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Let's say I've got some subspace
V, which tends to be
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our favorite letter for
subspaces, and it's equal to
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the span of two vectors in R4.
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Let's say that the first vector
is 1 0 0 1, and the
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second vector is 0 1 0 1.
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That is my subspace V.
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And you can see that these
are going to be a basis.
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That these are linearly
independent.
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Two vectors that are linear-- or
any set of vectors that are
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linearly independent and that
span a subspace are a basis
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for that subspace.
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You can see they are linearly
independent.
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This guy's has a 1 here.
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There's no way you can take some
combination of this guy
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to somehow get a 1 there.
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And this guy has a 1 here.
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There's no way you can get some
linear combination of
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these zeroes here
to a 1 there, so
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they're linearly dependent.
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You can also call this
a basis for V.
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Now, given that, let's see
if we can find out the
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transformation matrix for the
projection of any arbitrary
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vector onto this subspace.
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So let's say that X-- we're
dealing in R4 here, right?
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Let's say that x is a member
of R4, and I want to figure
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out a transformation
matrix for the
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projection onto V of x.
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Now, in the last video, we came
up with a general way to
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figure this out.
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We said if A is a transformation
matrix-- sorry.
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If A is a matrix who's columns
are the basis for the
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subspace, so let's say A is
equal to 1 0 0 1, 0 1 0 1.
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So A is a matrix whose columns
are the basis for our
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subspace, then the projection
of x onto V would be equal
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to-- and this is kind of hard.
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The first time you look at, it
gives you a headache, but
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there's a certain pattern or
symmetry or a way of-- you
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could say it's A times, you're
gonna have something in the
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middle, and then you have A
transpose times your vector x.
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And the way I remember it is in
the middle, you have these
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two guys switched around.
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So then you have A transpose
A, and you take
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the inverse of it.
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You probably won't be using
this in your everyday life
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five or ten years from now, so
it's OK if you don't memorize
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it, but temporarily, put this
in your medium-term memory
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because it's a good thing
to know for doing these
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projection problems.
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So if we want to find the
general matrix for this
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transformation, we just have to
determine what this matrix
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is equal to, and that's just a
bunch of matrix operations.
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So that's A.
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What is A transpose?
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A transpose is going to be equal
to just all the rows
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turn into columns.
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So the first column becomes
the first row.
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So it becomes 1 0 0 1.
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The second column becomes
the second row 0 1 0 1.
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That's what A transpose is.
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Now, what's A transpose A?
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To figure out that, I want to
figure out what A transpose
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times A is.
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So let me multiply A
transpose times A.
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So I'll rewrite A right here.
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1 0 0 1, 0 1 0 1.
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This is giving us some
good practice on
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matrix-matrix products.
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This is going to be
equal to what?
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Well, first of all, this is
a 2-by-4 matrix, and I'm
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multiplying it by a 4-by-2
matrix, so it's going to be a
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2-by-2 matrix.
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So the first entry is
essentially the dot product of
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that row with that column.
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So it's 1 times 1 plus 0
times 0 plus 0 times 0
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plus 1 times 1.
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So it's just going to be 2 for
that first entry right there.
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And then you take the dot
product of this guy with this
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guy right here.
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So it's 1 times 0, which is 0,
plus 0 times 1, which is 0,
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plus 0 time 0, which is 0, plus
1 times 1, which is 1.
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Now, we do this guy dotted with
this column right there.
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0 times 1 is 0 plus 1 time 0 is
0 plus 0 times 0 is 0 plus
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1 times 1 is 1.
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And then finally,
this row dotted
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with this second column.
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Second row, second column.
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0 times 0 is 0, 1 times 1
is 1, 0 times 0 is 0,
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1 times 1 is 1.
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So we have 1 times
1 plus 1 times 1.
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It's going to be 2.
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It's going to be equal to 2.
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So this right here
is A transpose A.
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But that's not good enough.
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We need to figure out what the
inverse of A transpose A is.
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This is A transpose A.
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But we need to figure out
A transpose A inverse.
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So what's the inverse of this?
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So let me write it here.
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The inverse A transpose
A inverse is going
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to be equal to what?
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It's 1 over the determinant
of this guy.
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What's the determinant here?
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It's going to be 1 over the
determinant of this.
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The determinant is 2 times 2,
which is 4, minus 1 times 1.
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So it's 4 minus 1, which is 3.
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So 1 over the determinant times
this guy, where if I
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swap these two, so I swap the
1's-- sorry, I swap the 2's.
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So this 2 goes here, and then
this orange 2 goes over here.
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And then I make these
1's negative.
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This becomes a minus 1 and
this becomes a minus 1.
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We learned that this is a
general solution for the
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inverse of a 2-by-2 matrix.
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I think it was 10 or 11 videos
ago, and you probably learned
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this in your Algebra II class,
frankly, but there you go.
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We have A transpose A inverse.
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So we have this guy.
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We have this whole guy here
is just this matrix.
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I could multiply the 1/3 into
it, but I don't have to do
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that just yet.
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But let's figure out the
whole matrix now.
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The whole A times this
guy, A transpose A
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inverse times A transpose.
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Let me write it this way.
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So the projection onto the
subspace V of x is going to be
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equal to A.
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1 0 0 1-- let me write a little
bit bigger like this.
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So 1 0 0 1, 0 1 0 1 times A
transpose A inverse, right?
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A times A transpose A inverse,
which is this guy right here.
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Let's just put the 1/3 out
front just because
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that's just a scalar.
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I'll put the 1/3 out front
times this guy.
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This A transpose A inverse
is 1/3 times 2
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minus 1, minus 1, 2.
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And then I'm going to multiply
it times A transpose.
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And all that times
our vector x.
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So A transpose is right there.
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It is 1 0 0 1, 0 1 0 1.
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And then all of that's going
to be times your vector x.
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So we still have some
nice matrix-matrix
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products ahead of us.
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Let see if we can do these.
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So the first one, let's just
multiply these two guys.
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I don't think there's any
simple way to do it.
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This is a 2-by-2 matrix and this
is a 2-by-4 matrix, so
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when I multiply them,
I'm going to end up
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with a 2-by-4 matrix.
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Let me write that 2-by-4
matrix right here.
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And then I can write this
guy right here.
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1 0 0 1, 0 1 0 1.
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And then I have the 1/3 that
was from A transpose A
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inverse, but I put the scaling
factor out there.
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And all of this is equal to the
projection of x onto V.
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So let's do this product.
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So this first entry is going to
be 2 times 1 plus minus 1
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times 0, so that is just 2.
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Then you're going to have 2
times 0 plus minus 1 times 1.
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Well, that's minus 1.
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Then you have 2 times 0
plus minus 1 times 0.
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Well, that's just 0.
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And then you're going to
have 2 times 1 plus
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minus 1 times 1.
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That's 2 minus 1.
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That's just 1, right?
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2 times 1 plus minus
1 times 1.
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Fair enough.
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Now, let's do the second row.
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Minus 1 times 1 plus 2 times
0, so that's just minus 1.
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Minus 1 times 0 plus
2 times 1.
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Well, that's just 2.
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Minus 1 times 0 plus
2 times 0.
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That's just 0.
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Minus 1 times 1 plus
2 times 1.
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Well, that's minus 1 plus
2, so that is 1.
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Almost there, and, of course, we
have to multiply it times x
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at the end.
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That's what the transformation
is.
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But this right here is our
transformation matrix.
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One more left to do.
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Let's hope I haven't made any
careless mistakes and that I
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won't make any when doing
this product.
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This is going to be a little
more complicated because this
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is a 4 by 2 times a 2 by 4.
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I'm going to end up with
a 4-by-4 matrix.
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Let me give myself some
breathing room here because
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I'm going to generate a 4-by-4
matrix right there.
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And so what am I going to get?
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So this first entry is going
to be 1 times 2 plus
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0 times minus 1.
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So it's just going
to be equal to 2.
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The next entry: 1 times-- this
row times any column here is
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just going to be the first entry
in the column because it
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gets zeroed out.
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So 1 times 2 plus 0 times
minus 1 is just 2.
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1 times minus 1 plus 0 times
2 is just minus 1.
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1 times 0 plus 0 times 0 is 0.
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1 times 1 plus 0 times
1 is just 1.
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When you take this row and you
multiply it times these
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columns, you literally just
got your first row there.
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Now, let's do this row
times these columns.
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Now, you've got a 0 here, so
you're going to have a 0 times
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the first entry of all
of these and a 1
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times the second one.
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So 0 times 2 plus 1 times
minus 1 is minus 1.
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0 times minus 1 plus
1 times 2 is 2.
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You're just going to get
the second row here.
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2 0 1.
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That actually makes sense,
because if you just look at
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this part of the matrix, it's
the 2-by-2 identity matrix.
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So, anyway, that's a little hint
why this looks very much
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like that, but we're just
going to go through this
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matrix product.
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Now, you multiply this-- let me
do it in a different color.
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You multiply this guy times
each of these columns.
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That guy dotted with that is
just going to 0 because this
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guy's essentially the 0 row
vector, so you're just going
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to get a bunch of zeroes.
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And then, finally, this last
row, it's 1 times the first
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entry plus 1 times
the second entry.
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So this guy's going to be 2
plus minus 1, which is 1.
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Minus 1 plus 2, which is 1.
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0 plus 0, which is 0.
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And then 1 plus 1, which is 2.
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And all that times x.
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And there you have it.
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This is exciting!
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The projection onto V of x
is equal to this whole
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matrix times x.
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So this thing right here, I
could multiply the 1/3 into
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it, but we don't have
to do that.
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That'll just make it a little
bit more messy.
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This thing right here is the
transformation matrix.
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As you can see, since we're
transforming-- remember, this
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projection onto V,
this is a linear
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transformation from R4 to R4.
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You give me some member of R4,
and I'll give you another
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member of R4 that's in my
subspace that is the
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projection.
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So this is going to be
a 4-by-4 You can
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see it right there.
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Anyway, hopefully, you found
that useful to actually see a
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tangible result.
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R4 is very abstract, so this
would even be beyond our
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three-dimensional programming
example.
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We're dealing with a more
abstract data set where we're
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interested in finding
a projection.