>> In this video, we'll be
checking whether a given value
is an eigenvalue for a matrix,
and also giving the vector,
checking that if that vector is
an eigenvector for the matrix.
So, let's first just
go over the definition
of eigenvector and eigenvalue.
An eigenvector of an n by n
matrix A is a non-zero vector V
that has the property that
A times V equals lambda V
for some scalar lambda.
Right? So, this means
that V is a vector,
such that A times V is just
a scalar multiple of V. So,
there's a few things
to note here.
One, A is going to
be a square matrix.
Eigen vectors are only
defined for square matrices.
Two, eigenvectors,
by definition,
must be non-zero vectors.
And three, this is an
existence question.
A vector, a non-zero vector
is called an eigenvector
if there exists some scalar
lambda, so that A times Z is
that scalar times V.
Now an eigenvalue,
a scalar lambda is called
an eigenvalue of an n
by n matrix A. if there's
a nontrivial solution
to the equation Ax
equals lambda x,
i.e. A minus lambda times the
identity times x equals 0 has a
nontrivial solution.
And again, some things to
notice, this is only defined
for square matrices and it's
just asking about the existence
of nontrivial solutions
to some equation.
So, let's use this to
check in some examples.
So, what if I ask you, if
lambda equals 2, an eigenvalue
of the matrix A equals
3, 2, 3 minus 8.
Now, this is just asking
if A minus 2 times the identity
has a nontrivial null space.
So, A minus 2 times the
identity, I can just compute
that and I get this
1, 2, 3, 6 matrix.
Now, we just have to set up the
augmented matrix and row reduce
to find the null space.
There's my pivot, which tells me
the equation A minus 2 I equals
x equals 0 has a one
dimensional solution set.
Alright, so, it has a
nontrivial null space
for that matrix, A minus 2I.
Which tells me that lambda
equals 2 is an eigenvalue for A.
What if we turn it around
and ask the other way?
Suppose I give you a vector
3, minus 2, 1 and ask you,
is this an eigenvector
of the following matrix?
And if so, find the
corresponding eigenvalue.
So, we just need to
compute A times V,
and that's the matrix
vector product.
We just compute it and
get minus 15, 10, minus 5.
Which we notice is minus
5 times the vector 3,
minus 2, 1 that we were given.
Right, so, it's five
times the vector itself,
which means V is an eigenvector
of A with eigenvalue 5.
Now, a few remarks
before we finish up here.
One, eigenvectors,
by definition,
cannot be the zero vector.
That said, eigenvalues
are allowed to be 0.
So, any vector V that's in the
null space of a square matrix,
if V is not 0 then V is going
to be an eigenvector of A
with eigenvalue equal to 0.
And finally, eigenvectors will
see these are nice in the sense
that they tell us
which directions
in which multiplication
by the matrix
which could do some pretty
confusing things, right,
complicated things,
multiplication
by A is actually going to act
like multiplication by a scalar.
Right, and so, these
eigendirections,
these eigenvectors are going
to tell us which directions
in which multiplication by
A is really easier to study.
So, one of the things
that we've seen here is
that if we're given an
actual scalar lambda,
it's easy to check if lambda is
an eigenvalue of A. And also,
if we're given an actual
vector V, it's easy to check
if V is an eigenvector
for A. Now,
if you're not given either
one, and you're just asked
to find all of the
eigenvalues and eigenvectors
of a given matrix, you
have to do more work.
And so, this is the study of
the characteristic equation.
I'll make another video
for that and link it here.
