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PROFESSOR: Hi, guys.
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My name is Nikola,
and in this video,
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we're going to
work out an example
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of an orthogonal
projection matrix.
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Specifically, we are gonna
compute the projection matrix
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onto the plane given by the
equation x plus y minus z
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equals 0.
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So before we start,
let me just recall
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what a projection matrix is.
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So you've seen this sketch
here a million times already.
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A projection matrix takes any
vector in three-space-- well,
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just in this case, we are
dealing with a three-space--
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and projects it
down onto the plane,
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a two-dimensional
subspace of R^3.
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So I'll give you a few
moments to consider
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the problem for yourselves.
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And then you'll
see my take on it.
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Hi again.
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So in lecture, Professor Strang
derived, in meticulous detail,
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the formula for the
projection matrix.
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So it was given by the
following slightly complicated
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expression.
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It's A times A
transpose A inverse A
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transpose where A
is a matrix that
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somehow encodes the subspace
we're projecting on.
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In particular, A has, as
its columns, a_1, a_2,
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I'm going to denote them--
a basis for the plane we're
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projecting on.
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So essentially,
what we need to do
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is find two such vectors
that span the plane and start
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computing with a matrix.
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This is fairly straightforward.
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One choice that works,
for example, is [1, -1, 0]
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for the first column, and [1, 0,
1] for the second column.
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And let me write out the
matrix A. So in the formula,
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the slightly more
complicated combination
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is A transpose A inverse, so let
me compute that first for you.
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So A transpose A
is a 2 by 2 matrix.
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And it's not so hard to figure
out that it's 2 and 1; 1, 2.
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Now we shall invert it
using the familiar formula.
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1 over the determinant.
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2 times 2 minus 1 is 3.
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And so we switch the
diagonal entries,
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and we flip the signs of
the off-diagonal ones.
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Right.
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And therefore, projection
matrix is given by the following
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product of matrices:
...1/3, [2, -1; -1,
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2] and then transpose of A,
which is [1, -1, 0; 1, 0, 1].
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I'm gonna carry out
this multiplication
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in inhumanly fast fashion.
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So I'm just gonna write down
the answer, which is 1/3
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[2, -1, 1; -1, 2, 2; 1, 1, 1].
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So what you can
do now is-- well,
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you can check whether this
answer actually makes sense.
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One thing you can
do is just-- well,
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a projection matrix is
supposed to take the normal
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to the plane down to 0.
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So you can just multiply P and
the normal vector [1, 1, -1].
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And if you get 0, maybe
you've done a good job.
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Another curious thing that I
would like to point out here
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is: so you see we had
lots of freedom choosing
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the matrix A. We could have
chosen any two columns that
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span the subspace,
that spans the plane.
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The beautiful thing about
it is that in the end,
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we'll get the same answer.
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So I hope there
will be many of you
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who would say, hey, there is an
easier way to do the problem.
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And I'll agree
with these people.
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So let's see what would
be an easier approach.
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Well, let's go back
to the sketch here.
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And let's make the
following observation,
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that any vector is a
sum of two components.
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The first component is its
projection onto the plane.
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And the other component
is its projection
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onto the orthogonal complement
of the plane, in this case,
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onto the normal vector
through the plane.
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So in the language
of linear algebra,
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this is just b equals
to its projection
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onto the plane plus
its projection-- I'm
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gonna call it P_N-- onto
the orthogonal complement
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of the plane.
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I'm gonna suggestively
write here the identity
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matrix so that you
can immediately
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read off a matrix equality.
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Associated with this
equality here, it's
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the identity equals P plus P_N.
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And therefore, the
projection matrix
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is just the identity minus
the projection matrix
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onto the normal vector.
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Now, this object here, P_N, is
much easier to compute, well,
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for two reasons.
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First one is that projecting
onto a one-dimensional subspace
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is infinitely easier
than projecting
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onto a higher-dimensional
subspace.
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And second, we already
have-- well, immediately we
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can read off from the
equation of the plane what
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the normal vector is.
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So we don't have
derive these guys.
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We don't have to do
what we did here.
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So essentially, P_N will be
N N transpose N inverse N
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transpose.
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And that's equal to [1, 1, -1]--
N transpose N inverse,
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this is just a number.
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It's 1 over the magnitude
of the normal vector,
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so that's-- then, the
magnitude squared, so that's 3.
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And 1, 1, -1.
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I'm gonna write the answer here.
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It's 1/3, 1, 1, -1; 1,
1, -1; and -1, 1, 1.
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And in order to get the
projection matrix-- yes?
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AUDIENCE: [INAUDIBLE].
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PROFESSOR: Oh.
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Thank you.
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Thank you.
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And in order to get
the projection matrix,
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we just subtract from the
identity this expression.
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And you can confirm that it's--
we get the same answer as here.
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I think we're done here.