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BEN HARRIS: Hi,
and welcome back.
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Today we're going to do
a problem about the four
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fundamental subspaces.
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So here we have a
matrix B. B is written
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as the product of a
lower triangular matrix
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and an upper triangular matrix.
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And we're going to find
a basis for, and compute
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the dimension of,
each of the four
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fundamental subspaces of B.
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I'll give you a minute to try
that on your own, to hit pause,
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and then I'll be right
back in just a minute
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and we can do it together.
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OK.
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We're back.
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Now, the first
thing to notice is
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that this is an LU
decomposition of B.
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So we have L here and U here.
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Now let's go one
space at a time.
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Let's start with
the column space.
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And first, let's just say what
the dimension of the column
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space is.
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OK, so let's look
at our U matrix.
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How many pivots do we have?
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We have two pivots, so the
column space has dimension 2.
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This is the number of pivots.
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Good.
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Now, how do we find a
basis for the column space?
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How do we find that basis?
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Well, in lecture,
Professor Strang
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had started with a matrix
B. He did elimination on it,
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and then he took the pivot
columns in the original matrix.
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And that's great.
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That works.
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You can also take the pivot
columns in the L matrix.
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You can see by multiplying
this out that it will amount
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to essentially the same thing.
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So a basis for
this column space,
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I can just take these two
pivot columns of my L matrix,
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[1, 2, -1] and [0, 1, 0].
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Good.
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OK.
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So there's the dimension
of and the basis
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for the column space of B.
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Next, let's do the
null space together.
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OK.
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What's the dimension
of the null space?
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Well, the dimension
of the null space
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is always the number of columns
minus the number of pivots.
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Right?
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It's the number
of free variables.
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So here, that's just one.
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Good.
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And how do we find this one
vector in the null space?
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Well, what we do
is we can just plug
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in 1 for our free
variable, and we can
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backsolve to get the other two.
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So this equation tells me
that my second number is -1,
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and this equation tells me
that that third variable -3/5,
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if I can fit it in here.
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That's a -3/5, if it's difficult
to see that on the tape.
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Now, let's move on,
next, to the row space.
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Next is the row space.
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So how do we find the
dimension of the row space?
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I'm going to write row space
as column space of B transpose.
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How do we find that?
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Well, remember that one of
our big facts in this class
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is that the dimension
of the row space
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is the same as the dimension
of the column space.
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It's just the number of pivots.
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So that's good.
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It's 2.
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And how do we find a
basis for the row space?
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There are a couple ways
of thinking about this.
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One way to think about it is:
we got this upper triangular
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matrix from B by
doing elimination.
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And elimination doesn't
change the row space.
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So I can just use the two
pivot rows of the matrix U.
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Basis for my row
space here is: I just
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put these two pivot
rows together,
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and I get a basis
for this row space.
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The last one is always the
toughest and the trickiest.
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We have to do the left null
space or the null space of B
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transpose.
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First, let's compute
its dimension.
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What's the dimension of
this left null space?
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Well, there's a similar
formula to the one
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we used when we were
computing the dimension
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of the null space.
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It's just the number of rows
minus the number of pivots.
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So there are three rows.
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Our matrix is 3 by 3.
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And there are two pivots.
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So this is just
one-dimensional, again.
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We need to compute, now,
this left null space.
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Let me go back to
our original matrix.
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The way to do this is to take
B equals L*U, and invert L,
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and get E*B equals U. So we need
to move L over to the left-hand
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side.
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If we do that-- I'm just
going to write that down here.
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So what's the inverse
of the L matrix?
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We just get 1, -2,
1, 0, 1, 1, times B,
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is our U matrix, this
upper triangular matrix.
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Now that I moved L
over to the other side,
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I can read off the vectors
in my left null space.
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Now, I'm looking at
not my pivot variables
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but my free variables, because
it's some sort of null space.
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but I want to look
at this E matrix.
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So the third row of this
E matrix, the third row
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corresponds to
the three row here
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and when I multiply this
by B, I just get zeros,
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so this is in the
left null space.
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A basis for this
left null space is--
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see if I can fit it here--
just this [1, 0, 1].
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Good.
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So we've found the dimension of
and basis for all of the four
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fundamental subspaces.
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Before I move on, I
just want to recall
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which of the L matrix
or the U matrix
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we used for each
of these subspaces.
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So for the column space,
we used the pivot columns
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of the L matrix.
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For the null space, we
looked at the U matrix.
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For the row space, we also
looked at the U matrix.
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And for the left null space, we
needed to invert the L matrix
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and look at the free row.
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We're done with the problem.
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But the last thing
that's useful is
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to draw a picture,
which I have right here.
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I know in lecture
Professor Strang
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has drawn you some sort of
cartoon pictures of what
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these subspaces look like.
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But here I want to try
to actually draw them
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in a special case.
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So if you can read my drawing
here, what do we have?
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We have the row space here,
and the null space here.
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Right?
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And so B maps this
picture into this picture.
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The null space here--
all the scalar multiples
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of this vector-- all go
to 0, because they're
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in the null space.
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That's exactly
what B takes to 0.
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B takes everything else,
including the row space,
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into this column space.
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And what does B transpose do?
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Well, B transpose kills
this left null space,
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kills this vector,
and it take everything
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else into the row
space, into the column
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space of B transpose.
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OK.
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Thanks for doing this
exercise together.
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I hope this picture is helpful.