WEBVTT
00:00:05.760 --> 00:00:06.820
ANA RITA PIRES: Hi there.
00:00:06.820 --> 00:00:08.510
Welcome to recitation.
00:00:08.510 --> 00:00:10.990
In lecture, you've been
learning about vector
00:00:10.990 --> 00:00:15.040
spaces whose vectors are
actually matrices or functions,
00:00:15.040 --> 00:00:17.640
and this is what our
problem today is about.
00:00:17.640 --> 00:00:22.020
We have a set of 2 by 3
matrices whose null space
00:00:22.020 --> 00:00:24.250
contains the vector [2, 1, 1].
00:00:24.250 --> 00:00:27.610
And I want you to show
that this set is actually
00:00:27.610 --> 00:00:32.970
a vector subspace of the
space of all 2 by 3 matrices.
00:00:32.970 --> 00:00:36.590
And then, I want you
to find a basis for it.
00:00:36.590 --> 00:00:39.150
When you're done, here is
an additional question.
00:00:39.150 --> 00:00:44.500
What about the set of those 2
by 3 matrices whose column space
00:00:44.500 --> 00:00:47.760
contains the vector [2, 1]?
00:00:47.760 --> 00:00:48.430
All right.
00:00:48.430 --> 00:00:50.670
Hit pause and work
on it yourself,
00:00:50.670 --> 00:00:53.824
and when you're ready, I'll come
back and show you how I did it.
00:01:01.390 --> 00:01:01.890
Hi.
00:01:01.890 --> 00:01:05.090
I hope you managed to solve it.
00:01:05.090 --> 00:01:07.070
Let's do it.
00:01:07.070 --> 00:01:11.360
So, how do we show that
something is a vector subspace?
00:01:11.360 --> 00:01:13.920
Well, there are only two
things that we need to check.
00:01:13.920 --> 00:01:18.420
One is that if two vectors,
in this case two matrices,
00:01:18.420 --> 00:01:21.670
are in that space, then
their sum is in that space.
00:01:21.670 --> 00:01:24.640
And if you take a vector,
in this case a matrix,
00:01:24.640 --> 00:01:28.440
and you multiply by a scalar
you'll still be in the space.
00:01:28.440 --> 00:01:37.230
So, suppose that the matrices
A and B are in this set
00:01:37.230 --> 00:01:39.590
that we want to
prove is a subspace.
00:01:39.590 --> 00:01:46.570
So that means that A
times the vector [2, 1, 1]
00:01:46.570 --> 00:01:48.710
is equal to the vector [0, 0].
00:01:48.710 --> 00:01:52.050
Notice that the dimensions
are right: A is 2 by 3,
00:01:52.050 --> 00:01:53.260
so this 3 by 1.
00:01:53.260 --> 00:01:55.470
I should get a 2 by 1.
00:01:55.470 --> 00:01:59.170
Suppose that [2, 1, 1] is
in the null space of A,
00:01:59.170 --> 00:02:07.310
and that [2, 1, 1] is also in
the null space of B. Then what
00:02:07.310 --> 00:02:15.190
is A plus B times [2, 1, 1]?
00:02:15.190 --> 00:02:18.480
Is [2, 1, 1] in the
null space of A plus B?
00:02:18.480 --> 00:02:20.680
Well, if you think
about what this means,
00:02:20.680 --> 00:02:23.000
you're just adding
entry by entry.
00:02:23.000 --> 00:02:27.710
And you can do it
slowly on your own
00:02:27.710 --> 00:02:35.900
and just check that
this is what happens.
00:02:35.900 --> 00:02:38.530
But by now you are familiar
enough with matrices
00:02:38.530 --> 00:02:40.810
that this should
not be a surprise.
00:02:40.810 --> 00:02:43.110
Well, this is [0, 0],
and this is [0, 0],
00:02:43.110 --> 00:02:46.220
so their sum is [0, 0].
00:02:46.220 --> 00:02:50.230
So, indeed, [2, 1, 1] is in
the null space of A plus B.
00:02:50.230 --> 00:02:56.260
So if A and B are in the set, A
plus B is in the set, as well.
00:02:56.260 --> 00:02:58.040
Let's check the other thing.
00:02:58.040 --> 00:03:03.340
The other thing is A times
[2, 1, 1] is [0, 0].
00:03:03.340 --> 00:03:06.050
So A is in the set,
because [2, 1, 1]
00:03:06.050 --> 00:03:10.675
is in the null space of A. And
also, let's let c be a scalar.
00:03:13.550 --> 00:03:16.690
That just means
that c is a number.
00:03:16.690 --> 00:03:21.670
Then we want to
check that [2, 1,
00:03:21.670 --> 00:03:24.840
1] is in the null space
of the matrix c*A.
00:03:24.840 --> 00:03:27.950
That matrix is just the
matrix A except every entry is
00:03:27.950 --> 00:03:30.360
multiplied by the number c.
00:03:30.360 --> 00:03:35.790
Well, again, this is
how matrices work.
00:03:35.790 --> 00:03:37.770
You can just pull
out the constant
00:03:37.770 --> 00:03:40.400
and do A times [2, 1, 1] first.
00:03:40.400 --> 00:03:43.240
Now this is the vector [0, 0].
00:03:43.240 --> 00:03:48.270
So, this will simply be c
times [0, 0], which is [0, 0].
00:03:48.270 --> 00:03:53.370
So the matrix c*A is also
contained in this set.
00:03:53.370 --> 00:03:55.760
So the set is closed
under addition
00:03:55.760 --> 00:03:58.040
and under multiplication
by scalar,
00:03:58.040 --> 00:04:01.206
so the set is, indeed,
a vector subspace.
00:04:01.206 --> 00:04:05.200
Well that takes care of the
first part of the question.
00:04:05.200 --> 00:04:08.880
The second part was: find
a basis for the subspace.
00:04:08.880 --> 00:04:12.710
So let's work on that now.
00:04:12.710 --> 00:04:16.110
So the condition for a
matrix to be in this subspace
00:04:16.110 --> 00:04:20.760
is that the vector [2, 1, 1]
is in the null space.
00:04:20.760 --> 00:04:27.130
So I must have A times
[2, 1, 1] is equal to [0, 0].
00:04:27.130 --> 00:04:28.210
So what is happening?
00:04:28.210 --> 00:04:31.650
Well A is a 2 by 3 matrix.
00:04:31.650 --> 00:04:34.410
So you can actually think
about what happens in each row
00:04:34.410 --> 00:04:35.520
separately.
00:04:35.520 --> 00:04:40.450
You'll have the first row of A
times [2, 1, 1] is equal to 0.
00:04:40.450 --> 00:04:44.397
And the second row of A times
[2, 1, 1] is equal to 0.
00:04:44.397 --> 00:04:45.605
So let's see what that means.
00:04:49.230 --> 00:05:01.660
Each row of the matrix A in this
vector subspace must be [a, b,
00:05:01.660 --> 00:05:06.410
c] [2; 1; 1] equal to 0.
00:05:06.410 --> 00:05:10.220
This is not really a good
sentence, but you understand.
00:05:10.220 --> 00:05:15.030
Which means that--
well let's see-- 2a
00:05:15.030 --> 00:05:19.460
plus b plus c is equal to 0.
00:05:19.460 --> 00:05:21.565
So I can actually write
it in this format.
00:05:25.320 --> 00:05:34.810
It must be of the
form a, b, and then
00:05:34.810 --> 00:05:42.380
c must be equal to -2a minus b.
00:05:42.380 --> 00:05:43.810
Right?
00:05:43.810 --> 00:05:51.000
So, furthermore, we
can say that it-- well,
00:05:51.000 --> 00:05:54.570
one thing that you can do-- let
me write this here to help you.
00:05:54.570 --> 00:06:02.920
It will be [a, 0, -2a]
plus [0, b, -bl.
00:06:02.920 --> 00:06:03.420
See?
00:06:03.420 --> 00:06:06.080
So what I'm doing is
I'm splitting this
00:06:06.080 --> 00:06:09.220
into the linear
combination of two vectors.
00:06:09.220 --> 00:06:11.750
I can pull out the
a out of this one,
00:06:11.750 --> 00:06:16.340
and pull the b out
of this one, and it
00:06:16.340 --> 00:06:20.660
must be a linear
combination-- that's
00:06:20.660 --> 00:06:24.570
what this means, linear
combination-- of the following:
00:06:24.570 --> 00:06:31.460
[1, 0, -2] and [0, 1, -1].
00:06:31.460 --> 00:06:33.000
Does that make sense?
00:06:33.000 --> 00:06:36.820
So this is what each
row of a must satisfy.
00:06:36.820 --> 00:06:39.710
So we can now put
everything together
00:06:39.710 --> 00:06:43.720
into what a basis for the
vector space has to be.
00:06:43.720 --> 00:06:50.650
The basis will be, well
it's 2 by 3 matrix,
00:06:50.650 --> 00:06:53.650
and each row must be a linear
combination of these two
00:06:53.650 --> 00:06:54.820
vectors.
00:06:54.820 --> 00:06:59.860
So let's write that-- 1, 0, -2.
00:06:59.860 --> 00:07:04.740
I'll keep the second row
with zeros-- 0, 1, -1,
00:07:04.740 --> 00:07:06.865
and I'll keep the
second row of zeros.
00:07:06.865 --> 00:07:12.560
And now the same, but keeping
the first row with zeros,
00:07:12.560 --> 00:07:19.710
I'm taking these vectors
on the second row.
00:07:19.710 --> 00:07:23.640
So this is a basis
for my vector space.
00:07:23.640 --> 00:07:26.220
One, two, three,
four; that also means
00:07:26.220 --> 00:07:30.560
that the dimension
of the subspace is 4.
00:07:30.560 --> 00:07:32.760
There was one further
question, which
00:07:32.760 --> 00:07:38.600
was what can you say about
the set of those matrices that
00:07:38.600 --> 00:07:41.430
contain the vector [2, 1]
in their column space?
00:07:44.610 --> 00:07:48.180
What about the set of those 2
by 3 matrices whose column space
00:07:48.180 --> 00:07:50.147
contains the vector [2, 1].
00:07:50.147 --> 00:07:51.230
Is that a vector subspace?
00:07:54.600 --> 00:07:58.020
Well one quick check
that you can always do
00:07:58.020 --> 00:08:03.070
is check that the zero vector,
in this case the zero matrix,
00:08:03.070 --> 00:08:05.390
belongs to the set.
00:08:05.390 --> 00:08:13.180
Does the zero 2 by 3
matrix-- [0, 0, 0; 0, 0, 0].
00:08:13.180 --> 00:08:15.800
Does this matrix
belong to this set?
00:08:15.800 --> 00:08:18.840
Does this matrix contain
the vector [2, 1]
00:08:18.840 --> 00:08:20.540
in its column space?
00:08:20.540 --> 00:08:25.490
It does not, so this set
cannot be a vector subspace.
00:08:25.490 --> 00:08:29.210
If you want to think about how
this 0 belonging in the set
00:08:29.210 --> 00:08:32.570
has anything to do with the
two conditions being closed
00:08:32.570 --> 00:08:35.770
under sum and closed under
multiplication by scalar,
00:08:35.770 --> 00:08:38.179
simply think that
you should always
00:08:38.179 --> 00:08:41.590
be able to multiply a
matrix by the scalar 0
00:08:41.590 --> 00:08:44.210
and have it still be in the set.
00:08:44.210 --> 00:08:46.040
That will be your zero matrix.
00:08:46.040 --> 00:08:49.250
Well the zero matrix
is not in the set,
00:08:49.250 --> 00:08:51.722
so the set is not
a vector subspace.
00:08:51.722 --> 00:08:52.430
All right?
00:08:52.430 --> 00:08:53.310
We're done.
00:08:53.310 --> 00:08:54.881
Thank you.