06:01.770 --> 00:06:04.570
That's the 0 vector in R3,
because we have three rows

00:06:04.570 --> 00:06:06.250
right there, and you
can figure it out.

00:06:06.250 --> 00:06:08.760
This times this has
to equal that 0.

00:06:08.760 --> 00:06:10.745
That dotted with that
essentially is going

00:06:10.745 --> 00:06:11.780
to equal that 0.

00:06:11.780 --> 00:06:13.600
That dotted with that
is equal to that 0.

00:06:13.600 --> 00:06:17.970
I say essentially because I
didn't define a row vector dot

00:06:17.970 --> 00:06:19.340
a column vector.

00:06:19.340 --> 00:06:21.410
I've only defined column vectors
dotted with other

00:06:21.410 --> 00:06:21.850
column vectors.

00:06:21.850 --> 00:06:24.320
But we've been over that in a
previous video, where you can

00:06:24.320 --> 00:06:27.720
say this is a transpose
of a column vector.

00:06:27.720 --> 00:06:29.800
So let's just take this,
and write a system of

00:06:29.800 --> 00:06:31.290
equations with this.

00:06:31.290 --> 00:06:34.480
So we get 1 times x1.

00:06:34.480 --> 00:06:37.840
So this times this is going
to be equal to that 0.

00:06:37.840 --> 00:06:41.740
So one times x1, that is x1.

00:06:41.740 --> 00:06:44.330
Plus 0 times x2.

00:06:44.330 --> 00:06:46.500
Let me just write that out.

00:06:46.500 --> 00:06:49.740
Plus 3 times x3.

00:06:49.740 --> 00:06:55.870
Plus 2 times x4 is
equal to that 0.

00:06:55.870 --> 00:06:59.730
And then -- I'll do it in
yellow right here -- I

00:06:59.730 --> 00:07:03.200
have 0 times x1.

00:07:03.200 --> 00:07:07.410
Plus 1 times x2.

00:07:07.410 --> 00:07:10.450
Minus 2 times x3.

00:07:10.450 --> 00:07:14.040
Minus x4 is equal to 0.

00:07:14.040 --> 00:07:15.440
And then this gives
me no information.

00:07:15.440 --> 00:07:17.400
0 times all this
is equal to 0.

00:07:17.400 --> 00:07:19.540
So it just turns into
0 equals 0.

00:07:19.540 --> 00:07:22.640
So let's see if we can solve for
our pivot entries, or our

00:07:22.640 --> 00:07:23.570
pivot variables.

00:07:23.570 --> 00:07:24.930
What are our pivot entries?

00:07:24.930 --> 00:07:26.160
This is a pivot entry.

00:07:26.160 --> 00:07:27.470
That's a pivot entry.

00:07:27.470 --> 00:07:29.730
That's what reduced row echelon
form is all about,

00:07:29.730 --> 00:07:31.740
getting these entries that are
1 and they're the only

00:07:31.740 --> 00:07:34.860
non-zero term in their
respective columns.

00:07:34.860 --> 00:07:37.060
And that every pivot entry
is to the right of a

00:07:37.060 --> 00:07:38.370
pivot entry above it.

00:07:38.370 --> 00:07:42.140
And then the columns that don't
have pivot entries?

00:07:42.140 --> 00:07:45.210
These columns represent
the free variables.

00:07:45.210 --> 00:07:48.260
So this column has
no pivot entry.

00:07:48.260 --> 00:07:51.610
And so when you take the dot
product, this column turned

00:07:51.610 --> 00:07:54.290
into this column in our
system of equations.

00:07:54.290 --> 00:07:57.190
So we know that x3 is
a free variable.

00:07:57.190 --> 00:07:58.890
x3 is free.

00:07:58.890 --> 00:08:00.730
We can set it equal
to anything.

00:08:00.730 --> 00:08:07.690
Likewise x4 is a
free variable.

00:08:07.690 --> 00:08:11.140
X1 and x2 are pivot variables,
because their corresponding

00:08:11.140 --> 00:08:14.360
columns in our reduced row
echelon form have pivot

00:08:14.360 --> 00:08:15.500
entries in them.

00:08:15.500 --> 00:08:16.140
Fair enough.

00:08:16.140 --> 00:08:17.580
So let's see if we
can simplify this

00:08:17.580 --> 00:08:18.920
into a form we know.

00:08:18.920 --> 00:08:21.440
And we've seen this before.

00:08:21.440 --> 00:08:25.080
So if I solve for x1 --
this 0 I can ignore.

00:08:25.080 --> 00:08:32.000
That 0 I can ignore -- I could
say that x1 is equal to minus

00:08:32.000 --> 00:08:36.650
3x3 minus 2x4.

00:08:36.650 --> 00:08:38.856
I just subtracted these two
from both sides of the

00:08:38.856 --> 00:08:49.720
equation and I can say that x2
is equal to 2x3 plus x4.

00:08:49.720 --> 00:08:52.750
And if we want to write our
solution set now, so if I

00:08:52.750 --> 00:09:00.540
wanted to find the null space of
A, which is the same thing

00:09:00.540 --> 00:09:05.400
as the null space of the reduced
row echelon form of A,

00:09:05.400 --> 00:09:09.730
is equal to all of the vectors
-- let me do a new color.

00:09:09.730 --> 00:09:15.050
Maybe I'll do blue -- is equal
to all of the vectors x1, x2,

00:09:15.050 --> 00:09:19.070
x3, x4 that are equal to --

00:09:19.070 --> 00:09:20.620
So what are they going
to be equal to?

00:09:20.620 --> 00:09:24.390
X1 has to be equal to
minus 3x3 minus 2x4.

00:09:24.390 --> 00:09:26.440
Just to be clear, these are free
variables because I can

00:09:26.440 --> 00:09:28.490
set these to be anything.

00:09:28.490 --> 00:09:30.560
And these are pivot variables
because I can't just set them

00:09:30.560 --> 00:09:31.450
to anything.

00:09:31.450 --> 00:09:34.390
When I determine what my x3's
and my x4's are, they

00:09:34.390 --> 00:09:37.060
determine what my x1's and
my x2's have to be.

00:09:37.060 --> 00:09:38.350
So these are pivoted
variables.

00:09:38.350 --> 00:09:39.470
These are free variables.

00:09:39.470 --> 00:09:40.750
I can make this guy pi.

00:09:40.750 --> 00:09:42.830
And I can make this
guy minus 2.

00:09:42.830 --> 00:09:44.680
We can set them to anything.

00:09:44.680 --> 00:09:47.805
So x1 is equal to -- let's see,
let me write it this way

00:09:47.805 --> 00:09:51.655
-- they're equal to x3 -- let me
do it in a different color

00:09:51.655 --> 00:09:53.850
-- do x3 like this.

00:09:53.850 --> 00:10:04.420
So it's equal to x3 times some
vector plus x4 times some

00:10:04.420 --> 00:10:06.080
other vector.

00:10:06.080 --> 00:10:08.910
So any solution set in my null
space is going to be a linear

00:10:08.910 --> 00:10:10.240
combination of these
two vectors.

00:10:10.240 --> 00:10:13.230
We can figure out what these
two vectors are just from

00:10:13.230 --> 00:10:16.830
these two constraints
right here.

00:10:16.830 --> 00:10:22.910
So -- let me do it in a neutral
color -- x1 is equal

00:10:22.910 --> 00:10:29.460
to minus 3 times x3
minus 2 times x4.

00:10:29.460 --> 00:10:30.980
Straightforward enough.

00:10:30.980 --> 00:10:37.500
x2 is equal to 2 times
x3 plus x4.

00:10:37.500 --> 00:10:39.410
What's x3 equal to?

00:10:39.410 --> 00:10:40.690
Well x3 is equal to itself.

00:10:40.690 --> 00:10:42.870
Whatever we set x3 equal to,
that's going to be x3.

00:10:42.870 --> 00:10:47.200
So x3 is going to be 1 times
x3 plus 0 times x4.

00:10:47.200 --> 00:10:48.580
It is not going to have
any x4 in it.

00:10:48.580 --> 00:10:51.460
X3 is going to be kind of
an independent variable.

00:10:51.460 --> 00:10:52.200
It's going to be free.

00:10:52.200 --> 00:10:53.390
We can set whatever it is.

00:10:53.390 --> 00:10:55.240
We set it and then that's
going to be x3 in

00:10:55.240 --> 00:10:56.900
our solution set.

00:10:56.900 --> 00:10:58.990
x4 is not going to have
any x3 in it.

00:10:58.990 --> 00:11:01.140
It's just going to
be 1 times x4.

00:11:01.140 --> 00:11:03.980
And so our null space is
essentially all of the linear

00:11:03.980 --> 00:11:06.410
combinations of these
two vectors.

00:11:06.410 --> 00:11:07.740
This can be any real number.

00:11:07.740 --> 00:11:12.420
This is just any real number and
x4 is just a member of the

00:11:12.420 --> 00:11:13.140
real space.

00:11:13.140 --> 00:11:18.700
So all of these, the set of all
of the valid solutions to

00:11:18.700 --> 00:11:21.220
Ax is equal to 0 -- where
did I write that.

00:11:21.220 --> 00:11:22.300
Did I even write that down?

00:11:22.300 --> 00:11:23.930
No I haven't even written
that anywhere.

00:11:23.930 --> 00:11:29.130
The set of all Ax is equal to
0, where this is my x, it

00:11:29.130 --> 00:11:33.250
equals all the linear
combinations of this vector

00:11:33.250 --> 00:11:34.590
and that vector right there.

00:11:34.590 --> 00:11:37.540
And we know what all of the
linear combinations mean.

00:11:37.540 --> 00:11:45.460
It means my null space is equal
to the span of these two

00:11:45.460 --> 00:11:51.230
guys, the span of minus
3, 2, 1, 0.

00:11:51.230 --> 00:12:00.500
And minus 2, 1, 0, 1.

00:12:00.500 --> 00:12:02.675
Now let me ask you a question.

00:12:08.030 --> 00:12:13.070
Are the columns in A, are they
a linearly independent set?

00:12:13.070 --> 00:12:16.980
Are they a linearly
independent set?

00:12:16.980 --> 00:12:22.100
So if we write these vectors
right there, these are the

00:12:22.100 --> 00:12:24.170
column vectors of A.

00:12:24.170 --> 00:12:25.220
So let me write that down.

00:12:25.220 --> 00:12:29.400
So are the column vectors of
A -- so what were they?

00:12:29.400 --> 00:12:29.790
Let's see.

00:12:29.790 --> 00:12:34.750
1, 3, 2.

00:12:34.750 --> 00:12:36.380
No it's 1, 2, 3.

00:12:40.480 --> 00:12:44.740
1, 1, 4.

00:12:44.740 --> 00:12:47.350
1, 4, 1.

00:12:47.350 --> 00:12:50.260
And 1, 3, 2.

00:12:50.260 --> 00:12:51.770
So this is just the column
vectors of A.

00:12:51.770 --> 00:12:54.600
I could just write A is just
this much of columns, but my

00:12:54.600 --> 00:12:57.940
question is, is this a linearly
independent set?

00:13:04.620 --> 00:13:08.690
And here you might immediately
start thinking, well when we

00:13:08.690 --> 00:13:11.010
said that something is linearly
independent -- so

00:13:11.010 --> 00:13:19.485
linearly independence implies
that there's only one solution

00:13:19.485 --> 00:13:22.010
-- we saw this I think two
videos ago, that there's only

00:13:22.010 --> 00:13:31.100
one solution -- one solution
to Ax is equal to 0.

00:13:31.100 --> 00:13:33.710
And that is the 0 solution,
that x is

00:13:33.710 --> 00:13:35.550
equal to the 0 vector.

00:13:35.550 --> 00:13:40.020
Or another way to say that is
that the null space of my

00:13:40.020 --> 00:13:44.840
matrix A is equal to
just the 0 vector.

00:13:44.840 --> 00:13:46.220
That's what linear independence
implies.

00:13:46.220 --> 00:13:47.430
And it goes both ways.

00:13:47.430 --> 00:13:50.160
If my null space is just a 0
vector, then I know it's

00:13:50.160 --> 00:13:51.430
linearly independent.

00:13:51.430 --> 00:13:54.800
If my null space includes other
vectors, then I am not

00:13:54.800 --> 00:13:56.180
linearly independent.

00:13:56.180 --> 00:13:58.400
Now my null space of A,
what does it include?

00:13:58.400 --> 00:14:00.080
Is it just the 0 vector?

00:14:00.080 --> 00:14:01.810
Well, no it includes
every linear

00:14:01.810 --> 00:14:03.200
combination of these guys.

00:14:03.200 --> 00:14:05.590
It includes actually an infinite
number of vectors,

00:14:05.590 --> 00:14:07.110
not just one solution.

00:14:07.110 --> 00:14:09.180
Obviously 0 vector is contained
here, if you just

00:14:09.180 --> 00:14:11.950
multiply both of these -- if you
pick 0 for that and that.

00:14:11.950 --> 00:14:14.820
It's contained, but you can get
a whole set of vectors.

00:14:14.820 --> 00:14:25.850
So because the null span of A,
the null space, sorry, the

00:14:25.850 --> 00:14:29.720
null space of A does not just
contain the 0 vector.

00:14:29.720 --> 00:14:33.730
So it has more than just 0.

00:14:37.510 --> 00:14:38.950
So what does that mean?

00:14:38.950 --> 00:14:40.170
Well that means that
there's more than

00:14:40.170 --> 00:14:41.300
one solution to this.

00:14:41.300 --> 00:14:44.220
And that means that this is
a linearly dependent set.

00:14:51.620 --> 00:14:52.950
And what does that mean?

00:14:52.950 --> 00:14:54.940
At the very beginning of the
video I said, what's the

00:14:54.940 --> 00:14:56.250
column space of A.

00:14:56.250 --> 00:14:59.400
And we said, the column space
of A is just the span of the

00:14:59.400 --> 00:15:00.490
column vectors.

00:15:00.490 --> 00:15:01.950
I just wrote it out like that.

00:15:01.950 --> 00:15:04.410
And I said, well it's not clear
whether this is a valid

00:15:04.410 --> 00:15:06.680
basis for the column
space of A.

00:15:06.680 --> 00:15:07.630
And what's a basis?

00:15:07.630 --> 00:15:13.590
A basis is a set of vectors that
span a subspace, and they

00:15:13.590 --> 00:15:14.970
are also linearly independent.

00:15:14.970 --> 00:15:19.230
And we just showed that these
guys are not linearly

00:15:19.230 --> 00:15:21.910
independent.

00:15:21.910 --> 00:15:28.070
So that means that they are
not a basis for the column

00:15:28.070 --> 00:15:30.360
space of A.

00:15:30.360 --> 00:15:33.720
They do span the column space
of A, by definition really.

00:15:33.720 --> 00:15:34.780
But they're not a basis.

00:15:34.780 --> 00:15:36.620
They need to be linearly
independent for

00:15:36.620 --> 00:15:38.020
them to be a basis.

00:15:38.020 --> 00:15:41.120
So let's see if we can figure
out what a basis for this

00:15:41.120 --> 00:15:42.040
column space would be.

00:15:42.040 --> 00:15:44.000
And to do that we just
have to get rid of

00:15:44.000 --> 00:15:45.380
some redundant vectors.

00:15:45.380 --> 00:15:49.340
If I can show you that this guy
can be represented by some

00:15:49.340 --> 00:15:50.810
combination of these two
guys, then I can

00:15:50.810 --> 00:15:51.650
get rid of that guy.

00:15:51.650 --> 00:15:53.480
He's not adding any
new information.

00:15:53.480 --> 00:15:54.370
Same with that guy.

00:15:54.370 --> 00:15:55.400
Who knows?

00:15:55.400 --> 00:15:59.890
So let's see if we can figure
this piece of the puzzle out.

00:15:59.890 --> 00:16:06.770
So we know already that x1, let
me write it this way, that

00:16:06.770 --> 00:16:11.010
x1 times -- Maybe I'll just kind
of leave you hanging and

00:16:11.010 --> 00:16:12.270
continue this in
the next video.

00:16:12.270 --> 00:16:18.780
But we know that x1
times 1, 2, 3.

00:16:18.780 --> 00:16:23.470
Plus x2 times 1, 1, 4.

00:16:23.470 --> 00:16:28.040
Plus x3 times 1, 4, 1.

00:16:28.040 --> 00:16:32.250
Plus x4 times 1, 3, 2.

00:16:32.250 --> 00:16:35.600
We know that this
is equal to 0.

00:16:35.600 --> 00:16:41.260
Now if we are able to solve for
x4 in terms of -- let me

00:16:41.260 --> 00:16:47.540
just think that I can solve
for the vectors that are

00:16:47.540 --> 00:16:49.300
associated with my
free variables

00:16:49.300 --> 00:16:50.600
using the other vectors.

00:16:50.600 --> 00:16:52.080
Let me see if I can do that.

00:16:52.080 --> 00:16:55.060
And you'll see it's actually
pretty straightforward.

00:16:55.060 --> 00:16:58.820
So let's say I want
to solve for x4.

00:16:58.820 --> 00:17:01.450
So if I subtract this from
both sides of this

00:17:01.450 --> 00:17:04.819
equation, I get what?

00:17:04.819 --> 00:17:07.690
Let me put it this way, let
me set x3 equal to 0.

00:17:07.690 --> 00:17:08.640
It was a free variable.

00:17:08.640 --> 00:17:09.290
I can do that.

00:17:09.290 --> 00:17:16.290
So if I set x3 is equal to 0,
then what do I get here?

00:17:16.290 --> 00:17:18.710
If I said x3 equals 0,
this guy disappears.

00:17:18.710 --> 00:17:21.780
And if I subtract this from both
sides of this equation, I

00:17:21.780 --> 00:17:27.230
get x1 times 1, 2, 3.

00:17:27.230 --> 00:17:31.940
Plus x2 times 1, 1, 4.

00:17:31.940 --> 00:17:34.430
Is equal to -- I'm just
setting x3 equal to 0.

00:17:34.430 --> 00:17:35.880
That was a free variable.

00:17:35.880 --> 00:17:36.930
So I'm setting x3 equal to 0.

00:17:36.930 --> 00:17:38.940
So this whole thing
disappears.

00:17:38.940 --> 00:17:47.490
So that is equal to minus
x4 times 1, 3, 2.

00:17:47.490 --> 00:17:48.900
Now I set x3 equal to 0.

00:17:48.900 --> 00:17:55.050
Let me set x4 to be
equal to minus 1.

00:17:55.050 --> 00:17:58.440
If x4 is equal to minus
1, what is minus x4?

00:17:58.440 --> 00:18:01.300
Well then this thing will
just be equal to 1.

00:18:01.300 --> 00:18:03.500
And I'll have x1
times 1, 2, 3.

00:18:03.500 --> 00:18:07.560
Plus x2 times 1, 1, 4 will
equal this fourth

00:18:07.560 --> 00:18:08.980
vector right here.

00:18:08.980 --> 00:18:11.350
And can I always find
things like this?

00:18:11.350 --> 00:18:13.250
Well sure I can actually find
the particular ones.

00:18:13.250 --> 00:18:17.340
If x3 is equal to 0, and x4 is
minus 1 -- Let me copy and

00:18:17.340 --> 00:18:35.790
paste this that I have up
here -- Let me scroll

00:18:35.790 --> 00:18:37.000
down a little bit.

00:18:37.000 --> 00:18:39.890
This is what we got when we
figured out our null space,

00:18:39.890 --> 00:18:41.440
right there.

00:18:41.440 --> 00:18:44.582
So if I'm setting -- remember
these are the free variables

00:18:44.582 --> 00:18:47.590
-- if I set x3 equal to
0 and x4 is equal to

00:18:47.590 --> 00:18:50.070
minus 1, what is x1?

00:18:50.070 --> 00:18:54.700
Then this will imply that x1 is
equal to minus 3 times x3,

00:18:54.700 --> 00:18:58.350
that's just 0, minus
2 times x4.

00:18:58.350 --> 00:19:01.450
If x4 is minus 1, minus
2 times minus 1,

00:19:01.450 --> 00:19:02.820
x1 will equal 2.

00:19:02.820 --> 00:19:05.350
And then what will
x2 be equal to?

00:19:05.350 --> 00:19:10.370
x2 is equal to 2 times x3,
which is 0, plus x4.

00:19:10.370 --> 00:19:12.010
So it's equal to minus 1.

00:19:12.010 --> 00:19:15.450
So I just showed you that if I
set this equal to 2 and this

00:19:15.450 --> 00:19:19.000
equal to minus 1, I have a
linear combination of this

00:19:19.000 --> 00:19:21.340
vector and this vector
that can add up

00:19:21.340 --> 00:19:22.190
to this fourth vector.

00:19:22.190 --> 00:19:23.140
And you can even verify it.

00:19:23.140 --> 00:19:25.430
2 times 1 minus 1
is equal to 1.

00:19:25.430 --> 00:19:29.290
2 times 2 minus 1
is equal to 3.

00:19:29.290 --> 00:19:32.340
2 times 3 is 6, minus
4 is equal to 2.

00:19:32.340 --> 00:19:33.520
So it checks out.

00:19:33.520 --> 00:19:37.820
So I just showed you using,
really, our definitions

00:19:37.820 --> 00:19:40.730
looking at what were
our free variables

00:19:40.730 --> 00:19:42.110
versus our pivot variables.

00:19:42.110 --> 00:19:46.210
We were able to show you, kind
of just very simply solve for

00:19:46.210 --> 00:19:48.310
this third, this fourth
vector, in terms

00:19:48.310 --> 00:19:49.540
of these first two.

00:19:49.540 --> 00:19:52.860
So we know, if we go back to
the set that this fourth

00:19:52.860 --> 00:19:58.430
vector is really unnecessary,
really not adding anything to

00:19:58.430 --> 00:20:00.330
the span of the set
of vectors.

00:20:00.330 --> 00:20:02.910
Because this guy can be written
as a combination of

00:20:02.910 --> 00:20:04.730
this guy and this guy.

00:20:04.730 --> 00:20:07.320
Now let's see if this guy, this
third guy, we can do the

00:20:07.320 --> 00:20:08.340
same exercise.

00:20:08.340 --> 00:20:10.350
This is also dictated
by a free variable.

00:20:10.350 --> 00:20:11.990
So let's see if I can write
him as a combination

00:20:11.990 --> 00:20:13.430
of these first two.

00:20:13.430 --> 00:20:15.450
Well we'll do the exact
same thing.

00:20:15.450 --> 00:20:22.030
Instead of setting x3 equal to 0
and x4 equal to minus 1, let

00:20:22.030 --> 00:20:27.445
us set x4 is equal to
0 because I want

00:20:27.445 --> 00:20:28.760
to cross that out.

00:20:28.760 --> 00:20:33.000
And let me set x3 is
equal to minus 1.

00:20:33.000 --> 00:20:38.360
If x3 is equal to minus
1, what does this

00:20:38.360 --> 00:20:39.900
equation reduce to?

00:20:39.900 --> 00:20:44.480
We get x1 times 1, 2, 3.

00:20:44.480 --> 00:20:48.760
Plus x2 times 1, 1, 4.

00:20:48.760 --> 00:20:51.640
Is equal to -- if this is
minus 1 times 1, 4, 1.

00:20:51.640 --> 00:20:54.620
And then we add it to both sides
of this equation, we get

00:20:54.620 --> 00:20:58.960
plus 1 times 1, 4, 1.

00:20:58.960 --> 00:21:01.710
And once again we can just
solve for our x1 and x2.

00:21:01.710 --> 00:21:06.940
If x4 is 0 and x3 is minus
1, then x1 x4 is 0.

00:21:06.940 --> 00:21:10.220
So x3 is just minus 3,
times x3, so x1 would

00:21:10.220 --> 00:21:11.430
be equal to 3, right?

00:21:11.430 --> 00:21:13.950
Minus 3 times minus 1.

00:21:13.950 --> 00:21:16.120
And what would x2 be equal to?

00:21:16.120 --> 00:21:17.730
x4 is 0, we can ignore that.

00:21:17.730 --> 00:21:19.700
x2 would be equal to minus 2.

00:21:19.700 --> 00:21:23.250
So this would be 3, and then
this would be minus 2.

00:21:23.250 --> 00:21:23.830
Let's see if it works out.

00:21:23.830 --> 00:21:26.500
3 times 1 minus 2 is 1.

00:21:26.500 --> 00:21:28.620
3 times 2 minus 2 is 4.

00:21:28.620 --> 00:21:30.620
3 times 3 minus 8 is 1.

00:21:30.620 --> 00:21:31.800
It checks out.

00:21:31.800 --> 00:21:35.120
So I'm able to write this
vector, that was associated

00:21:35.120 --> 00:21:36.980
with the free variable,
as a linear

00:21:36.980 --> 00:21:39.180
combination of these two.

00:21:39.180 --> 00:21:41.990
So we can get rid of
him from our set.

00:21:41.990 --> 00:21:44.290
So now I've shown that this guy
can be written as a linear

00:21:44.290 --> 00:21:45.240
combination of these two.

00:21:45.240 --> 00:21:46.880
This guy can be written
as a linear

00:21:46.880 --> 00:21:47.690
combination of these two.

00:21:47.690 --> 00:21:51.710
So the span of all of those guys
should be equal to the

00:21:51.710 --> 00:21:54.120
span -- So let me write
it this way.

00:21:56.690 --> 00:22:00.630
The column space of A,
I can now re-write.

00:22:00.630 --> 00:22:02.520
Before it was the span of
all of those vectors.

00:22:02.520 --> 00:22:05.150
It was the span of all of
the column vectors,

00:22:05.150 --> 00:22:10.850
v1, v2, v3, and v4.

00:22:10.850 --> 00:22:13.470
Now I just showed you that v3
and v4 can be rewritten in

00:22:13.470 --> 00:22:15.240
terms of v1 and v2.

00:22:15.240 --> 00:22:16.160
So they're redundant.

00:22:16.160 --> 00:22:20.270
So that is equal to the span
of v1 and v2 which are just

00:22:20.270 --> 00:22:23.140
those two vectors.

00:22:23.140 --> 00:22:29.450
Vector 1, 2, 3, and
vector 1, 1, 4.

00:22:29.450 --> 00:22:32.800
Now are any of these
guys redundant?

00:22:32.800 --> 00:22:35.150
Can I express one of
them as a linear

00:22:35.150 --> 00:22:36.690
combination of the other?

00:22:36.690 --> 00:22:38.250
Essentially when I'm talking
about the linear combination

00:22:38.250 --> 00:22:39.550
of only one other
vector it's just

00:22:39.550 --> 00:22:40.930
multiplying it by a scalar.

00:22:40.930 --> 00:22:43.330
Well let's think about that.

00:22:43.330 --> 00:22:45.510
There are multiple ways you can
show this, but the easiest

00:22:45.510 --> 00:22:48.730
way is well look, to go from
this entry to that entry I'm

00:22:48.730 --> 00:22:50.320
just multiplying by 1.

00:22:50.320 --> 00:22:53.525
But if I multiply this whole
vector times 1, then I'm going

00:22:53.525 --> 00:22:55.980
to get a 2 here and I'm
going to get a 3 here.

00:22:55.980 --> 00:22:57.230
So it won't work.

00:22:59.840 --> 00:23:01.940
If I want to represent this guy
as a scalar multiple of

00:23:01.940 --> 00:23:05.960
that guy, so any scalar multiple
of 1, 2, 3 is going

00:23:05.960 --> 00:23:14.810
to be equal to 1c, 2c, 3c.

00:23:14.810 --> 00:23:15.460
Right?

00:23:15.460 --> 00:23:18.030
And so we're saying this guy has
to be represented somehow

00:23:18.030 --> 00:23:21.110
like that, if we say that this
guy is somehow a scalar,

00:23:21.110 --> 00:23:22.580
somehow can be represented
by that guy.

00:23:22.580 --> 00:23:26.600
So that would have to
be equal to 1, 1, 4.

00:23:26.600 --> 00:23:29.250
When you look at this top entry
it implies that c would

00:23:29.250 --> 00:23:31.940
have to be equal to 1.

00:23:31.940 --> 00:23:34.150
But when you look at this second
entry you think that c

00:23:34.150 --> 00:23:37.020
would have to be equal to 1/2.

00:23:37.020 --> 00:23:38.030
So you get a contradiction.

00:23:38.030 --> 00:23:40.390
Over here c would have
to be equal to 4/3.

00:23:40.390 --> 00:23:42.530
So there's no c where
this will work.

00:23:42.530 --> 00:23:43.740
There's no multiple of c.

00:23:43.740 --> 00:23:45.650
And you can work
that both ways.

00:23:45.650 --> 00:23:48.240
So there's no way that you can
represent one of these guys as

00:23:48.240 --> 00:23:49.560
a linear combination
of the other.

00:23:49.560 --> 00:23:52.040
And you can actually prove
other ways, maybe more

00:23:52.040 --> 00:23:54.120
formally, that this is
linearly independent.

00:23:54.120 --> 00:23:55.810
But given that this is linearly
independent -- I

00:23:55.810 --> 00:24:01.710
think you're satisfied with that
-- we can then say that

00:24:01.710 --> 00:24:19.820
the set of vectors 1, 2, 3, and
1, 1, 4, this is a basis

00:24:19.820 --> 00:24:23.720
for the column span of A.

00:24:23.720 --> 00:24:26.000
Now I'm going to let you go in
this video because I think

00:24:26.000 --> 00:24:27.200
I've gone well over time.

00:24:27.200 --> 00:24:28.870
But what I'm going to do in
the next few videos is now

00:24:28.870 --> 00:24:32.180
that I've established that this
is a basis for the column

00:24:32.180 --> 00:24:34.780
span of A, we can attempt
to visualize it.

00:24:34.780 --> 00:24:39.140
Because we can say that the
column span of A is equal to

00:24:39.140 --> 00:24:42.100
the span of these two vectors.

00:24:42.100 --> 00:24:43.840
And we can think about
what the span of

00:24:43.840 --> 00:24:44.850
those two vectors are.

00:24:44.850 --> 00:24:47.280
We're going to see that
it's a plane in R3.

00:24:47.280 --> 00:24:49.180
Span of 1, 1, 4.

00:24:49.180 --> 00:24:50.910
And this is a quick reminder,
I've said a couple times.

00:24:50.910 --> 00:24:53.760
When I say it's a basis all I'm
saying is that these guys,

00:24:53.760 --> 00:24:56.510
they both span the column
space of A.

00:24:56.510 --> 00:24:58.275
When I had four vectors,
they also spanned the

00:24:58.275 --> 00:24:59.190
column space of A.

00:24:59.190 --> 00:25:02.460
But what makes them a basis is
that these guys are linearly

00:25:02.460 --> 00:25:02.980
independent.

00:25:02.980 --> 00:25:05.910
There's no extra information, or
redundant vectors that can

00:25:05.910 --> 00:25:09.260
be represented by other vectors
within the basis.

00:25:09.260 --> 00:25:10.360
They are linearly independent.

00:25:10.360 --> 00:25:12.510
Anyway, I'll let
you go for now.