We'll start looking at eigenvalues and 
eigenvectors in this video.
Let A be a complex n-by-n matrix,
x be an n-vector with complex entries
such that x is not the zero vector, and
lambda be a complex number.
We say that x is an eigenvector of A
with eigenvalue lambda
if the following hold:
A x equal to lambda times x.
In other words,
if x is an eigenvector of A, then A times x
is simply a scalar multiple of x.
Now, notice that by definition,
the zero vector is never an eigenvector
because we require
the vector to be non-zero.
Let's look at an example.
Say A is the matrix 1 2, 1 0,
and x is the vector 2 1.
Now if you multiply
A and x, you get in the first entry
1 times 2 plus 2 times 1, 
so we have 2 plus 2,
and in the second entry, 
you get 1 times 2, that's 2,
plus 0 times 1, that's 0, so that's just 2.
And that's 4 2.
And you can't express this
as 2 times 2 1, which is 2 times x.
So in this case,
2 1 is an eigenvector of this matrix A
and the eigenvalue will be 2.
So the question that we're going 
to address now is
how to find eigenvalues and eigenvectors.
And for that, we're going to
look at characteristic polynomials.
Suppose that x is an eigenvector of A 
with eigenvalue lambda.
That is, A times x is equal to lambda times x.
But this can be rewritten as
A times x minus
lambda times x.
But lambda times x can be written as
lambda times I and then times x.
So we can rewrite this as 
A minus lambda times I
times x equal to 0.
Now because
x is a nonzero vector, this tells us that
x is a nonzero vector in the nullspace
of A minus lambda I.
In this example here, if you look at
A minus 2 I, the matrix is
-1 2, 1 -2.
And the nullspace contains the vector 2 1.
So again x is in the nullspace of 
A minus lambda I.
But that means that this matrix 
A minus lambda I is singular,
so its determinant is going to be 0.
And the determinant of A minus lambda I
is a polynomial in lambda.
And this polynomial is denoted by
P subscript A,
called the characteristic polynomial of A.
So every eigenvalue of A
is a root of this polynomial.
And conversely,
every root of this polynomial is going
to be an eigenvalue of A.
So let's look at our example again.
And this time, we want to form the
characteristic polynomial of A.
So P subscript A
is given by the determinant
of 1 2, 1 0, minus
lambda times the identity matrix,
and that's the determinant of
1 minus lambda 2, 1 and
minus lambda.
And expanding this using the formula for
 determinants of 2-by-two matrices,
we get that this is 1 minus lambda
times minus lambda
minus 2 times 1, and
that will give us lambda squared minus
lambda minus 2.
Clearly, if lambda is equal to 2, this is 0.
Now, there's also another root.
If you set
lambda equal to -1, you also get zero.
So -1 squared is 1,
minus -1 is plus 1, so that's 2, 2 -2 is 0.
So -1 is also an eigenvalue.
And what would be an eigenvector 
with this eigenvalue?
Well, all you need to do is look at this matrix
and find a non-zero vector in its nullspace.
So this matrix is 2 2, 1 1,
and it's not hard to see that
if you take the vector 1 -1,
this is in the nullspace of this matrix.
So 1 -1 is en eigenvector of
A with eigenvalue -1.
Notice that there can be no other eigenvalues for A
because if you look at this
characteristic polynomial,
it's of degree two and so there can be
no more than two distinct eigenvalues
and we have found two of them,
namely, 2 and -1.
So those are all the possible eigevalues for A.
To summarize, in order to find all the
eigenvalues of a matrix,
first you form the characteristic
polynomial of the matrix,
and then find all the roots of that polynomial.
And to find an eigenvector for 
a particular eigenvalue,
you just find a non-zero vector in the
nullspace of A minus the eigenvalue times I.
