Hello, in this video were going to
look at specific example for finding eignevalues
and eigenvectors. We're gonna use that
example that we are already checked out
when we were constructing an
an alternate basis and looking for
a reflection. So we want to find the eigenvalues and eigenvectors
for the linear transformation of reflection of
vectors about the line
y=-x.
From another video you may recall that
this was
T(x)=A*x where A was
(0, -1, -1, 0).
So we find the eigenvalues
by solving in this equation.  We want to
find where the determinant
A-lambda*I is equal to 0.
We're going to solve that for lambda.
Here's our
A, our standard matrix, and
subtracting off this lambda*I really
results in subtracting off
lambdas from the diagonal. Here's our
setup.
I have A-minus lambda times I.
These lambdas come into the diagonal
times zero just leave them there.
I have zero -1*lambda
(-1, 0, -1,0)
and
zero minus lambda. I want to
take that determinant of that so I'm going
to multiply by down the diagonal,
lambda squared, subtract up the off diagonal,
subtract off -1 times -1 which is one.
I can factor that out and I get
(lambda -1)
times (lambda+1) = 0, that will give me
eigenvalues of plus and minus one.
Now to find the corresponding eigenvectors.
To find the corresponding eigenvectors,
I'm gonna find the nonzero nullspace of
A-lambda*I.
for each lambda. Let's start
with one, when I take
one and plug it back into that
a-minus lambda*I,
which is right here, I get (-1, -1, -1, -1)
and I augment that with zero so
this is easy enough to solve, but I'm going to use the calculator to sort a show you how
you might do this.  I have that matrix stored in A and
I'm going to do a 2ND MATRIX I'm
gonna do an RREF,
which we do a lot in this class, and go
back to MATRIX
and "hit up" A.
We should get a matrix here that has a
free variable
because that will give us our nonzero nullspace.
So reduced row echelon form is (1, 1,
0,0). I have 1,1
augmented with zero and
0, 0 augmented with 0. That means
I would have, if I call this
x1 and x2,
x1 is equal to a -x2, you
bring in this
x2 over, and x2 two is free.  So a good
eigenvector I can get from this, if I just factor out that
-x2, is going to be (-1,1).
That will be my first
eigenvector corresponding  to 1. remember
you can have
any multiple of this so that would work
as well.
Alright, now we'll find the one corresponding to -1
and that is stored in my matrix B.  So
I'm going to
go back to 2ND MATRIX, RREF,
2ND MATRIX, B.
For this one I get (1, -1,
0, 0), and it may be obvious to you
what your eigenvalues, 
vectors
I should say, will be. And if it is,
especially for these 2X2's, you
can go ahead and solve them that way.
In this case I'll have x1, if I recycle these variables
is equal to positive x2 and x2
is free.
Factoring out the x2
I'm left with (1,1), but again
it can be any multiple of this factor.
So it could be
(2,2), (-pi,-pi),
you get the jist. So that's it,
therefore what we have
for lambda equal 1, we have this
eigenvector (1, -1) or
(-1, 1), see I even used a multiple down here.
And this is called the eigenspace for A
associated with the eigenvalue 1.
And for -1 I have the vector
(1,1), so this is the eigenvalue and eigenvector
associated with the eigenspace A
called E sub lambda
equals -1.  We're gonna follow this
example throughout and see how this will
tie in with
some more theory later. Hope this has
helped.
Thank you!
