Hello, in this video we're going to be
looking at the theory behind
eigenvalues and eigenvectors. So we'll
start out with a definition
and that will be about the eigenvalue
and eigenvectors a matrix A or transformation T.
Essentially our definition says
that eigenvalues sometimes I call these lower case
ev's associated with the matrix
A or linear transformation T(x)
where that T(x) is equal to A*x
is a scalar lambda such that
A*v is equal to lambda
times that vector be for some nonzero
vector v
called an eigenvector for that matrix
or transformation.
So essentially when we're looking at eigenvalues and eigenvectors,
if I have a matrix and if I can take that
matrix
A times a vector v and a get
a scalar multiple of that vector. So
essentially that matrix multiplication
gives me a scalar multiple
where that transformation of the vector
gets me a scalar multiple.
Then I call those the eigenvalues and eigenvectors.
Now notice that eigenvectors can never be zero because
that's sorta, you know, the trivial case.
A*0 will always equal lambda time 0 so we can't have that.
Okay, so how do we find these?
Well, to find eigenvalue of A,
essentially we're gonna be using this
equation sorta manipulating it just a
little bit.
So I start out with this equation
A*v is equal to lambda*v and subtract lambda*v
from both sides so I have a
matrix-vector
minus this other vector, that's a vector 0
over here. What I wanna do as I wanna
factor out the v.
You know I probably have to do that on
the right hand side, but
the problem is I have a matrix minus a vector there
and that can't work. So I'm going to
change this into a matrix without
ruining the essence of that scalar. So
I'm going to put in this
identity matrix, I, here. So I have
A-lamda*I, and the same dimension of A so its I sub n
times v is equal to 0.  And this should still be the zero vector over
here.
So how do I find an eigenvalue of A?
Well eigenvalues are scalars so what I
wanna do,
since v has to be nonzero,
I need for this
to be singular.  I need to find
all nonzero solutions of this.  If this was
just zero
then this would be nonsingular, this
would be invertible, but we know this is
going to be singular or not invertible for this to work.
So what I wanna do is, I wanna find where
the determinant
of A-minus lambda*I sub n
is equal to scalar 0.
So this is my equation that I really
wanna solve
to get those eigenvalues.
Okay, so it's a gonna turn out to be a
polynomial.
In fact, we should get
n zeros from this polynomial. Now they
could be repeated they could be complex
and they could be real so
we're gonna have a polynomial here and
hopefully we'll get n
zeros from that, those will be our eigenvalues.  One thing to notice
about our eigenvalues is they are going to
be unique.
And what that means is essentially that
when you find them they can't be,
you know, different. They have to be those
eigenvalues that you find from that
polynomial from solving that polynomial,
and there can't be any multiples
or anything else like we'll see with eignevectors.
And speaking of eigenvectors, well how do you find those?
Well to find the corresponding eigenvectors
of the matrix A above we're gonna go
ahead
and use that same equation,
A*v is equal to lambda*v.
We're going to manipulate that,
our equation star, so essentially what
we're going to do
is solve this equation
for eigenvectors, v.
What that means is that we have a matrix,
so this is a matrix, times a vector
equal to 0.
We know that 0 work, but 0
can't be an eigenvector of this matrix,
it's trivial.
What we're going to do is we're going to
find
the rest of the nullspace, so the rest of the nullspace
will be our eigenvectors.
So that's going to be finding
the nonzero nullspace of
A-minus lambda*I sub n.
where A is that nxn matrix and
so we're gonna be using this equation
to find our eigenvectors.
This will give us the
eigenvectors of A and
eigenvectors as we will see
are not unique.
What that means is if I find
an eigenvalue, lambda,
then the corresponding
eigenvector
v can be
all multiples
of v as well. So
we can find an eigenvalue, all of us
can find and eigenvalue lambda,
but we can all have different correct eigenvectors,
v, for that eigenvalue. We'll look an
example of this in our next video,
but I wanted to start you off with the
theory here.
Ok, I hope this has helped. Thank you!
