DAVID SHIROKOFF: Hi, everyone.
So for this problem,
we're just going
to take a look at computing some
eigenvalues and eigenvectors
of several matrices.
And this is just a review
problem for exam number three.
So specifically, we're
given a projection
matrix which has the form
of a a transpose divided
by a transpose a, where
a is the vector 3 and 4.
The second problem is
for a rotation matrix
Q, which is the numbers 0.6,
negative 0.8, 0.8, and 0.6.
And then the third one is
for a reflection matrix
which is 2P minus the identity.
So I'll let you work these out.
And then I'll come
back in a second,
and I'll fill in my solutions.
Hi, everyone.
Welcome back.
OK, so for the
first problem, we're
given a matrix P, which
is a projection matrix.
And from earlier
on in the course,
we probably already know that
the eigenvalues of a projection
matrix are either 0 or 1.
And I'll just recall,
how do you know that?
Well if x is an
eigenvector of P,
then it satisfies the
equation P*x equals lambda*x.
But for a projection matrix,
P squared is equal to P.
So if P is a projection,
we have P squared equals P.
And specifically, what this
means is P squared x is equal
to lambda*x.
So we have P acting on P
of x is equal to lambda*x.
And on the left-hand side, P*
is going to give me a lambda*x.
P* again will give
me a lambda x.
So we get lambda squared
x equals lambda*x.
And if I bring everything
to the left-hand side,
I get lambda times lambda
minus 1 x equals 0.
And because x is not a zero
vector, what that means is
lambda has to be either 0 or 1.
So this is just a quick
proof that the eigenvalue
of a projection matrix
is either 0 or 1.
So we already know
that P is going
to have eigenvalues of 0 or 1.
Now specifically,
how do I identify
which eigenvectors
correspond to 0
and which eigenvectors
correspond to 1?
Well, in this case, P
has a specific form,
which is a times a transpose
divided by a transpose a.
So I'll just write out
explicitly what this is.
So a transpose a, 1
divided by a transpose a
is going to be 9 plus
16 on the denominator.
Then we're going to have
3 and 4 and 3 and 4.
Now when we have a
matrix of this form,
it's always going to be
the case that the vector
a is going to be an
eigenvector with eigenvalue 1.
So let's check.
What is P acting on a?
Well, we end up with: the
matrix P is 1/25 [3; 4] [3, 4].
This is the matrix P. And if
we acted on the vector [3; 4],
notice how this piece right
here, we can multiply out.
This is going to be a transpose,
and this is going to be a.
And if we multiply
these two pieces out,
we get 25, which is exactly
the denominator a transpose a.
So at the end of the day, we get
[3, 4]; Because the 25 divides
out with the 25.
Now note that this is
exactly what we started with.
This is exactly a.
So note here that the
vector a corresponds
to an eigenvalue of 1.
Meanwhile, for an
eigenvalue of 0,
well, it always turns
out to be the case
that if I take any vector
perpendicular to a,
P acting on that vector
is going to be 0.
So what's a vector,
which I'll call b,
that's perpendicular to a?
Well, note that a is
just a two by two vector.
So that means there's only
going to be one direction that's
perpendicular to a.
Now just by eyeballing
it, I can see
that a vector that's going
to be perpendicular to a
is negative 4 and 3.
So let's quickly check
that this is an eigenvector
of P with eigenvalue 0.
So what we need to show is that
P acting on this vector, b,
is 0.
So P acting on b is
going to be 1/25.
It's going to be [3; 4]
[3, 4], multiplied by [4, 3].
And note how when I multiply
out this row on this column,
I get negative 3 times
4 plus 3 times 4,
which is going to be 0.
OK?
So this shows that this vector
b has an eigenvalue of 0
because note that we
can write this as 0*b.
OK.
For the second part, Q,
what are the eigenvectors
and eigenvalues
of this matrix, Q?
Well, Q is a rotation matrix.
So I'll just write out Q again,
0.6, negative 0.8, 0.8, 0.6.
So note that we can identify the
diagonal elements with a cosine
of some angle theta.
And we can associate
the off-diagonal parts
as sine theta and
negative sine theta.
And the reason we can
do that is because 0.6
squared plus 0.8 squared is 1.
So this is a rotation matrix.
Now, to work out
the eigenvalues,
I take a look at the
characteristic equation.
So this is going
to give me, if I
take a look at the
characteristic equation,
it's going to be 0.6
minus lambda, squared.
Then we have minus times
0.8 times negative 0.8.
So that's going to
be plus 0.8 squared.
And we want this to be 0.
So if I rewrite this, I get
lambda is 0.6 plus or minus
0.8i, where i is the
imaginary number.
So notice how the eigenvalues
come in complex conjugate
pairs.
And this is always the case
when we have a real matrix.
So we can find, first off, just
the eigenvalue that corresponds
to 0.6 plus 0.8i.
And then at the end,
we'll be able to find
the second eigenvector by just
taking the complex conjugate
of the first one.
So let's compute
Q minus lambda*I.
And if we have this acting
on some eigenvector u,
we want this to be 0.
Now Q minus lambda*I
is going to be,
for the case lambda
is 0.6 plus 0.8i,
this is going to give me
a quantity of minus 0.8i,
minus 0.8, 0.8, and minus 0.8i.
And I'm going to
write down components
of u, which are u_1 and u_2.
And we want this to vanish.
And we note that the second
row is a constant multiple
of the first row.
Specifically, if I multiplied
this first row through by i,
we would get negative i
squared, which is just 1.
And then the second part
would be negative i,
so we would just get the
second row back, which is good.
So we just need to
find u_1, u_2 that are
orthogonal to this first row.
And again, just by inspection,
I can pick 1 and negative i.
So note that that would give
me negative 0.8i plus 0.8i,
and this vanishes.
So this is the eigenvector that
corresponds to the eigenvalue
lambda 0.6 plus 0.8i.
In the meantime, if I take
the second eigenvalue,
which is negative 0.8i, I
can take u which is just
the complex conjugate
of this u up here.
So it'll be 1, plus i.
So this concludes the
eigenvalues and eigenvectors
of this matrix Q.
OK.
Now lastly, number three, we're
looking at a reflection matrix
which has the form 2P minus
I, where P is the same matrix
that we had in part one.
Now at first glance,
it looks like we
might have to diagonalize
this entire matrix.
However, note that
by shifting 2P by I,
we only shift the eigenvalues.
And we don't actually
change the eigenvectors.
So note that this matrix
R, which is 2P minus I,
it's going to have the
same eigenvectors as P.
It's just going to have
different eigenvalues.
So first off, we're going
to have one eigenvector.
So the first eigenvector
is going to be a.
So we have one
eigenvector which is a.
So we have one
eigenvector which is a.
And note that for the
vector a, it corresponds
to the eigenvalue of 1.
So what eigenvalue does
this correspond to?
This is going to give
me a lambda which
is 2 times 1 minus 1.
So it's 1.
So note that a, the
vector a, not only
has an eigenvalue of 1 for P,
but it has an eigenvalue of 1
for R as well.
The second case was b.
And remember that b has
an eigenvalue of 0 for P.
So when we act R acting on b,
we'll have 2 times 0 minus 1 b.
So this is going to
give us negative b.
So the eigenvalue for b
is going to be negative 1.
OK.
And this is actually a general
case for reflection matrices,
is that they typically
have eigenvalues
of plus 1 or negative 1.
OK, so we've just taken a
look at several matrices
that come up in practice.
We've looked at projection
matrices, reflection matrices,
and rotation matrices.
And we've seen a little
bit of the properties
of their eigenvalues
and eigenvectors.
So I'll just conclude here,
and good luck on your test.
