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ANA RITA PIRES: Hi there.
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Welcome to recitation.
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In lecture, you've been
learning about vector
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spaces whose vectors are
actually matrices or functions,
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and this is what our
problem today is about.
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We have a set of 2 by 3
matrices whose null space
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contains the vector [2, 1, 1].
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And I want you to show
that this set is actually
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a vector subspace of the
space of all 2 by 3 matrices.
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And then, I want you
to find a basis for it.
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When you're done, here is
an additional question.
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What about the set of those 2
by 3 matrices whose column space
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contains the vector [2, 1]?
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All right.
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Hit pause and work
on it yourself,
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and when you're ready, I'll come
back and show you how I did it.
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Hi.
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I hope you managed to solve it.
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Let's do it.
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So, how do we show that
something is a vector subspace?
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Well, there are only two
things that we need to check.
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One is that if two vectors,
in this case two matrices,
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are in that space, then
their sum is in that space.
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And if you take a vector,
in this case a matrix,
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and you multiply by a scalar
you'll still be in the space.
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So, suppose that the matrices
A and B are in this set
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that we want to
prove is a subspace.
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So that means that A
times the vector [2, 1, 1]
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is equal to the vector [0, 0].
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Notice that the dimensions
are right: A is 2 by 3,
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so this 3 by 1.
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I should get a 2 by 1.
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Suppose that [2, 1, 1] is
in the null space of A,
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and that [2, 1, 1] is also in
the null space of B. Then what
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is A plus B times [2, 1, 1]?
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Is [2, 1, 1] in the
null space of A plus B?
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Well, if you think
about what this means,
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you're just adding
entry by entry.
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And you can do it
slowly on your own
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and just check that
this is what happens.
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But by now you are familiar
enough with matrices
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that this should
not be a surprise.
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Well, this is [0, 0],
and this is [0, 0],
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so their sum is [0, 0].
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So, indeed, [2, 1, 1] is in
the null space of A plus B.
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So if A and B are in the set, A
plus B is in the set, as well.
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Let's check the other thing.
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The other thing is A times
[2, 1, 1] is [0, 0].
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So A is in the set,
because [2, 1, 1]
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is in the null space of A. And
also, let's let c be a scalar.
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That just means
that c is a number.
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Then we want to
check that [2, 1,
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1] is in the null space
of the matrix c*A.
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That matrix is just the
matrix A except every entry is
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multiplied by the number c.
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Well, again, this is
how matrices work.
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You can just pull
out the constant
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and do A times [2, 1, 1] first.
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Now this is the vector [0, 0].
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So, this will simply be c
times [0, 0], which is [0, 0].
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So the matrix c*A is also
contained in this set.
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So the set is closed
under addition
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and under multiplication
by scalar,
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so the set is, indeed,
a vector subspace.
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Well that takes care of the
first part of the question.
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The second part was: find
a basis for the subspace.
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So let's work on that now.
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So the condition for a
matrix to be in this subspace
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is that the vector [2, 1, 1]
is in the null space.
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So I must have A times
[2, 1, 1] is equal to [0, 0].
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So what is happening?
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Well A is a 2 by 3 matrix.
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So you can actually think
about what happens in each row
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separately.
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You'll have the first row of A
times [2, 1, 1] is equal to 0.
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And the second row of A times
[2, 1, 1] is equal to 0.
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So let's see what that means.
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Each row of the matrix A in this
vector subspace must be [a, b,
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c] [2; 1; 1] equal to 0.
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This is not really a good
sentence, but you understand.
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Which means that--
well let's see-- 2a
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plus b plus c is equal to 0.
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So I can actually write
it in this format.
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It must be of the
form a, b, and then
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c must be equal to -2a minus b.
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Right?
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So, furthermore, we
can say that it-- well,
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one thing that you can do-- let
me write this here to help you.
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It will be [a, 0, -2a]
plus [0, b, -bl.
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See?
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So what I'm doing is
I'm splitting this
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into the linear
combination of two vectors.
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I can pull out the
a out of this one,
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and pull the b out
of this one, and it
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must be a linear
combination-- that's
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what this means, linear
combination-- of the following:
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[1, 0, -2] and [0, 1, -1].
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Does that make sense?
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So this is what each
row of a must satisfy.
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So we can now put
everything together
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into what a basis for the
vector space has to be.
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The basis will be, well
it's 2 by 3 matrix,
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and each row must be a linear
combination of these two
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vectors.
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So let's write that-- 1, 0, -2.
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I'll keep the second row
with zeros-- 0, 1, -1,
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and I'll keep the
second row of zeros.
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And now the same, but keeping
the first row with zeros,
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I'm taking these vectors
on the second row.
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So this is a basis
for my vector space.
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One, two, three,
four; that also means
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that the dimension
of the subspace is 4.
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There was one further
question, which
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was what can you say about
the set of those matrices that
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contain the vector [2, 1]
in their column space?
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What about the set of those 2
by 3 matrices whose column space
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contains the vector [2, 1].
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Is that a vector subspace?
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Well one quick check
that you can always do
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is check that the zero vector,
in this case the zero matrix,
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belongs to the set.
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Does the zero 2 by 3
matrix-- [0, 0, 0; 0, 0, 0].
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Does this matrix
belong to this set?
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Does this matrix contain
the vector [2, 1]
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in its column space?
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It does not, so this set
cannot be a vector subspace.
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If you want to think about how
this 0 belonging in the set
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has anything to do with the
two conditions being closed
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under sum and closed under
multiplication by scalar,
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simply think that
you should always
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be able to multiply a
matrix by the scalar 0
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and have it still be in the set.
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That will be your zero matrix.
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Well the zero matrix
is not in the set,
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so the set is not
a vector subspace.
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All right?
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We're done.
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Thank you.