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DAVID SHIROKOFF: Hi, everyone.

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So for this problem,
we're just going

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to take a look at computing some
eigenvalues and eigenvectors

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of several matrices.

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And this is just a review
problem for exam number three.

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So specifically, we're
given a projection

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matrix which has the form
of a a transpose divided

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by a transpose a, where
a is the vector 3 and 4.

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The second problem is
for a rotation matrix

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Q, which is the numbers 0.6,
negative 0.8, 0.8, and 0.6.

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And then the third one is
for a reflection matrix

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which is 2P minus the identity.

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So I'll let you work these out.

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And then I'll come
back in a second,

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and I'll fill in my solutions.

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Hi, everyone.

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Welcome back.

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OK, so for the
first problem, we're

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given a matrix P, which
is a projection matrix.

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And from earlier
on in the course,

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we probably already know that
the eigenvalues of a projection

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matrix are either 0 or 1.

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And I'll just recall,
how do you know that?

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Well if x is an
eigenvector of P,

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then it satisfies the
equation P*x equals lambda*x.

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But for a projection matrix,
P squared is equal to P.

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So if P is a projection,
we have P squared equals P.

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And specifically, what this
means is P squared x is equal

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to lambda*x.

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So we have P acting on P
of x is equal to lambda*x.

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And on the left-hand side, P*
is going to give me a lambda*x.

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P* again will give
me a lambda x.

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So we get lambda squared
x equals lambda*x.

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And if I bring everything
to the left-hand side,

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I get lambda times lambda
minus 1 x equals 0.

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And because x is not a zero
vector, what that means is

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lambda has to be either 0 or 1.

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So this is just a quick
proof that the eigenvalue

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of a projection matrix
is either 0 or 1.

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So we already know
that P is going

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to have eigenvalues of 0 or 1.

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Now specifically,
how do I identify

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which eigenvectors
correspond to 0

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and which eigenvectors
correspond to 1?

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Well, in this case, P
has a specific form,

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which is a times a transpose
divided by a transpose a.

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So I'll just write out
explicitly what this is.

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So a transpose a, 1
divided by a transpose a

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is going to be 9 plus
16 on the denominator.

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Then we're going to have
3 and 4 and 3 and 4.

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Now when we have a
matrix of this form,

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it's always going to be
the case that the vector

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a is going to be an
eigenvector with eigenvalue 1.

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So let's check.

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What is P acting on a?

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Well, we end up with: the
matrix P is 1/25 [3; 4] [3, 4].

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This is the matrix P. And if
we acted on the vector [3; 4],

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notice how this piece right
here, we can multiply out.

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This is going to be a transpose,
and this is going to be a.

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And if we multiply
these two pieces out,

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we get 25, which is exactly
the denominator a transpose a.

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So at the end of the day, we get
[3, 4]; Because the 25 divides

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out with the 25.

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Now note that this is
exactly what we started with.

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This is exactly a.

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So note here that the
vector a corresponds

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to an eigenvalue of 1.

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Meanwhile, for an
eigenvalue of 0,

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well, it always turns
out to be the case

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that if I take any vector
perpendicular to a,

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P acting on that vector
is going to be 0.

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So what's a vector,
which I'll call b,

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that's perpendicular to a?

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Well, note that a is
just a two by two vector.

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So that means there's only
going to be one direction that's

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perpendicular to a.

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Now just by eyeballing
it, I can see

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that a vector that's going
to be perpendicular to a

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is negative 4 and 3.

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So let's quickly check
that this is an eigenvector

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of P with eigenvalue 0.

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So what we need to show is that
P acting on this vector, b,

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is 0.

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So P acting on b is
going to be 1/25.

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It's going to be [3; 4]
[3, 4], multiplied by [4, 3].

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And note how when I multiply
out this row on this column,

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I get negative 3 times
4 plus 3 times 4,

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which is going to be 0.

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OK?

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So this shows that this vector
b has an eigenvalue of 0

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because note that we
can write this as 0*b.

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OK.

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For the second part, Q,
what are the eigenvectors

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and eigenvalues
of this matrix, Q?

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Well, Q is a rotation matrix.

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So I'll just write out Q again,
0.6, negative 0.8, 0.8, 0.6.

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So note that we can identify the
diagonal elements with a cosine

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of some angle theta.

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And we can associate
the off-diagonal parts

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as sine theta and
negative sine theta.

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And the reason we can
do that is because 0.6

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squared plus 0.8 squared is 1.

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So this is a rotation matrix.

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Now, to work out
the eigenvalues,

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I take a look at the
characteristic equation.

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So this is going
to give me, if I

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take a look at the
characteristic equation,

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it's going to be 0.6
minus lambda, squared.

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Then we have minus times
0.8 times negative 0.8.

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So that's going to
be plus 0.8 squared.

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And we want this to be 0.

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So if I rewrite this, I get
lambda is 0.6 plus or minus

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0.8i, where i is the
imaginary number.

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So notice how the eigenvalues
come in complex conjugate

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pairs.

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And this is always the case
when we have a real matrix.

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So we can find, first off, just
the eigenvalue that corresponds

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to 0.6 plus 0.8i.

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And then at the end,
we'll be able to find

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the second eigenvector by just
taking the complex conjugate

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of the first one.

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So let's compute
Q minus lambda*I.

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And if we have this acting
on some eigenvector u,

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we want this to be 0.

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Now Q minus lambda*I
is going to be,

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for the case lambda
is 0.6 plus 0.8i,

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this is going to give me
a quantity of minus 0.8i,

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minus 0.8, 0.8, and minus 0.8i.

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And I'm going to
write down components

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of u, which are u_1 and u_2.

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And we want this to vanish.

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And we note that the second
row is a constant multiple

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of the first row.

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Specifically, if I multiplied
this first row through by i,

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we would get negative i
squared, which is just 1.

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And then the second part
would be negative i,

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so we would just get the
second row back, which is good.

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So we just need to
find u_1, u_2 that are

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orthogonal to this first row.

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And again, just by inspection,
I can pick 1 and negative i.

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So note that that would give
me negative 0.8i plus 0.8i,

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and this vanishes.

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So this is the eigenvector that
corresponds to the eigenvalue

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lambda 0.6 plus 0.8i.

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In the meantime, if I take
the second eigenvalue,

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which is negative 0.8i, I
can take u which is just

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the complex conjugate
of this u up here.

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So it'll be 1, plus i.

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So this concludes the
eigenvalues and eigenvectors

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of this matrix Q.

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OK.

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Now lastly, number three, we're
looking at a reflection matrix

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which has the form 2P minus
I, where P is the same matrix

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that we had in part one.

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Now at first glance,
it looks like we

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might have to diagonalize
this entire matrix.

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However, note that
by shifting 2P by I,

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we only shift the eigenvalues.

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And we don't actually
change the eigenvectors.

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So note that this matrix
R, which is 2P minus I,

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it's going to have the
same eigenvectors as P.

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It's just going to have
different eigenvalues.

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So first off, we're going
to have one eigenvector.

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So the first eigenvector
is going to be a.

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So we have one
eigenvector which is a.

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So we have one
eigenvector which is a.

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And note that for the
vector a, it corresponds

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to the eigenvalue of 1.

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So what eigenvalue does
this correspond to?

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This is going to give
me a lambda which

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is 2 times 1 minus 1.

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So it's 1.

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So note that a, the
vector a, not only

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has an eigenvalue of 1 for P,
but it has an eigenvalue of 1

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for R as well.

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The second case was b.

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And remember that b has
an eigenvalue of 0 for P.

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So when we act R acting on b,
we'll have 2 times 0 minus 1 b.

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So this is going to
give us negative b.

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So the eigenvalue for b
is going to be negative 1.

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OK.

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And this is actually a general
case for reflection matrices,

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is that they typically
have eigenvalues

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of plus 1 or negative 1.

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OK, so we've just taken a
look at several matrices

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that come up in practice.

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We've looked at projection
matrices, reflection matrices,

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and rotation matrices.

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And we've seen a little
bit of the properties

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of their eigenvalues
and eigenvectors.

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So I'll just conclude here,
and good luck on your test.