Unique rowspace solution to Ax = b

WEBVTT
Kind: captions
Language: en

00:00:00.650 --> 00:00:03.830
Let's say I've got an
m-by-n matrix A.

00:00:03.830 --> 00:00:05.500
That's my matrix right there.

00:00:05.500 --> 00:00:09.340
And I could just write it as a
series of n column vectors, so

00:00:09.340 --> 00:00:15.520
it could be a1, a2,
all the way to an.

00:00:15.520 --> 00:00:19.590
Now, let's say that I have
some other vector b.

00:00:19.590 --> 00:00:25.540
Let's say b is a member of
the column space of A.

00:00:25.540 --> 00:00:27.880
Remember, the column space is
just the set of all of the

00:00:27.880 --> 00:00:29.830
vectors that can be represented
as a linear

00:00:29.830 --> 00:00:36.200
combination of the columns of A,
so that means that b can be

00:00:36.200 --> 00:00:38.230
represented as a linear
combination of

00:00:38.230 --> 00:00:39.280
the columns of A.

00:00:39.280 --> 00:00:43.840
So I'll just write the constant
factors as x1 times

00:00:43.840 --> 00:00:58.800
a1 plus x2 times a2, all the way
to plus xn times an, where

00:00:58.800 --> 00:01:03.830
x1, x2, xn, they're all just
arbitrary real numbers.

00:01:03.830 --> 00:01:11.550
Another way to state this is
that that means that a, which

00:01:11.550 --> 00:01:17.560
I could write as a1, a2, all
the way to an, times some

00:01:17.560 --> 00:01:26.400
vector x1, x2, all the way
to xn, is equal to b.

00:01:26.400 --> 00:01:28.940
These two statements
are equivalent.

00:01:28.940 --> 00:01:31.110
We know that b is a member
of the column space.

00:01:31.110 --> 00:01:33.120
That means that b can be
represented as a linear

00:01:33.120 --> 00:01:36.660
combination of the columns of
A, and that this statement

00:01:36.660 --> 00:01:39.320
right here can be rewritten
this way.

00:01:39.320 --> 00:01:48.940
So you can write that the
equation Ax equals b has at

00:01:48.940 --> 00:01:58.910
least one solution x that
is a member of Rn.

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And the entries of x would
represent the weights on the

00:02:03.070 --> 00:02:06.210
column vectors of A to get your
linear combination b.

00:02:06.210 --> 00:02:08.970
This is all a bit of review.

00:02:08.970 --> 00:02:11.490
Now, let's draw Rn.

00:02:11.490 --> 00:02:14.050
Any solution to this equation
right here is going to be a

00:02:14.050 --> 00:02:14.940
member of Rn.

00:02:14.940 --> 00:02:16.920
Remember, this is an
m-by-n matrix.

00:02:16.920 --> 00:02:20.320
We had n columns and this has
to be a member of Rn right

00:02:20.320 --> 00:02:22.380
there, so let's draw Rn.

00:02:22.380 --> 00:02:29.650
So Rn maybe looks like
that, so that is Rn.

00:02:29.650 --> 00:02:33.070
And let's look at some of the
subspaces that we have in Rn.

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We have the null space.

00:02:34.290 --> 00:02:35.740
That's going to be in Rn.

00:02:35.740 --> 00:02:39.600
The null space is all of the
solutions to the equation Ax

00:02:39.600 --> 00:02:41.830
is equal to 0.

00:02:41.830 --> 00:02:42.980
That's going to be in Rn.

00:02:42.980 --> 00:02:46.020
It's all of the x's that satisfy
that equation, so let

00:02:46.020 --> 00:02:48.240
me draw that right here.

00:02:48.240 --> 00:02:53.140
So let's say I have the null
space right there, so that is

00:02:53.140 --> 00:02:55.720
the null space of A.

00:02:55.720 --> 00:02:57.870
And then what else
do we have in Rn?

00:02:57.870 --> 00:03:00.340
Well, we have the orthogonal
complement of the

00:03:00.340 --> 00:03:01.960
null space of A.

00:03:01.960 --> 00:03:04.550
Let me draw that.

00:03:04.550 --> 00:03:06.702
So we have the orthogonal
complement-- let me doing it

00:03:06.702 --> 00:03:09.260
in a different color.

00:03:09.260 --> 00:03:12.750
We have the orthogonal
complement of the null space

00:03:12.750 --> 00:03:19.130
of A, which we can also
call-- we learned

00:03:19.130 --> 00:03:20.160
this in the last video.

00:03:20.160 --> 00:03:25.700
This is also going to be equal
to the row space of A, which

00:03:25.700 --> 00:03:27.090
is also just the column space.

00:03:31.190 --> 00:03:34.780
The row space of A is the column
space of A transpose.

00:03:34.780 --> 00:03:37.580
So we have two spaces here.

00:03:37.580 --> 00:03:42.780
That is the row space of A.

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So I have two subsets of Rn.

00:03:44.730 --> 00:03:47.320
I have the null space and then
I have the null space's

00:03:47.320 --> 00:03:49.810
complement, orthogonal
complement, which is the row

00:03:49.810 --> 00:03:51.080
space of A.

00:03:51.080 --> 00:03:54.150
Now, we've seen in several
videos now, and I proved it I

00:03:54.150 --> 00:03:58.490
think two videos ago, that
any vector in Rn can be

00:03:58.490 --> 00:04:04.750
represented as a sum of a member
of our null space,

00:04:04.750 --> 00:04:08.620
let's call that vector n, and
let's say some vector in our

00:04:08.620 --> 00:04:13.430
row space, let's call
that vector r.

00:04:13.430 --> 00:04:16.470
Any vector in Rn can be
represented as a sum of some

00:04:16.470 --> 00:04:20.920
vector in our null space and
some vector in our row space.

00:04:20.920 --> 00:04:24.590
So any solution to this equation
is a member of Rn so

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it must be able to be
represented by some member of

00:04:29.190 --> 00:04:31.630
our null space and some member
of our row space.

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So let's write that down.

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So let's say x is a solution
to Ax equals b, which also

00:04:47.630 --> 00:04:54.670
means that x is a member of Rn,
so because it's a member

00:04:54.670 --> 00:04:57.950
of Rn, we can represent it as
a combination of one vector

00:04:57.950 --> 00:04:58.970
here and one vector there.

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So let's say that x is equal
to some vector r0, plus n0,

00:05:11.460 --> 00:05:26.190
where r0 is a member of our row
space and n0 is a member

00:05:26.190 --> 00:05:28.140
of the row space's orthogonal
complement.

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They are the orthogonal
complements of each other, so

00:05:30.510 --> 00:05:36.760
n0 is a member of
our null space.

00:05:36.760 --> 00:05:38.730
Fair enough.

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Now, one thing we might wonder
is, clearly this vector isn't

00:05:44.140 --> 00:05:45.950
a solution to Ax equals b.

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This vector is a solution
to Ax is equal to 0.

00:05:49.430 --> 00:05:52.670
But we might be curious as to
whether this solution right

00:05:52.670 --> 00:06:00.870
here, this member of our row
space is a solution to Ax is

00:06:00.870 --> 00:06:01.780
equal to b.

00:06:01.780 --> 00:06:04.440
This is kind of what we're
focused on in this.

00:06:04.440 --> 00:06:07.260
So let's solve for
r0 right here.

00:06:07.260 --> 00:06:10.560
So if we solve for r0, if we
subtract n0 from both sides,

00:06:10.560 --> 00:06:16.360
we get r0 is equal
to x minus n0.

00:06:16.360 --> 00:06:18.840
All I did was subtract n0 from
both sides and I switched

00:06:18.840 --> 00:06:19.410
things around.

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I solved for r0.

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Now, if we multiply, A times
r0 is equal to A times this

00:06:28.180 --> 00:06:31.030
whole thing-- let me switch
colors-- that's equal to A

00:06:31.030 --> 00:06:47.770
times x0 minus n0, which is
equal to Ax minus An0.

00:06:47.770 --> 00:06:49.550
And what is this equal to?

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Well, A times x, we already said
that x is a solution to

00:06:52.910 --> 00:06:55.450
Ax equals b, so this
right here is going

00:06:55.450 --> 00:06:56.410
to be equal to b.

00:06:56.410 --> 00:06:59.790
And n0 is a member of our null
space, which means it

00:06:59.790 --> 00:07:03.070
satisfies this solution right
here, that A times any member

00:07:03.070 --> 00:07:05.850
of our null space is going to
be equal to the zero vector.

00:07:05.850 --> 00:07:08.150
So that's going to be equal
to the zero vector.

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So you have the vector b minus
the zero vector, and you're

00:07:11.350 --> 00:07:13.760
just going to have
the vector b.

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So we just found out that A
times this member of our row

00:07:17.510 --> 00:07:20.400
space-- let's call that
r0, that's that

00:07:20.400 --> 00:07:22.390
guy right there maybe.

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A times r0 is equal to b.

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So this is a solution.

00:07:27.350 --> 00:07:42.020
So r0 is a solution to
Ax is equal to b.

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So far, it 's kind of an
interesting result that we

00:07:45.790 --> 00:07:46.810
have already.

00:07:46.810 --> 00:07:50.720
If you give me any vector here
b that is a member of our

00:07:50.720 --> 00:07:55.320
column space, then there is
going to be some member of our

00:07:55.320 --> 00:08:03.370
row space right here that is a
solution to Ax is equal to b.

00:08:03.370 --> 00:08:06.140
Now, the next question you might
wonder is, is this the

00:08:06.140 --> 00:08:09.720
only guy in our row space
that is a solution to

00:08:09.720 --> 00:08:11.280
Ax is equal to b?

00:08:11.280 --> 00:08:12.650
And to prove that,
let's assume that

00:08:12.650 --> 00:08:14.300
there's another guy here.

00:08:14.300 --> 00:08:23.380
Let's say that r1 is a member
of our row space and a

00:08:23.380 --> 00:08:30.440
solution to Ax is equal to b.

00:08:30.440 --> 00:08:34.620
Now, the row space is a valid
subspace, so if I take the sum

00:08:34.620 --> 00:08:37.409
or the difference of any two
vectors in the row space, I'll

00:08:37.409 --> 00:08:39.240
get another member
of the row space.

00:08:39.240 --> 00:08:46.320
That's one of the requirements
for being a valid subspace.

00:08:46.320 --> 00:08:47.520
So let's see this.

00:08:47.520 --> 00:08:52.180
So if I take two members of our
subspace, so if I take r1

00:08:52.180 --> 00:08:54.710
minus r0 and I take their
difference, which is just the

00:08:54.710 --> 00:08:57.940
sum-- well, you multiply 1 times
the negative and that

00:08:57.940 --> 00:08:59.700
has to be a member of the
subspace when you're summing

00:08:59.700 --> 00:09:02.380
them, so this has to be a
member of our subspace.

00:09:02.380 --> 00:09:09.000
So this must also be a member
of our of our row space.

00:09:09.000 --> 00:09:11.630
That's because our row space
is a valid subspace.

00:09:11.630 --> 00:09:13.960
You get two of its members, you
take its difference, that

00:09:13.960 --> 00:09:15.230
also has to be a member.

00:09:15.230 --> 00:09:16.170
Fair enough.

00:09:16.170 --> 00:09:17.270
Now, let's see what
happens when you

00:09:17.270 --> 00:09:20.170
multiply this guy by A.

00:09:20.170 --> 00:09:26.570
So if I take A times r1 minus
r0, what do I get?

00:09:26.570 --> 00:09:31.300
I get A times r1 minus
A times r0.

00:09:31.300 --> 00:09:35.770
We already figured out, or for
r1 we assumed that it is a

00:09:35.770 --> 00:09:39.230
solution to Ax is equal to b,
and r0, we already found out,

00:09:39.230 --> 00:09:40.890
it is a solution to
Ax equal to b.

00:09:40.890 --> 00:09:42.670
So either of these, when
you multiply them by

00:09:42.670 --> 00:09:44.580
A, it equals b.

00:09:44.580 --> 00:09:47.520
So this equals b and that equals
b, so you get b minus

00:09:47.520 --> 00:09:50.140
b, which is the zero vector.

00:09:50.140 --> 00:09:51.710
Now, this is interesting.

00:09:51.710 --> 00:09:57.440
This tells us that r minus r0 is
a solution to the equation

00:09:57.440 --> 00:10:00.770
Ax is equal to 0, right?

00:10:00.770 --> 00:10:05.110
When I put r1 minus r0 in the
place of x right there and I

00:10:05.110 --> 00:10:08.550
multiplied it times
the A, I got 0.

00:10:08.550 --> 00:10:16.060
I got 0, which implies that r1
minus r0, that this vector is

00:10:16.060 --> 00:10:18.440
a member of our null space.

00:10:18.440 --> 00:10:24.220
So I have a vector here that's
a member of my row space, and

00:10:24.220 --> 00:10:27.540
we got that from the fact that
both of these are members of

00:10:27.540 --> 00:10:30.150
our row space and the row space
is closed under addition

00:10:30.150 --> 00:10:34.630
and subtraction, and the vector
r1 minus r0 is a member

00:10:34.630 --> 00:10:35.910
of my null space.

00:10:35.910 --> 00:10:38.040
And we've seen this several
times already.

00:10:38.040 --> 00:10:44.100
If I have a vector that is in
a subspace and it's also in

00:10:44.100 --> 00:10:47.302
the orthogonal complement of the
subspace, the null space

00:10:47.302 --> 00:10:52.570
is also the orthogonal
complement of the row space,

00:10:52.570 --> 00:10:54.750
then the only possible
vector that that can

00:10:54.750 --> 00:10:56.130
be is the zero vector.

00:10:56.130 --> 00:11:00.300
That's the only vector that's
inside of a subspace and it's

00:11:00.300 --> 00:11:03.250
orthogonal complement or a
subspace and it's orthogonal

00:11:03.250 --> 00:11:03.810
complement.

00:11:03.810 --> 00:11:05.270
These two guys are
the orthogonal

00:11:05.270 --> 00:11:06.300
complements of each other.

00:11:06.300 --> 00:11:08.040
We drew it up here.

00:11:08.040 --> 00:11:15.190
So we get that r1 one
minus r0 must be

00:11:15.190 --> 00:11:16.410
equal to the zero vector.

00:11:16.410 --> 00:11:18.830
That's the only vector that's
in a subspace and its

00:11:18.830 --> 00:11:23.910
orthogonal complement,
which implies that r1

00:11:23.910 --> 00:11:25.955
must be equal to r0.

00:11:25.955 --> 00:11:28.700
When we take the difference,
we get the zero vector.

00:11:28.700 --> 00:11:30.925
So we have a couple of
neat results here.

00:11:33.430 --> 00:11:35.020
What do we know so far?

00:11:35.020 --> 00:11:39.270
We know that if we have some
vector b that is a member of

00:11:39.270 --> 00:11:50.420
our column space of A,
then there exists a

00:11:50.420 --> 00:11:54.920
unique member, right?

00:11:54.920 --> 00:11:56.180
We just proved the uniqueness.

00:11:56.180 --> 00:12:03.050
There exists a unique member
of the row space of A.

00:12:03.050 --> 00:12:05.190
Let me write it.

00:12:05.190 --> 00:12:06.910
Let me do it in a
different color.

00:12:06.910 --> 00:12:16.720
Of the row space of A, so this
is the row space of A such

00:12:16.720 --> 00:12:22.320
that a unique member of the
row space of A, let

00:12:22.320 --> 00:12:25.450
me call that r0.

00:12:25.450 --> 00:12:27.100
Let me do it in a
different color.

00:12:27.100 --> 00:12:30.120
I want to make this really
stand out in your brain.

00:12:30.120 --> 00:12:38.030
So we know that r0 is a member
of the row space of A such

00:12:38.030 --> 00:12:57.400
that r0 is a solution
to Ax is equal to b.

00:12:57.400 --> 00:12:59.530
It's a little bit of a complex
statement here, but it's

00:12:59.530 --> 00:13:00.040
interesting.

00:13:00.040 --> 00:13:02.917
You give me any b that's a
member of the column space of

00:13:02.917 --> 00:13:06.730
A, then there will exist a
unique member of the row space

00:13:06.730 --> 00:13:10.300
of A, that's my unique member of
the row space of A, that is

00:13:10.300 --> 00:13:14.710
a solution to Ax
is equal to b.

00:13:14.710 --> 00:13:17.250
Now, we can go further
with this.

00:13:17.250 --> 00:13:18.010
We can go further.

00:13:18.010 --> 00:13:23.660
We wrote up here that any
solution to this equation Ax

00:13:23.660 --> 00:13:28.110
is equal to b can be written as
a sum of r0 plus n0, where

00:13:28.110 --> 00:13:31.250
r0 is a member of our row space
and n0 is a member of

00:13:31.250 --> 00:13:33.450
our null space, and that's
because we have a subspace and

00:13:33.450 --> 00:13:34.405
its orthogonal complement.

00:13:34.405 --> 00:13:38.990
So any member of Rn can be
represented as a sum of a

00:13:38.990 --> 00:13:41.480
subspace and a member of the
subspace's orthogonal

00:13:41.480 --> 00:13:42.190
complement.

00:13:42.190 --> 00:13:45.790
Let me rewrite that down here.

00:13:45.790 --> 00:13:59.310
So we already said that any
solution x to Ax is equal to b

00:13:59.310 --> 00:14:10.330
can be written as a
combination-- let me write it

00:14:10.330 --> 00:14:16.520
this way-- as a combination
of r0 plus n0.

00:14:16.520 --> 00:14:17.950
Fair enough.

00:14:17.950 --> 00:14:23.560
Now, what happens if I wanted
to take the square of the

00:14:23.560 --> 00:14:25.620
length of x on both
sides of that.

00:14:25.620 --> 00:14:27.700
Let me write this down, and
you'll see why I'm writing

00:14:27.700 --> 00:14:29.890
this, because I have
another interesting

00:14:29.890 --> 00:14:32.040
result to show you.

00:14:32.040 --> 00:14:36.810
So if I were take the square
of any solution to this

00:14:36.810 --> 00:14:41.200
equation right here, well,
that's going to be the same

00:14:41.200 --> 00:14:45.640
thing as x dot x, which is the
same thing as this thing dot

00:14:45.640 --> 00:14:55.590
itself, same thing as r0
plus n0 dot r0 plus n0.

00:14:55.590 --> 00:14:57.330
And what is this equal to?

00:14:57.330 --> 00:15:12.550
This is equal to r0 dot r0 plus
n0 dot r0 plus n0 dot r0

00:15:12.550 --> 00:15:18.270
again plus n0 dot n0.

00:15:21.000 --> 00:15:23.870
I just kind of foiled it out and
we can do that because we

00:15:23.870 --> 00:15:28.110
know the dot product exhibits
the distributive property.

00:15:28.110 --> 00:15:30.900
So this thing right here
is equal to the

00:15:30.900 --> 00:15:34.310
length of r0 squared.

00:15:34.310 --> 00:15:41.070
Now, we're going to have--
what is n0 dot r0?

00:15:41.070 --> 00:15:44.340
We don't even have to simplify
this much more. n0 is a member

00:15:44.340 --> 00:15:45.580
of our null space.

00:15:45.580 --> 00:15:49.340
r0 is a member of
our row space.

00:15:49.340 --> 00:15:51.860
Each of them is in a subspace
that is the orthogonal

00:15:51.860 --> 00:15:54.120
complement of the other, which
means that everything here

00:15:54.120 --> 00:15:56.610
dotted with anything in
here is equal to 0.

00:15:56.610 --> 00:16:00.940
So r0 dot n0 is going
to be equal to 0.

00:16:00.940 --> 00:16:03.340
These guys are orthogonal to
each other, so that's going to

00:16:03.340 --> 00:16:06.300
be equal to 0, that's going to
be equal to 0, and then you

00:16:06.300 --> 00:16:12.790
get plus-- what's this? n0 dot
n0 is just the length of the

00:16:12.790 --> 00:16:14.160
vector n0 squared.

00:16:14.160 --> 00:16:15.210
These are all vectors.

00:16:15.210 --> 00:16:21.360
And so we get the length of the
vector x squared is equal

00:16:21.360 --> 00:16:25.490
to the length of our member of
our row space squared, our

00:16:25.490 --> 00:16:28.150
unique member of our row space
squared, plus that member of

00:16:28.150 --> 00:16:29.670
our null space squared.

00:16:29.670 --> 00:16:34.670
Now, this is definitely going
to be a positive number.

00:16:34.670 --> 00:16:38.140
It's at minimum 0, but it has to
be something larger than 0,

00:16:38.140 --> 00:16:40.200
so we can say that this
quantity right here is

00:16:40.200 --> 00:16:47.730
definitely greater than or
equal to just r0 squared.

00:16:47.730 --> 00:16:50.500
Or another way to think about
it is, you give me any

00:16:50.500 --> 00:16:59.490
solution to the equation Ax is
equal to b, and the square of

00:16:59.490 --> 00:17:01.620
its length is going to be
greater than or equal to the

00:17:01.620 --> 00:17:05.020
square of r0's length.

00:17:05.020 --> 00:17:08.910
Or since both of the lengths are
always positive, you can

00:17:08.910 --> 00:17:11.099
take kind of the positive square
root and you know you

00:17:11.099 --> 00:17:15.270
won't have to switch signs
there, that the length of any

00:17:15.270 --> 00:17:18.440
solution to Ax equals b is going
to be greater than or

00:17:18.440 --> 00:17:20.730
equal to the length of r0.

00:17:20.730 --> 00:17:23.810
So that makes r0 kind of
a special solution.

00:17:23.810 --> 00:17:26.200
So now let's write our entire
statement, everything that

00:17:26.200 --> 00:17:27.450
we've learned in this video.

00:17:30.340 --> 00:17:35.580
So if b is a member of the
column space of A, then there

00:17:35.580 --> 00:17:54.670
exists a unique r0 that is a
member of the row space of A,

00:17:54.670 --> 00:18:07.370
such that r0 is a solution
to Ax is equal to b.

00:18:07.370 --> 00:18:10.880
And not only is it a solution,
it's a special solution.

00:18:10.880 --> 00:18:22.500
r0 is the solution with the
least, or no solution has a

00:18:22.500 --> 00:18:24.560
smaller length than r0.

00:18:24.560 --> 00:18:26.330
Let me write it that way.

00:18:26.330 --> 00:18:28.800
Maybe some other solution could
be equal but could have

00:18:28.800 --> 00:18:30.190
the same length.

00:18:30.190 --> 00:18:47.600
And no other solution can
have a smaller length.

00:18:47.600 --> 00:18:51.035
Maybe we could write that if you
give me any vector b that

00:18:51.035 --> 00:18:53.860
is a member of the column space
of A, then there exists

00:18:53.860 --> 00:19:00.300
a unique member of the row space
that is essentially the

00:19:00.300 --> 00:19:02.800
smallest solution.

00:19:02.800 --> 00:19:06.210
You can write small as having
the least length to Ax is

00:19:06.210 --> 00:19:08.200
equal to b, which is a
pretty neat outcome.

00:19:08.200 --> 00:19:09.480
In the next video, we'll
explore this a

00:19:09.480 --> 00:19:11.640
little bit more visually.


Orthogonal complement of the orthogonal complement

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:04.480
Let's say I have some subspace
of rn called v.

00:00:04.480 --> 00:00:05.840
Let me draw it like this.

00:00:05.840 --> 00:00:10.550
So that it is r n.

00:00:10.550 --> 00:00:14.770
And I have some subspace of it
we'll call v right here.

00:00:14.770 --> 00:00:17.790
So that is my subspace v.

00:00:17.790 --> 00:00:25.170
We know that the orthogonal
complement v is equal to the

00:00:25.170 --> 00:00:27.840
set of all of the
members of rn.

00:00:30.590 --> 00:00:34.460
So x is a member of rn.

00:00:34.460 --> 00:00:45.510
Such that x dot v is equal to 0
for every v that is a member

00:00:45.510 --> 00:00:47.280
of r subspace.

00:00:47.280 --> 00:00:50.220
So our orthogonal complement of
our subspace is going to be

00:00:50.220 --> 00:00:53.090
all of the vectors that
are orthogonal to

00:00:53.090 --> 00:00:54.830
all of these vectors.

00:00:54.830 --> 00:00:57.120
And we've seen before that they
only overlap-- there's

00:00:57.120 --> 00:00:58.890
only one vector that's
a member of both.

00:00:58.890 --> 00:01:00.220
That's the zero vector.

00:01:00.220 --> 00:01:01.300
It's right there.

00:01:01.300 --> 00:01:03.530
Let's take the orthogonal
complement.

00:01:03.530 --> 00:01:07.190
Let's say it's this set right
here in pink, so that's the

00:01:07.190 --> 00:01:08.650
orthogonal complement.

00:01:08.650 --> 00:01:09.710
Fair enough.

00:01:09.710 --> 00:01:12.910
Now, what if we were to think
about the orthogonal

00:01:12.910 --> 00:01:17.590
complement of the orthogonal
complement?

00:01:17.590 --> 00:01:20.560
So, that's the orthogonal
complement in pink.

00:01:20.560 --> 00:01:23.720
We want the orthogonal
complement of that.

00:01:23.720 --> 00:01:28.050
So this is going to be all
of the x's-- let's just

00:01:28.050 --> 00:01:28.830
write it like this.

00:01:28.830 --> 00:01:37.350
All of the x's that are members
of rn such that x dot

00:01:37.350 --> 00:01:41.210
w is equal to 0.

00:01:41.210 --> 00:01:46.620
For every w that is a member
of the orthogonal

00:01:46.620 --> 00:01:47.660
complement of v.

00:01:47.660 --> 00:01:49.600
That's what that thing
is saying.

00:01:49.600 --> 00:01:53.080
So it's all of the vectors in
rn that are orthogonal to

00:01:53.080 --> 00:01:55.690
everything here.

00:01:55.690 --> 00:01:58.670
Obviously, all of the things in
v are going to be a member

00:01:58.670 --> 00:02:01.120
of that because these guys are
orthogonal to everything in

00:02:01.120 --> 00:02:01.970
these guys.

00:02:01.970 --> 00:02:05.470
But maybe this is just a subset
of the orthogonal

00:02:05.470 --> 00:02:07.360
complement of the orthogonal
complement.

00:02:07.360 --> 00:02:11.440
So maybe this thing in blue
right here looks like this.

00:02:11.440 --> 00:02:13.280
Maybe it's a slightly
larger set than v.

00:02:13.280 --> 00:02:16.450
Maybe there are some things,
these things that I'm shading

00:02:16.450 --> 00:02:21.350
in blue, maybe there are some
vectors that are orthogonal to

00:02:21.350 --> 00:02:24.490
the orthogonal complement of v
but that are outside of v.

00:02:24.490 --> 00:02:26.490
We don't know that yet.

00:02:26.490 --> 00:02:30.070
We don't know whether this
area right here exists.

00:02:30.070 --> 00:02:32.570
Or maybe the orthogonal
complement of the orthogonal

00:02:32.570 --> 00:02:33.020
complement.

00:02:33.020 --> 00:02:34.540
Maybe that takes us back to v.

00:02:34.540 --> 00:02:37.285
Maybe it's like the transpose
or an inverse function where

00:02:37.285 --> 00:02:39.990
it just goes back to our
original subspace.

00:02:39.990 --> 00:02:41.880
Let's see if we can
think about that

00:02:41.880 --> 00:02:44.550
a little bit better.

00:02:44.550 --> 00:02:47.780
Let's say that I have some
member of the orthogonal

00:02:47.780 --> 00:02:50.050
complement of the orthogonal
complement.

00:02:50.050 --> 00:02:56.880
So let's say I have some vector
x that is a member of

00:02:56.880 --> 00:03:00.790
the orthogonal complement of
the orthogonal complement.

00:03:00.790 --> 00:03:05.140
Now, we saw on the last video
that any vector in rn can be

00:03:05.140 --> 00:03:09.460
represented by a sum of some
vector in a subspace and the

00:03:09.460 --> 00:03:12.380
subspace's complement.

00:03:12.380 --> 00:03:15.930
So we know that x can be
represented-- we can say that

00:03:15.930 --> 00:03:20.000
x can be represented as the
sum of two vectors.

00:03:20.000 --> 00:03:22.980
One that's in v and one that's
in the orthogonal

00:03:22.980 --> 00:03:24.500
complement of v.

00:03:24.500 --> 00:03:28.130
So one, let's call that the
vector that's in v and let's

00:03:28.130 --> 00:03:30.400
call w the vector that's
in the orthogonal

00:03:30.400 --> 00:03:31.380
complement of V.

00:03:31.380 --> 00:03:34.400
Let me write it like this.

00:03:34.400 --> 00:03:41.700
Where v is a member of the
subspace v and the vector w is

00:03:41.700 --> 00:03:45.140
a member of the orthogonal
complement of v.

00:03:45.140 --> 00:03:45.790
Right?

00:03:45.790 --> 00:03:48.380
So this is some member.

00:03:48.380 --> 00:03:49.860
It could be some guy out here.

00:03:49.860 --> 00:03:51.080
It could be some
guy over here.

00:03:51.080 --> 00:03:53.370
He's a member of the orthogonal
complement of the

00:03:53.370 --> 00:03:54.120
orthogonal complement.

00:03:54.120 --> 00:03:55.930
Which is this whole area here.

00:03:55.930 --> 00:03:58.860
Which v is a subset of, but
we're not sure whether v

00:03:58.860 --> 00:04:00.360
equals that thing.

00:04:00.360 --> 00:04:04.170
But we say, look, anything
that's in the orthogonal

00:04:04.170 --> 00:04:06.730
complement of your orthogonal
complement, is going to be a

00:04:06.730 --> 00:04:07.645
member of rn.

00:04:07.645 --> 00:04:10.770
And anything in rn can be
represented as a sum of a

00:04:10.770 --> 00:04:13.120
vector in v and a vector
in the orthogonal

00:04:13.120 --> 00:04:14.890
complement of v.

00:04:14.890 --> 00:04:16.839
So that's all I wrote
right there.

00:04:16.839 --> 00:04:25.620
Now, what happens if I take the
dot product of x with w?

00:04:25.620 --> 00:04:28.460
What is this going
to be equal to?

00:04:28.460 --> 00:04:30.650
This is the orthogonal
complement of the orthogonal

00:04:30.650 --> 00:04:32.160
complement.

00:04:32.160 --> 00:04:36.300
Would you take the dot product
of any vector in this with any

00:04:36.300 --> 00:04:38.750
vector in the orthogonal
complement,

00:04:38.750 --> 00:04:40.030
which this vector is.

00:04:40.030 --> 00:04:40.280
Right?

00:04:40.280 --> 00:04:41.690
It's a member of the orthogonal
complement.

00:04:41.690 --> 00:04:43.660
You're going to get
0 by definition.

00:04:43.660 --> 00:04:45.490
These are all of the vectors.

00:04:45.490 --> 00:04:49.210
This factor is definitely
orthogonal to anything in just

00:04:49.210 --> 00:04:50.590
v perp right?

00:04:50.590 --> 00:04:52.780
Anything in v perp perp
is orthogonal to

00:04:52.780 --> 00:04:54.030
anything in v perp.

00:04:54.030 --> 00:04:57.420
So, this thing is going
to be equal to 0.

00:04:57.420 --> 00:04:59.560
But what's another way
of writing x dot w?

00:04:59.560 --> 00:05:00.950
We could write it like this.

00:05:00.950 --> 00:05:07.050
This is the same thing
as v plus w dot w.

00:05:07.050 --> 00:05:14.940
Which is the same thing as
v dot w plus w dot w.

00:05:14.940 --> 00:05:17.700
Now, what is v dot w?

00:05:17.700 --> 00:05:22.300
v is a member of our
original subspace.

00:05:22.300 --> 00:05:24.690
And if you take the dot product
of anything in our

00:05:24.690 --> 00:05:28.290
original subspace anything in
its orthogonal complement,

00:05:28.290 --> 00:05:30.740
you're going to get 0.

00:05:30.740 --> 00:05:33.970
So this term right here is going
to be 0, and you're just

00:05:33.970 --> 00:05:37.450
going to get this term which
is the same thing as the

00:05:37.450 --> 00:05:40.770
length of our vector
w squared.

00:05:40.770 --> 00:05:42.520
Now, that has to equal 0.

00:05:42.520 --> 00:05:45.330
Remember we just
wrote x dot w.

00:05:45.330 --> 00:05:48.080
x is a member of the orthogonal
complement of your

00:05:48.080 --> 00:05:49.250
orthogonal complement.

00:05:49.250 --> 00:05:51.600
So, you dot that with anything
in the orthogonal complement,

00:05:51.600 --> 00:05:53.020
that's got to be equal 0.

00:05:53.020 --> 00:05:55.710
But, if we write it the other
way, if we write it as the sum

00:05:55.710 --> 00:05:58.910
of v plus w and distribute this
w, we say that's the same

00:05:58.910 --> 00:06:01.570
thing as the magnitude
of w squared.

00:06:01.570 --> 00:06:04.810
So the magnitude of w squared
has got to be equal to 0.

00:06:04.810 --> 00:06:09.360
The magnitude of w squared, or
the length of w squared, has

00:06:09.360 --> 00:06:10.950
got to be equal to 0.

00:06:10.950 --> 00:06:14.810
Which tells us that w
is the zero vector.

00:06:14.810 --> 00:06:18.800
That's the only factor in rn
when you take its length and,

00:06:18.800 --> 00:06:21.030
especially when you square
it, you get 0.

00:06:21.030 --> 00:06:22.670
But you could just
take its length.

00:06:22.670 --> 00:06:24.080
So what does that mean?

00:06:24.080 --> 00:06:28.460
That means that our original
vector x is equal to v plus w.

00:06:28.460 --> 00:06:31.460
But w is just equal to 0.

00:06:31.460 --> 00:06:35.990
So that implies that
our original vector

00:06:35.990 --> 00:06:44.050
x is equal to v.

00:06:44.050 --> 00:06:47.180
And v is a member of
our subspace v.

00:06:47.180 --> 00:06:48.000
Right?

00:06:48.000 --> 00:06:55.460
So that tells us that x is a
member of our subspace v.

00:06:55.460 --> 00:06:59.550
So we just showed that if
something is a member of the

00:06:59.550 --> 00:07:03.870
orthogonal complement of the
orthogonal complement then

00:07:03.870 --> 00:07:06.830
that same vector has
to be a member of

00:07:06.830 --> 00:07:08.640
the original subspace.

00:07:08.640 --> 00:07:12.400
So there is no such thing as
something being in the

00:07:12.400 --> 00:07:15.070
orthogonal complement of the
orthogonal complement and not

00:07:15.070 --> 00:07:18.120
being a member of our
original subspace.

00:07:18.120 --> 00:07:23.220
All of this has to be inside
of this right there.

00:07:23.220 --> 00:07:26.430
So there is no outside
blue space like that.

00:07:26.430 --> 00:07:29.370
All of that is our original
subspace if you want to view

00:07:29.370 --> 00:07:30.250
it that way.

00:07:30.250 --> 00:07:32.770
Now I just at the beginning of
the video, anything in our

00:07:32.770 --> 00:07:34.770
subspace is going to be a
member of our orthogonal

00:07:34.770 --> 00:07:35.080
complement.

00:07:35.080 --> 00:07:37.130
And then you can kind of reason
that in your head.

00:07:37.130 --> 00:07:39.680
Let's use the same argument to
just be a little bit more

00:07:39.680 --> 00:07:40.930
rigorous about it.

00:07:45.880 --> 00:07:47.760
Right now we say if anything
is in the orthogonal

00:07:47.760 --> 00:07:49.970
complement of the orthogonal
complement, then it's going to

00:07:49.970 --> 00:07:51.460
be the original subspace.

00:07:51.460 --> 00:07:53.130
Let's go the other way.

00:07:53.130 --> 00:07:55.320
Let's say that something
is in the original

00:07:55.320 --> 00:08:00.990
subspace just like that.

00:08:00.990 --> 00:08:03.550
Let me draw another graph
right here because

00:08:03.550 --> 00:08:06.390
this might be useful.

00:08:06.390 --> 00:08:09.780
Let me draw rn again.

00:08:09.780 --> 00:08:13.310
Let me draw all of
rn like that.

00:08:13.310 --> 00:08:16.480
Now, we have the orthogonal
complement.

00:08:16.480 --> 00:08:23.600
Let me just draw that first.
So v perp And then you have

00:08:23.600 --> 00:08:26.650
the orthogonal complement of
the orthogonal complement

00:08:26.650 --> 00:08:30.200
which could be this
set right here.

00:08:30.200 --> 00:08:31.500
Right?

00:08:31.500 --> 00:08:34.100
This is v perp.

00:08:34.100 --> 00:08:36.350
I haven't even drawn
the subspace v.

00:08:36.350 --> 00:08:39.690
All I've shown is, I have some
subspace here, which I happen

00:08:39.690 --> 00:08:40.950
to call v perp.

00:08:40.950 --> 00:08:44.140
And then I have the orthogonal
complement of that subspace.

00:08:44.140 --> 00:08:47.200
So this means that anything in
rn can be represented as the

00:08:47.200 --> 00:08:51.400
sum of a vector that's here
and a vector that's here.

00:08:51.400 --> 00:08:56.300
So, if I say that w-- let
me do it in purple.

00:08:56.300 --> 00:09:00.400
If I say the vector w-- let
me write it this way.

00:09:00.400 --> 00:09:06.110
The vector v can be represented
as the sum of the

00:09:06.110 --> 00:09:20.620
vector w and the vector x where
w is a member of the

00:09:20.620 --> 00:09:25.580
orthogonal complement of v or
v perp And x is a member of

00:09:25.580 --> 00:09:28.560
its orthogonal complement.

00:09:28.560 --> 00:09:31.540
Notice, all I'm saying, I could
have called this set s.

00:09:31.540 --> 00:09:33.530
And then this would have been
s and its orthogonal

00:09:33.530 --> 00:09:34.410
complement.

00:09:34.410 --> 00:09:37.230
And we learned that anything in
rn could be represented as

00:09:37.230 --> 00:09:40.900
the sum of something in a
subspace and the subspace's

00:09:40.900 --> 00:09:42.350
orthogonal complement.

00:09:42.350 --> 00:09:45.220
So it doesn't matter that v is
somehow related to this.

00:09:45.220 --> 00:09:47.720
It can be represented as
a sum of a vector here

00:09:47.720 --> 00:09:49.270
plus a vector there.

00:09:49.270 --> 00:09:50.480
Fair enough.

00:09:50.480 --> 00:09:55.950
Now, what happens if
I dot v with w?

00:09:55.950 --> 00:09:59.340
I'm doing the exact same
argument that I did before.

00:09:59.340 --> 00:10:03.200
Well, if you take anything
that's a member of our

00:10:03.200 --> 00:10:05.560
original subspace, and you dot
it with anything in its

00:10:05.560 --> 00:10:10.120
orthogonal complement, that's
going to give us 0.

00:10:10.120 --> 00:10:13.900
What else is that going
to be equal to?

00:10:13.900 --> 00:10:17.680
If we write v in this way, v
dot w is the same thing as

00:10:17.680 --> 00:10:19.650
this thing dot w.

00:10:19.650 --> 00:10:29.600
So w plus x dot-- and this is
going to be equal to w dot w

00:10:29.600 --> 00:10:33.020
plus x dot w.

00:10:33.020 --> 00:10:36.340
And what's x dot w? x is in the
orthogonal complement of

00:10:36.340 --> 00:10:37.980
your orthogonal complement.

00:10:37.980 --> 00:10:39.710
And w is in the orthogonal
complement.

00:10:39.710 --> 00:10:40.950
So if you take the dot
product, you're

00:10:40.950 --> 00:10:41.600
going to get 0.

00:10:41.600 --> 00:10:43.210
They're orthogonal
to each other.

00:10:43.210 --> 00:10:49.020
So this is just equal to w dot
w or the length of w squared.

00:10:49.020 --> 00:10:52.340
And since since has to equal
0, we just have a bunch of

00:10:52.340 --> 00:10:56.060
equals here, that tells us that
once again the vector w

00:10:56.060 --> 00:10:58.570
has to be equal to 0.

00:10:58.570 --> 00:11:02.840
So that tells us v is
equal to w plus x.

00:11:02.840 --> 00:11:08.400
But if w is equal to 0, then v
is going to be equal to x.

00:11:08.400 --> 00:11:15.940
So we've just shown that if v is
a member of the subspace v,

00:11:15.940 --> 00:11:21.650
then v is a member of the
orthogonal complement of the

00:11:21.650 --> 00:11:22.590
orthogonal complement.

00:11:22.590 --> 00:11:24.900
Right? v is equal to x, which is
a member of the orthogonal

00:11:24.900 --> 00:11:27.170
complement or the orthogonal
complement.

00:11:27.170 --> 00:11:31.020
So we've proven it both ways.

00:11:31.020 --> 00:11:34.580
If you look at the original
statement, we wrote here that

00:11:34.580 --> 00:11:36.290
if you're a member of the
orthogonal complement of the

00:11:36.290 --> 00:11:37.840
orthogonal complement,
you're a member of

00:11:37.840 --> 00:11:39.730
the original subspace.

00:11:39.730 --> 00:11:44.190
So, we've proven this and,
earlier in the video, we

00:11:44.190 --> 00:11:49.160
proved that if x is a member of
the orthogonal complement

00:11:49.160 --> 00:11:53.870
of the orthogonal complement,
then x is a

00:11:53.870 --> 00:11:55.840
member of our subspace.

00:11:55.840 --> 00:11:57.300
So these two things
are equivalent.

00:11:57.300 --> 00:11:58.460
Anything that's in
the subspace is a

00:11:58.460 --> 00:12:00.150
member of v perp perp.

00:12:00.150 --> 00:12:03.150
Anything in v perp perp is
a member of our subspace.

00:12:03.150 --> 00:12:06.520
So, our subspace in v perp
perp are the same set.

00:12:06.520 --> 00:12:08.400
And of course it overlaps.

00:12:08.400 --> 00:12:09.670
This equals this.

00:12:09.670 --> 00:12:13.100
And of course it overlaps with
V perp and its orthogonal

00:12:13.100 --> 00:12:17.200
complement only at the zero
vector right there.


Orthogonal complement of the nullspace

WEBVTT
Kind: captions
Language: en

00:00:00.580 --> 00:00:03.010
I've got some matrix A.

00:00:03.010 --> 00:00:05.810
We learned several videos ago
that it's row space is the

00:00:05.810 --> 00:00:10.960
same thing as the column space
of it's transpose.

00:00:10.960 --> 00:00:19.720
So that right there is
the row space of A.

00:00:19.720 --> 00:00:22.990
That this thing's orthogonal
complement, so the set of all

00:00:22.990 --> 00:00:25.820
of the vectors that are
orthogonal to this, so its

00:00:25.820 --> 00:00:31.230
orthogonal complement is equal
to the nullspace of A.

00:00:31.230 --> 00:00:34.760
And, essentially, the same
result if you switch A and A

00:00:34.760 --> 00:00:38.750
transpose, we also learned that
the orthogonal complement

00:00:38.750 --> 00:00:46.450
of the column space of A
is equal to the left

00:00:46.450 --> 00:00:48.680
nullspace of A.

00:00:48.680 --> 00:00:52.450
Which is the same thing as the
nullspace of A transposed.

00:00:52.450 --> 00:00:55.410
We could write this, just to
understand the terminology,

00:00:55.410 --> 00:00:59.550
that's the left nullspace, which
is the same thing as the

00:00:59.550 --> 00:01:01.410
nullspace of A transposed.

00:01:01.410 --> 00:01:05.890
Now, what is the orthogonal
complement of the

00:01:05.890 --> 00:01:07.140
nullspace of A?

00:01:11.270 --> 00:01:13.910
Well, you might guess that it's
the row space of A, but

00:01:13.910 --> 00:01:15.740
we didn't have the tools
until the last video

00:01:15.740 --> 00:01:16.500
to figure that out.

00:01:16.500 --> 00:01:22.120
In the last video, we saw that
if we take the orthogonal

00:01:22.120 --> 00:01:25.400
complement -- let me write it
this way -- if we were to take

00:01:25.400 --> 00:01:28.110
the orthogonal complement of the
orthogonal complement, it

00:01:28.110 --> 00:01:30.620
equals the original sub space.

00:01:30.620 --> 00:01:32.030
So now, what are we doing?

00:01:32.030 --> 00:01:35.200
We're taking the orthogonal
complement of the

00:01:35.200 --> 00:01:36.510
nullspace of A.

00:01:36.510 --> 00:01:41.120
Well, the nullspace of A is just
this thing right here.

00:01:41.120 --> 00:01:44.520
So this is equal to taking the
orthogonal complement of the

00:01:44.520 --> 00:01:46.490
nullspace of A.

00:01:46.490 --> 00:01:48.630
But the nullspace of
A is this thing.

00:01:48.630 --> 00:01:54.790
It's the row space's orthogonal
complement.

00:01:54.790 --> 00:01:57.590
Now, we're essentially the
orthogonal complement of the

00:01:57.590 --> 00:01:59.020
orthogonal complement.

00:01:59.020 --> 00:02:01.760
We can use this property, which
we just proved in the

00:02:01.760 --> 00:02:05.160
last video, to say that this
is equal to just the

00:02:05.160 --> 00:02:07.270
row space of A.

00:02:07.270 --> 00:02:11.730
Which is the same thing as the
column space of A transposed.

00:02:11.730 --> 00:02:14.090
So the orthogonal complement
of the row space is the

00:02:14.090 --> 00:02:17.580
nullspace and the orthogonal
complement of the nullspace is

00:02:17.580 --> 00:02:18.550
the row space.

00:02:18.550 --> 00:02:19.920
We can apply that
same property on

00:02:19.920 --> 00:02:23.030
this side right here.

00:02:23.030 --> 00:02:28.840
What is the orthogonal
complement of the left

00:02:28.840 --> 00:02:32.750
nullspace of A?

00:02:32.750 --> 00:02:33.890
What is this?

00:02:33.890 --> 00:02:35.920
Well, this is going to be
equal to the orthogonal

00:02:35.920 --> 00:02:38.170
complement of this thing.

00:02:38.170 --> 00:02:40.740
Because that's what the left
nullspace of A is equal to.

00:02:40.740 --> 00:02:43.390
So it's equal to the orthogonal
complement of the

00:02:43.390 --> 00:02:47.090
orthogonal complement
of the column space.

00:02:47.090 --> 00:02:49.190
And we just learned in the last
video, if you take the

00:02:49.190 --> 00:02:51.180
orthogonal complement of the
orthogonal complement it

00:02:51.180 --> 00:02:53.070
equals the original subspace.

00:02:53.070 --> 00:02:56.680
So this is just equal to
the column space of A.

00:02:56.680 --> 00:02:59.460
So we now see some
nice symmetry.

00:02:59.460 --> 00:03:03.060
The nullspace is the orthogonal
complement of the

00:03:03.060 --> 00:03:06.580
row space, and then we see
that the row space is the

00:03:06.580 --> 00:03:09.000
orthogonal complement
of the nullspace.

00:03:09.000 --> 00:03:12.235
Similarly, the left nullspace
is the orthogonal complement

00:03:12.235 --> 00:03:14.310
of the column space.

00:03:14.310 --> 00:03:18.310
And the column space is the
orthogonal complement of the

00:03:18.310 --> 00:03:19.610
left nullspace.

00:03:19.610 --> 00:03:22.850
So we have some nice symmetry
that we're able to essentially

00:03:22.850 --> 00:03:26.500
prove given what we saw
in the last video.


Representing vectors in rn using subspace members

WEBVTT
Kind: captions
Language: en

00:00:00.670 --> 00:00:05.350
Let's say I have some subspace
V, that is a subset of Rn.

00:00:08.100 --> 00:00:10.690
And let's say that we also
have its orthogonal

00:00:10.690 --> 00:00:13.820
complement, we write
that as V perp.

00:00:13.820 --> 00:00:17.410
That'll also be a
subset of Rn.

00:00:17.410 --> 00:00:19.520
A couple of videos ago, it might
have even been the last

00:00:19.520 --> 00:00:25.820
video if I remember properly, we
learned that the dimension

00:00:25.820 --> 00:00:31.040
of V, plus the dimension of the
orthogonal complement of

00:00:31.040 --> 00:00:33.850
V, which is also another
subspace, is going

00:00:33.850 --> 00:00:35.220
to be equal to n.

00:00:35.220 --> 00:00:40.880
Remember dimension is just the
number of linearly independent

00:00:40.880 --> 00:00:43.470
vectors you need to have
a basis for V.

00:00:43.470 --> 00:00:46.320
And the dimension here is the
number of linearly independent

00:00:46.320 --> 00:00:48.960
vectors you need to have a
basis for the orthogonal

00:00:48.960 --> 00:00:50.460
complement of V.

00:00:50.460 --> 00:00:53.410
Now given this, let's see if we
can come up with some other

00:00:53.410 --> 00:00:57.340
interesting ways in which these
two subspaces relate to

00:00:57.340 --> 00:00:57.760
each other.

00:00:57.760 --> 00:01:01.360
Or how they might relate to
all of the vectors in Rn.

00:01:01.360 --> 00:01:06.250
So the first question is, do
these two subspaces have

00:01:06.250 --> 00:01:07.340
anything in common?

00:01:07.340 --> 00:01:10.200
Are there any vectors that are
in common with the two.

00:01:10.200 --> 00:01:13.580
And to test whether there is,
let's just assume there is,

00:01:13.580 --> 00:01:17.830
and see what the properties of
that vector would have to be.

00:01:17.830 --> 00:01:25.470
Let's assume right here that I
have some vector x that is a

00:01:25.470 --> 00:01:28.230
member of my subspace V.

00:01:28.230 --> 00:01:32.630
Let's also assume that x is
a member of the orthogonal

00:01:32.630 --> 00:01:33.880
complement of V.

00:01:39.180 --> 00:01:41.590
Now what does this second
statement mean?

00:01:41.590 --> 00:01:51.250
Membership in the orthogonal
complement means that x dot v,

00:01:51.250 --> 00:01:58.050
for any v that is a member of
our subspace, is going to be

00:01:58.050 --> 00:01:58.590
equal to 0.

00:01:58.590 --> 00:02:00.770
Let me write it this
way actually.

00:02:00.770 --> 00:02:07.660
x dot v is equal to 0
for any v that is a

00:02:07.660 --> 00:02:08.860
member of our subspace.

00:02:08.860 --> 00:02:11.620
That's what it means to be a
member of V's orthogonal

00:02:11.620 --> 00:02:12.770
complement.

00:02:12.770 --> 00:02:17.130
Now we assume that x is
also a member of V.

00:02:17.130 --> 00:02:20.320
So that means that we can
stick x here as well.

00:02:20.320 --> 00:02:22.270
For any member of V.

00:02:22.270 --> 00:02:23.600
x is also a member of V.

00:02:23.600 --> 00:02:29.310
So that implies that x
dot x is equal to 0.

00:02:29.310 --> 00:02:36.990
Another way to write that is
that the length of x squared

00:02:36.990 --> 00:02:37.970
is equal to 0.

00:02:37.970 --> 00:02:39.560
Or the length of x
is equal to 0.

00:02:39.560 --> 00:02:41.160
And that's only true
for one vector.

00:02:41.160 --> 00:02:43.110
You can even try it out
with the different

00:02:43.110 --> 00:02:46.030
constituents of x.

00:02:46.030 --> 00:02:49.750
The only vector that that's
true for is the 0 vector.

00:02:49.750 --> 00:02:52.770
So x has to be equal
to the 0 vector.

00:02:52.770 --> 00:02:55.320
That's the only vector in Rn
that when you dot it with

00:02:55.320 --> 00:02:58.210
itself you get 0, or whose
the square of its

00:02:58.210 --> 00:02:59.750
length is equal to 0.

00:02:59.750 --> 00:03:02.510
And we've shown that many,
many, many videos ago.

00:03:02.510 --> 00:03:08.310
What this tells us is at that
the intersection between V and

00:03:08.310 --> 00:03:11.490
V complement-- this kind of
upside down U just means

00:03:11.490 --> 00:03:13.840
intersection, it just means
where do these two sets

00:03:13.840 --> 00:03:19.680
overlap-- the only place that
these overlap is with a subset

00:03:19.680 --> 00:03:20.740
of the 0 vector.

00:03:20.740 --> 00:03:23.070
So if I were to draw all
of Rn like this.

00:03:26.770 --> 00:03:28.020
Let's say that this is Rn.

00:03:30.460 --> 00:03:32.820
And let's say I draw
the subspace V.

00:03:39.870 --> 00:03:42.300
And let's say I draw the
orthogonal complement to V.

00:03:42.300 --> 00:03:45.440
It's all of these vectors
right here.

00:03:45.440 --> 00:03:47.870
This is the orthogonal
complement to V right there.

00:03:47.870 --> 00:03:49.805
So this is V perp.

00:03:54.620 --> 00:03:56.920
These are all of the vectors
that when I dot it with any

00:03:56.920 --> 00:03:58.765
vector here, I'm going
to get equal to 0.

00:03:58.765 --> 00:04:00.050
So this is V perp.

00:04:00.050 --> 00:04:03.410
The intersection, their overlap,
the only vector that

00:04:03.410 --> 00:04:10.030
is a member of both
is the 0 vector.

00:04:10.030 --> 00:04:12.640
That's their only
intersection.

00:04:12.640 --> 00:04:13.490
So that's fair enough.

00:04:13.490 --> 00:04:15.900
The only vector that's a member
of a subspace and its

00:04:15.900 --> 00:04:18.500
orthogonal complement
is the 0 vector.

00:04:18.500 --> 00:04:20.220
Nothing too profound there.

00:04:20.220 --> 00:04:22.760
Let's see if we can come up with
some other interesting

00:04:22.760 --> 00:04:26.010
relations between the subspace
and its orthogonal complement.

00:04:26.010 --> 00:04:30.240
Maybe some arbitrary
vectors in Rn.

00:04:30.240 --> 00:04:33.830
So let's just write down--
well we know that the

00:04:33.830 --> 00:04:41.910
dimension of our subspace
V is equal to k.

00:04:44.470 --> 00:04:47.400
If its equal to k, we know that
its dimension plus its

00:04:47.400 --> 00:04:50.230
orthogonal complement has to be
equal to n, because we're

00:04:50.230 --> 00:04:51.480
dealing in Rn.

00:04:54.570 --> 00:04:57.165
And we also know that the
orthogonal complement of V is

00:04:57.165 --> 00:05:00.470
a subset of Rn, I drew
it right here.

00:05:00.470 --> 00:05:03.030
The dimension of V
is equal to k.

00:05:03.030 --> 00:05:04.610
That's a k right there.

00:05:04.610 --> 00:05:13.290
And what's the dimension of the
orthogonal complement of V

00:05:13.290 --> 00:05:14.470
going to be?

00:05:14.470 --> 00:05:16.420
Well, when you add them
together-- I wrote that up

00:05:16.420 --> 00:05:18.720
here-- they have to equal n.

00:05:18.720 --> 00:05:22.100
So this guy's going to
have to be n minus k.

00:05:22.100 --> 00:05:24.380
If you have k here.

00:05:24.380 --> 00:05:27.310
This guy's dimension is k, this
guy's dimension right

00:05:27.310 --> 00:05:30.840
here if, it's n minus k, that
when you add these two up, k

00:05:30.840 --> 00:05:33.930
plus n minus k is going
to be equal to n.

00:05:33.930 --> 00:05:36.760
So this guy will have a
dimension of n minus k.

00:05:36.760 --> 00:05:38.100
Now what does dimension mean?

00:05:38.100 --> 00:05:41.300
It means that that's the number
of linearly independent

00:05:41.300 --> 00:05:44.690
vectors you need to
form a basis.

00:05:44.690 --> 00:05:48.720
I have k vectors as
a basis for V.

00:05:48.720 --> 00:05:56.540
I have v1, v2, all
the way to vk.

00:05:56.540 --> 00:06:03.940
And this is a basis for V, which
just means they're all

00:06:03.940 --> 00:06:05.210
linearly independent.

00:06:05.210 --> 00:06:06.500
And they span V.

00:06:06.500 --> 00:06:09.850
Any member of V right here can
be represented as a linear

00:06:09.850 --> 00:06:13.070
combination of these vectors.

00:06:13.070 --> 00:06:16.750
Now the dimension of the
orthogonal complement of

00:06:16.750 --> 00:06:17.965
V is n minus k.

00:06:17.965 --> 00:06:20.550
So we could have n
minus k vectors.

00:06:20.550 --> 00:06:31.030
Let's call them w1, w2, all
the way to wn minus k.

00:06:31.030 --> 00:06:33.600
We have n minus k of
these characters.

00:06:33.600 --> 00:06:41.020
And this set is a basis
for the orthogonal

00:06:41.020 --> 00:06:42.260
complement of V.

00:06:42.260 --> 00:06:45.250
So any vector in here can be
represented as a linear

00:06:45.250 --> 00:06:47.560
combination of these
guys right here.

00:06:47.560 --> 00:06:49.355
And all of these guys are
linearly independent.

00:06:49.355 --> 00:06:53.030
So you don't have any redundant
vectors there.

00:06:53.030 --> 00:06:55.110
Now let's explore.

00:06:55.110 --> 00:06:56.640
And I'll tell you where
I'm trying to go.

00:06:56.640 --> 00:07:00.830
I'm trying to see if I combine
these two sets, whether I get

00:07:00.830 --> 00:07:02.860
a basis for all of Rn.

00:07:02.860 --> 00:07:05.390
That's what I'm trying
to understand.

00:07:05.390 --> 00:07:20.100
Let's just say that for some
constants c1 times v1 plus c2

00:07:20.100 --> 00:07:30.720
times v2 plus all the way to
ck times vk plus-- for the

00:07:30.720 --> 00:07:38.990
constants on these guys I'll use
d-- plus d1 times w1 plus

00:07:38.990 --> 00:07:47.960
d2 times w2, all the way to
plus dn minus k times the

00:07:47.960 --> 00:07:51.010
basis vector wn minus k.

00:07:51.010 --> 00:07:54.500
Let's say that I'm curious about
setting this sum equal

00:07:54.500 --> 00:08:02.160
to 0, equaling the 0 vector
for some scalers.

00:08:02.160 --> 00:08:06.120
The scalers are these
c's and these d's.

00:08:06.120 --> 00:08:09.080
And we know that there's at
least one solution set of

00:08:09.080 --> 00:08:10.700
scalers for which
this is true.

00:08:10.700 --> 00:08:15.490
We could multiply all of these
constants-- c1, c2, ck, d1,

00:08:15.490 --> 00:08:17.260
d2, all the way to dn minus k.

00:08:17.260 --> 00:08:19.030
They could all be 0.

00:08:19.030 --> 00:08:20.790
Or there might be more
than one solution.

00:08:20.790 --> 00:08:25.690
In fact, if the only solution is
that all of these constants

00:08:25.690 --> 00:08:29.240
have to be equal to 0, then we
know that all of these vectors

00:08:29.240 --> 00:08:32.030
are linearly independent with
respect to each other.

00:08:32.030 --> 00:08:33.990
And if they're all linearly
independent with respect to

00:08:33.990 --> 00:08:39.919
each other, then we know that
they can be a basis for Rn.

00:08:39.919 --> 00:08:40.830
But we don't know that yet.

00:08:40.830 --> 00:08:43.460
We don't know that the only
solution to this is all of the

00:08:43.460 --> 00:08:45.000
constants being equal to 0.

00:08:45.000 --> 00:08:46.240
So let's see if we can
experiment with

00:08:46.240 --> 00:08:48.270
this a little bit.

00:08:48.270 --> 00:08:50.820
If we take this equation, which
I just wrote down, we

00:08:50.820 --> 00:08:53.885
know that one solution is all of
the constants, the c's and

00:08:53.885 --> 00:08:55.110
d's equalling to 0, but
we don't know that

00:08:55.110 --> 00:08:56.460
that's the only one.

00:08:56.460 --> 00:09:00.100
Let's just subtract all of the
w vectors from both sides of

00:09:00.100 --> 00:09:01.900
this equation.

00:09:01.900 --> 00:09:02.880
So what are we going to get?

00:09:02.880 --> 00:09:09.990
We're going to get c1, v1
plus c2, v2 all the

00:09:09.990 --> 00:09:13.975
way to plus ck, vk.

00:09:13.975 --> 00:09:15.750
And we're going to subtract
this from both

00:09:15.750 --> 00:09:17.020
sides of the equation.

00:09:17.020 --> 00:09:20.180
It's going to be equal
to the 0 vector.

00:09:20.180 --> 00:09:22.510
Which is really just 0, I don't
even have to write it

00:09:22.510 --> 00:09:24.600
down, but maybe I'll write
it down there just so you

00:09:24.600 --> 00:09:25.040
understand.

00:09:25.040 --> 00:09:28.380
I'm just taking this equation,
I'm subtracting these guys

00:09:28.380 --> 00:09:29.140
from both sides.

00:09:29.140 --> 00:09:40.870
So 0 vector minus d1, w1 plus
d2, w2 plus all the way to dn

00:09:40.870 --> 00:09:45.920
minus k, wn minus k.

00:09:48.810 --> 00:09:52.510
All I did is I subtract these
terms right here from both

00:09:52.510 --> 00:09:54.200
sides of this equation.

00:09:54.200 --> 00:09:55.900
I don't even have to write
this is 0 here,

00:09:55.900 --> 00:09:57.800
that's a bit redundant.

00:09:57.800 --> 00:10:02.290
So what I have here is I have
some combination of the basis

00:10:02.290 --> 00:10:05.000
vector of V.

00:10:05.000 --> 00:10:08.880
So if I look at this, this is
some linear combination of the

00:10:08.880 --> 00:10:11.650
basis vectors in V.

00:10:11.650 --> 00:10:17.030
If I call this a vector-- let
me call this some vector x.

00:10:17.030 --> 00:10:22.960
Let's say x is equal to
c1, v1 plus c2, v2 all

00:10:22.960 --> 00:10:28.650
the way to ck, vk.

00:10:28.650 --> 00:10:30.950
We know that it's a linear
combination of our basis

00:10:30.950 --> 00:10:36.630
vectors of V, so x
is a member of V.

00:10:36.630 --> 00:10:39.330
By definition, any linear
combination of the basis

00:10:39.330 --> 00:10:41.380
vectors for a subspace
is going to be a

00:10:41.380 --> 00:10:43.500
member of that subspace.

00:10:43.500 --> 00:10:46.210
Well, similarly, what do we have
on the right-hand side of

00:10:46.210 --> 00:10:47.560
this equation?

00:10:47.560 --> 00:10:51.960
On the right-hand side of this
equation, I have some linear

00:10:51.960 --> 00:10:59.140
combination of the basis vectors
of V complement You

00:10:59.140 --> 00:11:01.490
could put just put a minus all
along that, but that won't

00:11:01.490 --> 00:11:15.550
change the fact that this is
some linear combination of V

00:11:15.550 --> 00:11:17.570
complement's basis vectors.

00:11:21.130 --> 00:11:29.900
So this vector over here is
going to be a member of-- we

00:11:29.900 --> 00:11:31.960
could also call this x.

00:11:31.960 --> 00:11:34.800
So x is equal to this, but its
also going to be equal to

00:11:34.800 --> 00:11:37.110
this, and since it can be
represented as a linear

00:11:37.110 --> 00:11:42.650
combination of the orthogonal
complement of V's basis

00:11:42.650 --> 00:11:45.580
vector, or V perp's basis
vector, we know that this also

00:11:45.580 --> 00:11:47.730
has to be a member of V perp.

00:11:47.730 --> 00:11:49.260
Let me just review this,
because it can be

00:11:49.260 --> 00:11:50.980
a little bit confusing.

00:11:50.980 --> 00:11:53.110
I just set up this equation
right here.

00:11:53.110 --> 00:11:54.950
We know that there's at least
one solution-- all of the

00:11:54.950 --> 00:11:56.920
constants equalling 0.

00:11:56.920 --> 00:11:58.470
Anyone could do this.

00:11:58.470 --> 00:12:01.660
Now I subtracted all of the
yellow terms from both sides,

00:12:01.660 --> 00:12:03.310
and I got this equality.

00:12:03.310 --> 00:12:05.430
The left=hand side of this
equality is linear

00:12:05.430 --> 00:12:08.970
combinations of the basis
vectors of V.

00:12:08.970 --> 00:12:14.390
So any linear combinations of
the basis vectors of V is

00:12:14.390 --> 00:12:16.030
going to be a member of V.

00:12:16.030 --> 00:12:18.720
That's the definition
of basis vectors.

00:12:18.720 --> 00:12:24.770
So if I set x equalling to this
letf-hand side, I can say

00:12:24.770 --> 00:12:26.560
that x is a member of V.

00:12:26.560 --> 00:12:28.690
Well, if x is equal to the
left-hand side, it's also

00:12:28.690 --> 00:12:30.540
equal to the right-hand side.

00:12:30.540 --> 00:12:34.040
The right-hand side is some
linear combination of V perps,

00:12:34.040 --> 00:12:37.160
or the orthogonal complement
of V's basis vectors.

00:12:37.160 --> 00:12:40.980
Which tells us that x is also
a member of V perp.

00:12:40.980 --> 00:12:42.960
Well, what does that mean?

00:12:42.960 --> 00:12:46.710
That means that x must
be equal to 0.

00:12:46.710 --> 00:12:48.240
I just showed you at the
beginning of the video, the

00:12:48.240 --> 00:12:52.190
only vector that's a member of
a subspace and its complement

00:12:52.190 --> 00:12:54.030
is the 0 vector.

00:12:54.030 --> 00:13:00.340
So we know that because these
are orthogonal complements, we

00:13:00.340 --> 00:13:04.545
know that x must
be equal to 0.

00:13:14.300 --> 00:13:18.870
So just to reiterate, we know
0's has to equal both of these

00:13:18.870 --> 00:13:20.040
sides of the equation.

00:13:20.040 --> 00:13:21.920
And these are the same
constants that we

00:13:21.920 --> 00:13:23.560
had to begin with.

00:13:23.560 --> 00:13:25.570
But what do we know about
these two sets?

00:13:34.680 --> 00:13:37.120
We know that the 0 vector
has to be equal to this.

00:13:37.120 --> 00:13:42.790
That's the only vector in Rn
that's a member both of V and

00:13:42.790 --> 00:13:45.200
of the orthogonal
complement of V.

00:13:45.200 --> 00:13:49.380
Now, this is a 0 vector and we
have this linear combination

00:13:49.380 --> 00:13:54.210
of V's being set equaling
to the 0 vector.

00:13:54.210 --> 00:13:56.920
What do we know about
these constants?

00:13:56.920 --> 00:14:00.910
What does c1, c2, all the
way to ck have to be?

00:14:00.910 --> 00:14:05.640
We know that v1 through
vk is a basis for V.

00:14:05.640 --> 00:14:08.890
That tells us that they span V
and that they are linearly

00:14:08.890 --> 00:14:10.140
independent.

00:14:12.080 --> 00:14:15.030
Linear independence by
definition means that the only

00:14:15.030 --> 00:14:21.330
solution to this equation right
here is that all of the

00:14:21.330 --> 00:14:24.010
constants have to be 0.

00:14:24.010 --> 00:14:31.190
So linear independence tells
us that c1, c2, all the way

00:14:31.190 --> 00:14:33.785
through ck must be 0.

00:14:37.000 --> 00:14:39.130
All of these guys right
here are 0.

00:14:39.130 --> 00:14:40.790
Which is the same as
all of these guys.

00:14:40.790 --> 00:14:43.380
All of these guys must be 0.

00:14:43.380 --> 00:14:44.310
Now let's look at
the right-hand

00:14:44.310 --> 00:14:45.590
side of this equation.

00:14:45.590 --> 00:14:47.370
We could put the minus
all the way, but the

00:14:47.370 --> 00:14:49.080
same argument holds.

00:14:49.080 --> 00:14:52.480
This linear combination
of V perp's basis

00:14:52.480 --> 00:14:54.330
vectors is equal to 0.

00:14:54.330 --> 00:14:59.280
The only solution to this--
because each of these w1's,

00:14:59.280 --> 00:15:05.510
w2's, and wn minus k's are
linearly independent-- being

00:15:05.510 --> 00:15:07.860
equal to 0 is all of
the constants have

00:15:07.860 --> 00:15:10.280
to be equal to 0.

00:15:10.280 --> 00:15:14.130
That falls out of linear
independence.

00:15:14.130 --> 00:15:16.610
If this negative is confusing
you a bit, if it makes it look

00:15:16.610 --> 00:15:19.180
different than that, you could
just multiply this negative

00:15:19.180 --> 00:15:23.960
out and say minus d1 would have
to be equal to 0, minus

00:15:23.960 --> 00:15:26.910
d2 would have to be 0,
minus d and minus k

00:15:26.910 --> 00:15:28.170
would have to be 0.

00:15:28.170 --> 00:15:30.060
But it's the exact
same argument.

00:15:30.060 --> 00:15:32.800
Linear independence, which falls
out of the fact that

00:15:32.800 --> 00:15:36.790
this is a basis set, implies
that the only solution to this

00:15:36.790 --> 00:15:39.660
being equal to 0 is each
of the constants

00:15:39.660 --> 00:15:41.330
being equal to 0.

00:15:41.330 --> 00:15:46.700
Well, that means that d1,
d2, all the way to dn

00:15:46.700 --> 00:15:50.590
minus k must be 0.

00:15:50.590 --> 00:15:53.290
Let's go back to what
I wrote up here.

00:15:53.290 --> 00:15:55.690
This was the original equation
that we were

00:15:55.690 --> 00:15:58.080
experimenting with.

00:15:58.080 --> 00:16:00.640
Just manipulating this
equation a bit and

00:16:00.640 --> 00:16:03.885
understanding that the only
intersection between V and V

00:16:03.885 --> 00:16:04.950
perp is a 0 vector.

00:16:04.950 --> 00:16:09.890
And that you only have linear
independence if the only

00:16:09.890 --> 00:16:15.080
solution to these vectors
equalling 0 is all of their

00:16:15.080 --> 00:16:16.600
constants equalling 0.

00:16:16.600 --> 00:16:20.210
Then we know that all of these
terms, c1 through ck, d1

00:16:20.210 --> 00:16:23.750
through dn minus k, they all
have to be equal to 0.

00:16:23.750 --> 00:16:26.710
That's the only solution to this
larger equation that I

00:16:26.710 --> 00:16:27.950
wrote up here.

00:16:27.950 --> 00:16:30.620
Well, the only solution to this
large equation that I

00:16:30.620 --> 00:16:33.960
wrote up here is that all of the
constants are equal to 0.

00:16:33.960 --> 00:16:42.360
That implies that if I were it
take the set right here of v1,

00:16:42.360 --> 00:16:49.270
v2, all the way to vk, and I
were to augment that with the

00:16:49.270 --> 00:17:02.360
basis vectors of V perp, which
are w1, w2, all the way to wn

00:17:02.360 --> 00:17:07.050
minus k, that this is a linearly
independent set.

00:17:13.770 --> 00:17:17.480
And I know that because the only
solution to this equation

00:17:17.480 --> 00:17:20.720
is each of these constants
having to be equal to 0.

00:17:20.720 --> 00:17:22.700
That's what linear independence
means.

00:17:22.700 --> 00:17:24.900
This implies this.

00:17:24.900 --> 00:17:26.760
Linear independence
implies that.

00:17:26.760 --> 00:17:29.620
We used the fact that linear
independence implies that all

00:17:29.620 --> 00:17:34.100
of these equal 0 to get the fact
that c1 all the way to ck

00:17:34.100 --> 00:17:36.450
was equal to 0.

00:17:36.450 --> 00:17:39.080
And then we use it again when
we set this thing also being

00:17:39.080 --> 00:17:40.190
equal to the 0 vector.

00:17:40.190 --> 00:17:42.760
We knew that all of the d's
had to be equal to 0.

00:17:42.760 --> 00:17:44.930
I don't know if you remember,
the 0 vector came out from the

00:17:44.930 --> 00:17:46.980
fact that that was the
only vector that is a

00:17:46.980 --> 00:17:48.420
member of both sets.

00:17:48.420 --> 00:17:49.945
I know I'm being a little bit
repetitive, but I really want

00:17:49.945 --> 00:17:53.270
you to understand that this
proof isn't some type of

00:17:53.270 --> 00:17:54.200
circular proof.

00:17:54.200 --> 00:17:57.210
That we just wrote this
equation, we wondered about

00:17:57.210 --> 00:18:00.530
what the solution set is to it,
we rearranged it, we said

00:18:00.530 --> 00:18:04.950
hey both sides of this equation
are members of both V

00:18:04.950 --> 00:18:05.900
and V perp.

00:18:05.900 --> 00:18:08.085
The only vector that's
a member of

00:18:08.085 --> 00:18:09.210
both is the 0 vector.

00:18:09.210 --> 00:18:11.530
So both of these sides
of the equation have

00:18:11.530 --> 00:18:12.710
to be equal to 0.

00:18:12.710 --> 00:18:16.170
The only solution to that is all
of these constants being

00:18:16.170 --> 00:18:18.280
equal to 0, because
each of these are

00:18:18.280 --> 00:18:19.960
linearly independent sets.

00:18:19.960 --> 00:18:21.450
So therefore all of these
constants have

00:18:21.450 --> 00:18:22.590
to be equal to 0.

00:18:22.590 --> 00:18:25.490
And then this augmented set,
where if you combined all of

00:18:25.490 --> 00:18:28.350
the basis vectors, that is
going to be linearly

00:18:28.350 --> 00:18:29.380
independent.

00:18:29.380 --> 00:18:33.420
Now many, many, many, many, many
videos ago, we learned

00:18:33.420 --> 00:18:53.320
that if we have some subspace
with dimension n, and we have

00:18:53.320 --> 00:19:01.380
n linearly independent vectors
that are members of your

00:19:01.380 --> 00:19:16.010
subspace, then those n linearly
independent vectors,

00:19:16.010 --> 00:19:24.660
or the set of your n vectors,
is a basis for the subspace.

00:19:29.780 --> 00:19:34.900
Now Rn is a subspace
of itself.

00:19:34.900 --> 00:19:38.370
Rn is an n dimensional
subspace.

00:19:45.050 --> 00:19:50.430
We could write the dimension
of Rn is equal to n.

00:19:50.430 --> 00:19:56.150
Now we have n linearly
independent vectors in Rn.

00:19:56.150 --> 00:20:00.010
So that tells us that these
guys right here

00:20:00.010 --> 00:20:09.910
are a basis for Rn.

00:20:09.910 --> 00:20:13.240
We have n linearly independent
vectors.

00:20:13.240 --> 00:20:15.790
We have n minus k that are
coming from V perp.

00:20:15.790 --> 00:20:18.040
We have n that are coming
from V from their

00:20:18.040 --> 00:20:21.570
basis for those subspace.

00:20:21.570 --> 00:20:23.270
So now we have a total
of n vectors.

00:20:23.270 --> 00:20:24.570
They're linearly independent.

00:20:24.570 --> 00:20:25.780
They're all members of Rn.

00:20:25.780 --> 00:20:27.800
So they are a basis for Rn.

00:20:27.800 --> 00:20:31.050
Which tells us that any vector
in Rn can be represented by

00:20:31.050 --> 00:20:35.950
linear combinations of these
guys, which is fascinating.

00:20:35.950 --> 00:20:38.020
So this is a basis for Rn.

00:20:38.020 --> 00:20:43.060
So that tells us that we can
take any vector-- let's say a

00:20:43.060 --> 00:20:46.000
is a member of Rn,
some vector.

00:20:46.000 --> 00:20:50.030
That means since this is a basis
for Rn that a can be

00:20:50.030 --> 00:20:52.110
represented to some linear
combination of

00:20:52.110 --> 00:20:53.930
all of these guys.

00:20:53.930 --> 00:21:02.160
So it can be represented as C1
times V1 plus C2 times V2 all

00:21:02.160 --> 00:21:06.830
the way to plus Ck
times with Vk.

00:21:06.830 --> 00:21:08.860
Let me use a different letter
just to make sure that you

00:21:08.860 --> 00:21:10.550
understand that this is a
different equation that I'm

00:21:10.550 --> 00:21:12.680
writing than I wrote earlier
in the video.

00:21:12.680 --> 00:21:14.390
So I can write this, and then
I can have some other

00:21:14.390 --> 00:21:21.050
constants that say plus e1 times
our V perp basis vector

00:21:21.050 --> 00:21:28.360
1 plus e2 times this guy plus
all the way to en minus k

00:21:28.360 --> 00:21:32.570
times the n minus k basis
vector for V perp.

00:21:32.570 --> 00:21:35.860
I can represent any vector
in Rn this way.

00:21:35.860 --> 00:21:37.930
Or another way to say it.

00:21:37.930 --> 00:21:39.180
What is this?

00:21:41.440 --> 00:21:48.420
This is some vector that is a
member of our subspace V.

00:21:48.420 --> 00:21:56.600
And then this is some vector
over here that is a member of

00:21:56.600 --> 00:21:58.380
the orthogonal complement
of V.

00:21:58.380 --> 00:22:00.380
This is just a linear
combination of V

00:22:00.380 --> 00:22:02.070
perp's basis vectors.

00:22:02.070 --> 00:22:03.330
This is just a linear
combination

00:22:03.330 --> 00:22:05.180
of V's basis vectors.

00:22:05.180 --> 00:22:12.380
So given that all of these
characters are a basis for Rn

00:22:12.380 --> 00:22:15.520
tells us that any member of Rn
can be represented as a linear

00:22:15.520 --> 00:22:16.900
combination of them.

00:22:16.900 --> 00:22:23.000
But that means that any member
of Rn can be represented as a

00:22:23.000 --> 00:22:30.800
sum of a member of our subspace
V plus a member of

00:22:30.800 --> 00:22:32.640
your subspace V perp.

00:22:32.640 --> 00:22:36.330
This is a member of V, and this
is a member of V perp.

00:22:36.330 --> 00:22:38.470
And that's a really, really
interesting idea.

00:22:38.470 --> 00:22:43.210
You give me a subspace and then
we can figure out its

00:22:43.210 --> 00:22:44.550
orthogonal complement.

00:22:44.550 --> 00:22:51.290
Any other vector in Rn can be
represented as a combination,

00:22:51.290 --> 00:22:55.720
or sum, of some vector in our
subspace and some vector in

00:22:55.720 --> 00:22:58.110
its orthogonal complement.

00:22:58.110 --> 00:23:00.230
Now the next question you
might be asking, is this

00:23:00.230 --> 00:23:02.430
representation unique?

00:23:02.430 --> 00:23:03.680
So is this unique?

00:23:09.636 --> 00:23:13.325
Well let's test it out by
assuming it's not unique.

00:23:21.470 --> 00:23:25.960
So that means that I have for
some vector a that is a member

00:23:25.960 --> 00:23:28.650
of Rn, I can represent
it two ways.

00:23:28.650 --> 00:23:34.530
I can represent it as equalling
some member of my

00:23:34.530 --> 00:23:39.250
subspace V, plus some member
of the orthogonal

00:23:39.250 --> 00:23:40.110
complement of V.

00:23:40.110 --> 00:23:41.580
I can represent it that way.

00:23:41.580 --> 00:23:44.250
Or I could represent it as
some other member of my

00:23:44.250 --> 00:23:48.290
subspace V plus some other
member of my orthogonal

00:23:48.290 --> 00:23:50.310
complement.

00:23:50.310 --> 00:23:55.930
So x1, x2 are members
of V perp.

00:23:55.930 --> 00:24:01.050
And then v1 and v2
are members of V.

00:24:01.050 --> 00:24:02.990
If we assume it's not unique,
there's two ways

00:24:02.990 --> 00:24:03.830
that I could do this.

00:24:03.830 --> 00:24:06.130
And I'm representing it
as these two vectors.

00:24:06.130 --> 00:24:09.730
Now clearly this side of this
equation is equal to that.

00:24:09.730 --> 00:24:11.780
These are both representations
of a.

00:24:11.780 --> 00:24:13.820
So we can rearrange
this a little bit.

00:24:13.820 --> 00:24:18.040
We could say that v1 minus v2--
if I subtract v2 from

00:24:18.040 --> 00:24:24.300
both sides, I get v1 minus
v2 is equal to-- that's

00:24:24.300 --> 00:24:26.790
subtracting v2 from both sides,
and if I subtract x1

00:24:26.790 --> 00:24:31.065
from both sides-- x2 minus x1.

00:24:34.850 --> 00:24:38.650
These are both members
of the subspace V.

00:24:38.650 --> 00:24:41.890
And any subspace is closed
under addition and

00:24:41.890 --> 00:24:45.405
subtraction, which is really
a special case of addition.

00:24:50.980 --> 00:24:54.840
The vector v1, let me write it
this way, let me call my sum

00:24:54.840 --> 00:24:58.850
vector z, being equal to both of
these guys which are equal

00:24:58.850 --> 00:24:59.810
to each other.

00:24:59.810 --> 00:25:02.720
z is the vector V1 minus V2.

00:25:02.720 --> 00:25:09.310
Any subspace is closed under
addition, if you take two

00:25:09.310 --> 00:25:11.880
vectors and find their
difference in a subspace, then

00:25:11.880 --> 00:25:13.800
that resulting difference
is also

00:25:13.800 --> 00:25:16.520
going to be in a subspace.

00:25:16.520 --> 00:25:22.260
z is going to be a member
of our subspace V.

00:25:22.260 --> 00:25:25.360
This vector right here-- which
is also the same thing we just

00:25:25.360 --> 00:25:29.490
said that to be equal our vector
z-- is going to be a

00:25:29.490 --> 00:25:31.800
member of our V perp.

00:25:31.800 --> 00:25:32.170
Why?

00:25:32.170 --> 00:25:37.800
Because both x1 and x2 are
members of the subspace V's

00:25:37.800 --> 00:25:39.230
orthogonal complement.

00:25:39.230 --> 00:25:42.630
And that is a subspace
as well.

00:25:42.630 --> 00:25:45.580
So it is closed under addition
and subtraction.

00:25:45.580 --> 00:25:48.270
So this is also going to be
a member of your subspace.

00:25:48.270 --> 00:25:52.800
So we could also say that z is
a member of V perp or the

00:25:52.800 --> 00:25:55.600
orthogonal complement of V.

00:25:55.600 --> 00:25:57.680
Well, we've done this
multiple times.

00:25:57.680 --> 00:25:59.140
This was the first thing
we showed in the video.

00:25:59.140 --> 00:26:02.300
The only vector that's a member
of a subspace and its

00:26:02.300 --> 00:26:05.450
orthogonal complement
is the 0 vector.

00:26:05.450 --> 00:26:07.780
So z has to be equal
to the 0 vector.

00:26:07.780 --> 00:26:10.760
So this is equal to
the 0 vector.

00:26:10.760 --> 00:26:13.810
Well, if both of these are equal
to the 0 vector, we know

00:26:13.810 --> 00:26:18.030
that v1 minus v2 is equal to
the 0 vector, which implies

00:26:18.030 --> 00:26:21.120
that v1 must be equal to v2.

00:26:21.120 --> 00:26:27.060
And we also know that x2 minus
x1 is equal to the 0 vector.

00:26:27.060 --> 00:26:29.900
Or x2 is equal to x1.

00:26:29.900 --> 00:26:33.440
So we try to say that, hey,
there's two ways to construct

00:26:33.440 --> 00:26:37.500
some arbitrary vector
a that's in Rn.

00:26:37.500 --> 00:26:39.300
And we wrote that down.

00:26:39.300 --> 00:26:44.140
But then we found out that no,
v1 must be equal to v2 and x1

00:26:44.140 --> 00:26:45.720
must be equal x2.

00:26:45.720 --> 00:26:50.780
So there's only a unique way to
write any member of Rn as a

00:26:50.780 --> 00:26:56.440
sum of a vector that's in our
subspace V and a vector that

00:26:56.440 --> 00:26:59.880
is in the orthogonal
complement of V.


Slope and Y-intercept Intuition

WEBVTT
Kind: captions
Language: en

00:00:00.720 --> 00:00:03.540
On the Khan Academy web app,
which I need to work on a

00:00:03.540 --> 00:00:05.570
little bit more to make it a
little bit faster, they have

00:00:05.570 --> 00:00:08.590
this one module that's called
the graph of the line.

00:00:08.590 --> 00:00:11.690
It has no directions on it, and
I thought I would make a little

00:00:11.690 --> 00:00:14.800
video here, at least to explain
how to do this module, and in

00:00:14.800 --> 00:00:17.870
the process, I think it'll help
people, even those of you who

00:00:17.870 --> 00:00:21.820
aren't using the module,
understand what the slope and

00:00:21.820 --> 00:00:24.590
the y-intercept of a line
is a little bit better.

00:00:24.590 --> 00:00:27.380
So this is a screen shot of
that module right here, and the

00:00:27.380 --> 00:00:31.550
idea is essentially to change
this line, and this line right

00:00:31.550 --> 00:00:34.640
here in orange is the line
specified by this equation

00:00:34.640 --> 00:00:36.840
right here, so right now it's
the equation of the

00:00:36.840 --> 00:00:38.510
line 1x plus 1.

00:00:38.510 --> 00:00:42.450
It has a slope of 1, you can
see that, for every amount it

00:00:42.450 --> 00:00:46.210
moved to the right it moves up
exactly 1, and has 1

00:00:46.210 --> 00:00:47.620
for its y-intercept.

00:00:47.620 --> 00:00:52.130
It intersects the y-axis
at exactly the point 0,1.

00:00:52.130 --> 00:00:55.920
Now, the goal of this exercise
is to change your slope and

00:00:55.920 --> 00:00:59.410
your y-intercept so that you go
through these two points, and

00:00:59.410 --> 00:01:02.300
this point's-- half of it's off
the screen, hopefully you can

00:01:02.300 --> 00:01:04.940
see them if you're watching
these in HD-- you can

00:01:04.940 --> 00:01:06.730
see these two points.

00:01:06.730 --> 00:01:10.700
Our goal is to make this line
go through them by essentially

00:01:10.700 --> 00:01:12.210
changing its equation.

00:01:12.210 --> 00:01:18.110
So it's a kind of a tactile way
of-- you know, as tactile as

00:01:18.110 --> 00:01:21.010
something on the computer can
get-- of trying to figure out

00:01:21.010 --> 00:01:23.560
the equation of the line that
goes through these two points.

00:01:23.560 --> 00:01:24.890
So how can we do that?

00:01:24.890 --> 00:01:27.650
So you can see here, when I
change the slope, if I make

00:01:27.650 --> 00:01:30.270
the slope higher, it
becomes more steep.

00:01:30.270 --> 00:01:31.410
Now the slope is 3.

00:01:31.410 --> 00:01:35.090
For every 1 I move to the
right, I have to go 3 up.

00:01:35.090 --> 00:01:38.210
My change in y is 3 for
every change in x of 1.

00:01:38.210 --> 00:01:39.470
Or that's my slope.

00:01:39.470 --> 00:01:41.130
My y-intercept is still 1.

00:01:41.130 --> 00:01:44.260
If I change my y-intercept, if
I make it go down, notice it

00:01:44.260 --> 00:01:45.920
just shifts the line down.

00:01:45.920 --> 00:01:48.180
It doesn't change its
inclination or its slope, it

00:01:48.180 --> 00:01:50.720
just shifts it down along
this line right there.

00:01:50.720 --> 00:01:54.850
So how do I make my line go
through those two points?

00:01:54.850 --> 00:01:58.890
Well it looks like, if I shift
it up enough-- let's shift up

00:01:58.890 --> 00:02:01.210
that point-- and then let's
say let's lower the slope.

00:02:01.210 --> 00:02:03.050
This looks like it has
a negative slope.

00:02:03.050 --> 00:02:07.130
So if I lower my slope, notice
I'm flattening out the line.

00:02:07.130 --> 00:02:08.880
That's a slope of 0.

00:02:08.880 --> 00:02:11.480
It looks like it has to be
even more negative than that.

00:02:11.480 --> 00:02:14.200
Let's see, maybe even more
negative than that, right?

00:02:14.200 --> 00:02:17.360
It has to look like a line that
goes bam, just down like that.

00:02:17.360 --> 00:02:19.750
Even more-- that looks close.

00:02:19.750 --> 00:02:22.810
Let me get my y-intercept
down to see if I can

00:02:22.810 --> 00:02:26.110
get closer to that.

00:02:26.110 --> 00:02:28.940
It still seems like my slope
is a little bit too high.

00:02:28.940 --> 00:02:29.870
That looks better.

00:02:29.870 --> 00:02:32.950
So let me get my y-intercept
down even further.

00:02:32.950 --> 00:02:35.000
It's now intersecting way
here, off the screen.

00:02:35.000 --> 00:02:37.050
You can't even see that.

00:02:37.050 --> 00:02:39.480
I just realized this is
copyright 2008 Khan

00:02:39.480 --> 00:02:41.100
Academy, it's now 2009.

00:02:41.100 --> 00:02:42.310
It's almost near
the end of 2009.

00:02:42.310 --> 00:02:43.680
I could just change that.

00:02:43.680 --> 00:02:45.470
Maybe I'll just
write 2010 there.

00:02:45.470 --> 00:02:45.840
OK.

00:02:45.840 --> 00:02:47.280
So y-intercept.

00:02:47.280 --> 00:02:49.940
Even more.

00:02:49.940 --> 00:02:52.210
So I lowered the y-intercept
but our slope is still

00:02:52.210 --> 00:02:53.080
not strong enough.

00:02:53.080 --> 00:02:54.870
The y-intercept is
actually off the chart.

00:02:54.870 --> 00:02:57.200
It's intersecting at minus 18.

00:02:57.200 --> 00:02:58.450
That's our current y-intercept.

00:02:58.450 --> 00:03:00.730
But the slope of minus 5
is still not enough, so

00:03:00.730 --> 00:03:03.240
let me lower the slope.

00:03:03.240 --> 00:03:05.760
So if I lower the slope, let's
see, if I lower the y-intercept

00:03:05.760 --> 00:03:09.830
a little bit more,
is that getting me?

00:03:09.830 --> 00:03:10.650
There you go.

00:03:10.650 --> 00:03:11.790
It got me to those points.

00:03:11.790 --> 00:03:13.960
So the equation of the line
that passes through both

00:03:13.960 --> 00:03:17.970
of those things is
minus 6x minus 22.

00:03:17.970 --> 00:03:20.430
Let's do another one.

00:03:20.430 --> 00:03:23.740
So, once again, it resets it,
so I just say the equation 1x

00:03:23.740 --> 00:03:26.540
plus 1, but it gives me these
two new points that I have

00:03:26.540 --> 00:03:28.060
to make it go through.

00:03:28.060 --> 00:03:31.480
And once again this is going to
be a negative slope, because

00:03:31.480 --> 00:03:34.370
for every x that I move
forward positive, my y

00:03:34.370 --> 00:03:36.140
is actually going down.

00:03:36.140 --> 00:03:38.030
So I'm going to have a negative
slope here, so let me lower

00:03:38.030 --> 00:03:39.890
the slope a little bit.

00:03:42.530 --> 00:03:44.920
It's actually doing fractions,
so this thing jumps

00:03:44.920 --> 00:03:45.600
around a little bit.

00:03:45.600 --> 00:03:47.510
I should probably change
that a little bit.

00:03:47.510 --> 00:03:50.210
That looks about right, so let
me shift the graph down a

00:03:50.210 --> 00:03:54.170
little bit by lowering
its y-intercept.

00:03:54.170 --> 00:04:00.420
By lowering its y-intercept,
can I hit those two points?

00:04:00.420 --> 00:04:01.410
There you go.

00:04:01.410 --> 00:04:03.890
This is the equation of that
line that goes to the points

00:04:03.890 --> 00:04:08.670
minus 5,1 and the
points 9,minus 9.

00:04:08.670 --> 00:04:11.340
You have a slope of minus 5/7.

00:04:11.340 --> 00:04:15.140
For every 7 you go to the
right, you go down 5.

00:04:15.140 --> 00:04:18.980
If you go 1, 2, 3, 4, 5,
6, 7, you're going to

00:04:18.980 --> 00:04:22.840
go down 1, 2, 3, 4, 5.

00:04:22.840 --> 00:04:25.040
And that, we definitely
see that on that line.

00:04:25.040 --> 00:04:27.980
And then the y-intercept is
minus 18 over 7, which is a

00:04:27.980 --> 00:04:30.270
little over 2, it's about a
little over-- it's what,

00:04:30.270 --> 00:04:32.010
a little over 2 and 1/2.

00:04:32.010 --> 00:04:34.050
And we see right there
that the y-intercept is

00:04:34.050 --> 00:04:35.420
a little over 2 and 1/2.

00:04:35.420 --> 00:04:37.340
That's the equation
for our line.

00:04:37.340 --> 00:04:38.860
Let's do another one.

00:04:38.860 --> 00:04:41.170
This is a fun module, because
there are no wrong answers.

00:04:41.170 --> 00:04:43.580
You can just keep messing with
it until you eventually get

00:04:43.580 --> 00:04:46.570
that line to go through both of
those points, but the idea is

00:04:46.570 --> 00:04:49.670
really give you that intuition
that the slope is just what the

00:04:49.670 --> 00:04:53.560
inclination of the line is, and
then the y-intercept is how far

00:04:53.560 --> 00:04:54.900
up and down it gets shifted.

00:04:54.900 --> 00:04:57.110
So this is going to be
a positive slope, but

00:04:57.110 --> 00:04:58.490
not as high as 1.

00:04:58.490 --> 00:05:06.510
It looks like, for every 1, 2,
3, 4, 5, 6, 7, 8, 9, 10, 11,

00:05:06.510 --> 00:05:08.440
12, for every 12 we go to the
right, we're going

00:05:08.440 --> 00:05:11.810
to go 1, 2, 3 up.

00:05:11.810 --> 00:05:15.260
So our slope is going to be 3
over 12, which is also 1 over

00:05:15.260 --> 00:05:17.070
4, and we can just look
at that visually.

00:05:17.070 --> 00:05:18.680
Let's lower our slope.

00:05:18.680 --> 00:05:20.750
That's 3/4, not low enough.

00:05:20.750 --> 00:05:22.580
1/2, not low enough.

00:05:22.580 --> 00:05:26.060
1/4, which I just figured out
it is, that looks right, and

00:05:26.060 --> 00:05:29.020
then we have to lower
the y-intercept.

00:05:29.020 --> 00:05:32.470
We're shifting it down,
and there we go.

00:05:32.470 --> 00:05:36.840
So the equation of this line,
its slope is 1/4, so the

00:05:36.840 --> 00:05:40.150
equation of the line
is 1/4x plus 1/4.

00:05:40.150 --> 00:05:43.220
So hopefully, for those of you
trying to do this module, that,

00:05:43.220 --> 00:05:46.060
1, explained how to do it, and
for those of you who don't even

00:05:46.060 --> 00:05:47.980
know what this module is, it
hopefully gives you a little

00:05:47.980 --> 00:05:52.070
intuition about what the slope
and the y-intercept do

00:05:52.070 --> 00:05:54.040
to an actual line.


Area of a circle

WEBVTT
Kind: captions
Language: en

00:00:00.690 --> 00:00:05.610
So we've got a circle here--
doesn't look like a perfect

00:00:05.610 --> 00:00:09.690
circle, but we can use our
imaginations --and let's say

00:00:09.690 --> 00:00:21.780
it's got a radius of 3 meters.

00:00:21.780 --> 00:00:24.580
My question, or the question
we're going to answer in this

00:00:24.580 --> 00:00:29.020
video is what is the
area of this circle?

00:00:29.020 --> 00:00:31.900
And remember, the area is just
how much space this circle

00:00:31.900 --> 00:00:34.760
takes up on a surface, or on
this computer screen that

00:00:34.760 --> 00:00:36.285
you're watching, or on
this piece of paper.

00:00:39.330 --> 00:00:42.100
If this was a room, it's how
much carpeting you would need

00:00:42.100 --> 00:00:44.240
to fill out this circular room.

00:00:44.240 --> 00:00:45.800
That's what the area is.

00:00:48.860 --> 00:00:51.370
Now, I'm not going to prove it
to you, and we'll do that

00:00:51.370 --> 00:00:54.950
later, but the area for circle
just takes on a fairly

00:00:54.950 --> 00:00:58.520
straightforward formula and I
want to just get you used to

00:00:58.520 --> 00:00:59.610
applying that formula.

00:00:59.610 --> 00:01:04.860
So the area of a circle
is equal to pi.

00:01:04.860 --> 00:01:08.710
Remember, pi was that number
that people figured out was the

00:01:08.710 --> 00:01:11.750
ratio between the circumference
and the diameter of the circle.

00:01:11.750 --> 00:01:15.420
It's 3.14159, keeps
going on and on and on.

00:01:15.420 --> 00:01:17.880
It's just the number, but
it's a very magical number.

00:01:17.880 --> 00:01:19.765
Pi times the radius squared.

00:01:22.600 --> 00:01:26.810
In fact another way of defining
pi:-- you could even rewrite

00:01:26.810 --> 00:01:32.690
this right here --the area over
your radius squared- so

00:01:32.690 --> 00:01:33.650
this is your radius.

00:01:33.650 --> 00:01:37.110
If you multiply the radius
times itself you could imagine

00:01:37.110 --> 00:01:41.100
that would be the area of a
cube that's like that --that

00:01:41.100 --> 00:01:44.180
the ratio between the area of
this entire circle and the

00:01:44.180 --> 00:01:49.235
ratio of this cube right
here-- or this square.

00:01:49.235 --> 00:01:50.330
I shouldn't say a cube.

00:01:50.330 --> 00:01:54.640
Cube would be if we went into
3D --but the ratio of the area

00:01:54.640 --> 00:01:59.260
of the circle to this square
right here is also

00:01:59.260 --> 00:02:00.230
equal to pie.

00:02:00.230 --> 00:02:01.940
That could be actually
an alternate way of

00:02:01.940 --> 00:02:03.730
defining what pi is.

00:02:03.730 --> 00:02:06.790
And if you were to measure it
very carefully using-- there's

00:02:06.790 --> 00:02:10.230
thousands of methods you could
do it --you would get 3.14159

00:02:10.230 --> 00:02:12.550
and keep going on
and on and on.

00:02:12.550 --> 00:02:14.340
But we're not going to delve
too deeply into that.

00:02:14.340 --> 00:02:17.100
Maybe one day I'll make a
whole play list on pi.

00:02:17.100 --> 00:02:19.010
But we just need to know that
the area is equal to pi times

00:02:19.010 --> 00:02:21.560
r squared, so let's just
apply the numbers here.

00:02:21.560 --> 00:02:27.595
So in our example, the area is
equal to pi times 3 meters

00:02:27.595 --> 00:02:34.360
squared, which is equal to pi
times 9 meters squared, or the

00:02:34.360 --> 00:02:38.100
conventional way to write this
is, equal to 9pi

00:02:38.100 --> 00:02:39.770
meters squared.

00:02:39.770 --> 00:02:43.210
And remembered 9pi, the
convention is just to leave it

00:02:43.210 --> 00:02:48.640
that way, but this is the same
thing as 9 times 3.14159, which

00:02:48.640 --> 00:02:52.140
is probably going to be, like,
28 point something

00:02:52.140 --> 00:02:52.950
meters squared.

00:02:52.950 --> 00:02:56.410
Just remember, this is just
some number and it's not 9.

00:02:56.410 --> 00:02:59.780
It's actually closer to 28
because it's going to be

00:02:59.780 --> 00:03:04.050
9 times 3.14159, but we
just leave it like that.

00:03:04.050 --> 00:03:06.310
And that normally will be
good enough for you to

00:03:06.310 --> 00:03:08.660
say, hey that is my area.

00:03:08.660 --> 00:03:10.830
That's my area: 9pi.

00:03:10.830 --> 00:03:15.510
Now let's go the other way:
let's say I have a circle and

00:03:15.510 --> 00:03:20.560
let's say that someone would
say that the area

00:03:20.560 --> 00:03:24.080
is equal to 16pi.

00:03:24.080 --> 00:03:32.770
What is the diameter of
that circle going to be?

00:03:32.770 --> 00:03:34.840
Well, we know that area
is equal to pi times

00:03:34.840 --> 00:03:35.510
the radius squared.

00:03:35.510 --> 00:03:37.980
So at least let's
figure out the radius.

00:03:37.980 --> 00:03:44.750
So the area, 16pi, is equal to
pi times our radius squared.

00:03:44.750 --> 00:03:47.070
I'm just applying this formula.

00:03:47.070 --> 00:03:48.790
We're just going to keep
applying this formula

00:03:48.790 --> 00:03:50.540
over and over again when
we're dealing with area.

00:03:50.540 --> 00:03:54.760
So area, which we've been
told is 16pi, is equal to

00:03:54.760 --> 00:03:57.400
pi times radius squared.

00:03:57.400 --> 00:04:01.340
Now, if we divide both sides of
this equation by pi, we get

00:04:01.340 --> 00:04:04.390
16 is equal to r squared.

00:04:04.390 --> 00:04:06.110
And then you take the square
root of both sides and

00:04:06.110 --> 00:04:08.410
you get 4 is equal to r.

00:04:08.410 --> 00:04:10.860
I guess r could also be equal
to negative 4, but we're

00:04:10.860 --> 00:04:12.780
dealing with distances here;
you can't have a

00:04:12.780 --> 00:04:14.090
negative radius.

00:04:14.090 --> 00:04:17.000
Or at least in the world
we're living in right now.

00:04:17.000 --> 00:04:18.830
Just keep things simple;
we just want to keep our

00:04:18.830 --> 00:04:19.900
distances positives.

00:04:19.900 --> 00:04:23.820
So let's say that this
has a radius of 4.

00:04:23.820 --> 00:04:26.550
Now if the radius is 4,
what is its diameter?

00:04:26.550 --> 00:04:29.140
Well, the diameter is always
going to be 2 times the radius.

00:04:29.140 --> 00:04:31.480
So this 4, we're going to
have another 4 there.

00:04:31.480 --> 00:04:35.690
The diameter is equal to 8.

00:04:35.690 --> 00:04:38.970
Now let's do a slightly harder
one that will kind of compound

00:04:38.970 --> 00:04:42.060
some of the other things that
we've learned in the past.

00:04:42.060 --> 00:04:45.400
So let's say that I
have a circle here.

00:04:45.400 --> 00:04:59.650
Let's say that its
circumference is equal to 20pi,

00:04:59.650 --> 00:05:01.430
and I want to know its area.

00:05:04.590 --> 00:05:06.540
So way you do all these
problems is just figure out

00:05:06.540 --> 00:05:09.020
everything you can, given what
they've given you, and then

00:05:09.020 --> 00:05:12.010
maybe you can work out the
thing they're asking for.

00:05:12.010 --> 00:05:14.380
So if I know that the
circumference is 25, what do

00:05:14.380 --> 00:05:16.440
I know about its radius?

00:05:16.440 --> 00:05:19.280
Well, we saw in the last video
that the circumference is equal

00:05:19.280 --> 00:05:23.400
to 2pi times the radius.

00:05:23.400 --> 00:05:25.650
So if the circumference is
equal to 20pi, we could write

00:05:25.650 --> 00:05:30.530
that 20pi is the circumference
is equal to 2pi

00:05:30.530 --> 00:05:32.270
times the radius.

00:05:32.270 --> 00:05:36.600
Now if you divide both sides of
this by pi, those cancel out.

00:05:36.600 --> 00:05:39.830
Then if you divide both sides
by 2, this becomes a 1, this

00:05:39.830 --> 00:05:43.430
becomes a 10, or you get
the radius is equal to 10.

00:05:43.430 --> 00:05:44.600
Which makes sense, right?

00:05:44.600 --> 00:05:48.560
2pi times 10 is going
to be equal to 20pi.

00:05:48.560 --> 00:05:50.220
So we've figured
out our radius.

00:05:50.220 --> 00:05:57.280
Now, we know that the area is
equal to pi times r squared.

00:05:57.280 --> 00:05:59.860
And lucky for us, using the
circumference, we were able

00:05:59.860 --> 00:06:01.420
to figure out the radius.

00:06:01.420 --> 00:06:04.690
Now using the radius, we
can figure out the area.

00:06:04.690 --> 00:06:11.500
So the area is going to be
equal to pi times r squared,--

00:06:11.500 --> 00:06:17.040
r is 10 --times 10 squared,
which is equal to pi times 100.

00:06:17.040 --> 00:06:21.000
Or it's equal to 100pi.

00:06:21.000 --> 00:06:21.780
Just like that.

00:06:21.780 --> 00:06:25.350
so your circumference was 20pi,
when you went around the

00:06:25.350 --> 00:06:29.910
circle, but your area of
your circle is 100pi.

00:06:29.910 --> 00:06:34.090
And if I gave you units it
would be 100pi units squared.

00:06:34.090 --> 00:06:37.640
That is your area
right there: 100pi.

00:06:37.640 --> 00:06:41.170
Anyway, I think that's pretty
good initial exposure to

00:06:41.170 --> 00:06:42.360
the area of a circle.

00:06:42.360 --> 00:06:44.450
I'll see in the next video.