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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.