As noted in the introduction to this chapter we’re going to
start with a square matrix A and try
to determine vectors x and scalars 
so that we will have,
In other words, when this happens, multiplying x by the matrix A will be equivalent of multiplying x by the scalar . This will not be possible for all vectors, x, nor will it be possible for all
scalars . The goal of this section is to determine the
vectors and scalars for which this will happen.
So, let’s start off with the following definition.
Note that eigenvalues and eigenvectors will always occur in
pairs. You can’t have an eigenvalue
without an eigenvector and you can’t have an eigenvector without an eigenvalue.
Okay, what we need to do is figure out just how we can
determine the eigenvalues and eigenvectors for a given matrix. This is actually easier to do than it might
at first appear to be. We’ll start with
finding the eigenvalues for a matrix and once we have those we’ll be able to
find the eigenvectors corresponding to each eigenvalue.
Let’s start with and rewrite it as follows,
Note that all we did was insert the identity matrix into the
right side. Doing this will allow us to
further rewrite this equation as follows,
Now, if is going to be an eigenvalue of A, this system must have a non-zero
solution, x, since we know that
eigenvectors associated with cannot be the zero vector. However, Theorem 3 from the previous
section or more generally Theorem
8 from the Fundamental Subspaces section tells us that this system will
have a non-zero solution if and only if
So, eigenvalues will be scalars, ,
for which the matrix will be singular, i.e. . Let’s get a couple more definitions out of
the way and then we’ll work some examples of finding eigenvalues.
Note that the coefficient of is 1 (one) and that is NOT a typo in the
definition. This also guarantees that
this polynomial will be of degree n. Also, from the Fundamental Theorem of Algebra
we now know that there will be exactly n
eigenvalues (possibly including repeats) for an matrix A. Note that because the Fundamental Theorem of
Algebra does allow for the possibility of repeated eigenvalues there will be at
most n distinct eigenvalues for an matrix.
Because an eigenvalue can repeat itself in the list of all eigenvalues
we’d like a way to differentiate between eigenvalues that repeat and those that
don’t repeat. The following definition
will do this for us.
Okay, let’s find some eigenvalues we’ll start with some matrices.
|
Example 2 Find
all the eigenvalues for the given matrices.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
(a)
We’ll do this one with a little more detail than we’ll do
the other two. First we’ll need the
matrix .
Next we need
the determinant of this matrix, which gives us the characteristic polynomial.
Now, set this
equal to zero and solve for the eigenvalues.
So, we have two
eigenvalues and since they occur only once in the list they are both simple
eigenvalues.
[Return to Problems]
(b)
Here is the
matrix and its characteristic polynomial.
We’ll leave it
to you to verify both of these. Now,
set the characteristic polynomial equal to zero and solve for the
eigenvalues.
Again, we get
two simple eigenvalues.
[Return to Problems]
(c)
Here is the
matrix and its characteristic polynomial.
We’ll leave it
to you to verify both of these. Now,
set the characteristic polynomial equal to zero and solve for the
eigenvalues.
In this case we have an eigenvalue of multiplicity
two. Sometimes we call this kind of
eigenvalue a double eigenvalue. Notice
as well that we used the notation to denote the fact that this was a double
eigenvalue.
[Return to Problems]
|
Now, let’s take a look at some matrices.
|
Example 3 Find
all the eigenvalues for the given matrices.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
(a)
As with the
previous example we’ll do this one in a little more detail than the remaining
two parts. First, we’ll need ,
Now, let’s take
the determinant of this matrix and get the characteristic polynomial for A.
We’ll use the “trick” that we reviewed in the previous section to take
the determinant. You could also use
cofactors if you prefer that method.
The result will be the same.
Next, set this
equal to zero.
Now, most of us
aren’t that great at finding the roots of a cubic polynomial. Luckily there is a way to at least get us
started. It won’t always work, but if
it does it can greatly reduce the amount of work that we need to do.
Suppose we’re
trying to find the roots of an equation of the form,
where the are all integers. If there are integer solutions to this (and
there may NOT be) then we know that they must be divisors of . This won’t give us any integer solutions,
but it will allow us to write down a list of possible integer solutions. The list will be all possible divisors of .
In this case
the list of possible integer solutions is all possible divisors of -30.
Now, that may
seem like a lot of solutions that we’ll need to check. However, it isn’t quite that bad. Start with the smaller possible solutions
and plug them in until you find one (i.e.
until the polynomial is zero for one of them) and then stop. In this case the smallest one in the list
that works is -1. This means that
must be a
factor in the characteristic polynomial.
In other words, we can write the characteristic polynomial as,
where is a quadratic polynomial. We find by performing long division on the
characteristic polynomial. Doing this
in this case gives,
At this point
all we need to do is find the solutions to the quadratic and nicely enough
for us that factors in this case. So,
putting all this together gives,
So, this matrix
has three simple eigenvalues.
[Return to Problems]
(b)
Here is and the characteristic polynomial for this
matrix.
Now, in this
case the list of possible integer solutions to the characteristic polynomial
are,
Again, if we
start with the smallest integers in the list we’ll find that -2 is the first
integer solution. Therefore, must be a factor of the characteristic
polynomial. Factoring this out of the
characteristic polynomial gives,
Finally,
factoring the quadratic and setting equal to zero gives us,
So, we have one
double eigenvalue ( ) and one simple eigenvalue ( ).
[Return to Problems]
(c)
Here is and the characteristic polynomial for this
matrix.
We have a very
small list of possible integer solutions for this characteristic polynomial.
The smallest
integer that works in this case is -1 and we’ll leave it to you to verify
that the complete factored form is characteristic polynomial is,
and so we can
see that we’ve got two eigenvalues (a multiplicity 2 eigenvalue) and (a simple eigenvalue).
[Return to Problems]
(d)
Here is and the characteristic polynomial for this
matrix.
Okay, in this
case the list of possible integer solutions is,
The smallest
integer that will work in this case is 3.
We’ll leave it to you to verify that the factored form of the
characteristic polynomial is,
and so we can
see that if we set this equal to zero and solve we will have one eigenvalue
of multiplicity 3 (sometimes called a triple eigenvalue),
[Return to Problems]
|
As you can see the work for these get progressively more
difficult as we increase the size of the matrix, for a general matrix larger
than we’d need to use the method of cofactors to
determine the characteristic polynomial.
There is one case kind of matrix however that we can pick
the eigenvalues right off the matrix itself without doing any work and the size
won’t matter.
Proof : We’ll
give the proof for an upper triangular matrix, and leave it to you to verify
the proof for a lower triangular and a diagonal matrix. We’ll start with,
Now, we can write down ,
Now, this is still an upper triangular matrix and we know
that the determinant of a triangular matrix is just the product of its main
diagonal entries. The characteristic
polynomial is then,
Setting this equal to zero and solving gives the eigenvalues
of,
|
Example 4 Find
the eigenvalues of the following matrix.
Solution
Since this is a
lower triangular matrix we can use the previous theorem to write down the
eigenvalues. It will simply be the
main diagonal entries. The eigenvalues
are,
|
We can now find the eigenvalues for a matrix. We next need to address the issue of finding
their corresponding eigenvectors. Recall
given an eigenvalue, ,
the eigenvector(s) of A that
correspond to will be the vectors x such that,
Also, recall that was chosen so that was a singular matrix. This in turn guaranteed that we would have a
non-zero solution to the equation above.
Note that in doing this we don’t just guarantee a non-zero solution, but
we also guarantee that we’ll have infinitely many solutions to the system. We have one quick definition that we need to
take care of before we start working problems.
Note that there will be one eigenspace of A for each distinct eigenvalue and so
there will be anywhere from 1 to n
eigenspaces for an matrix depending upon the number of distinct
eigenvalues that the matrix has. Also,
notice that we’re really just finding the null space for a system and we’ve
looked at that in several sections in the previous chapter.
Let’s take a look at some eigenspaces for some of the
matrices we found eigenvalues for above.
|
Example 5 For
each of the following matrices determine the eigenvectors corresponding to
each eigenvalue and determine a basis for the eigenspace of the matrix
corresponding to each eigenvalue.
(a) [Solution]
(b) [Solution]
Solution
We determined the eigenvalues for each of these in Example
2 above so refer to that example for the details in finding them. For each eigenvalue we will need to solve
the system,
to determine
the general form of the eigenvector.
Once we have that we can use the general form of the eigenvector to
find a basis for the eigenspace.
(a)
We know that the eigenvalues for this matrix are and .
Let’s first find the eigenvector(s) and eigenspace for . Referring to Example 2 for the formula for and plugging into this we can see that the system we need
to solve is,
We’ll leave it
to you to verify that the solution to this system is,
Therefore, the
general eigenvector corresponding to is of the form,
The eigenspace is all vectors of this form and so we can see that a basis for the eigenspace
corresponding to is,
Now, let’s find
the eigenvector(s) and eigenspace for . Plugging into the formula for from Example 2 gives the following system we
need to solve,
The solution to
this system is (you should verify this),
The general
eigenvector and a basis for the eigenspace corresponding to is then,
Note that if we
wanted to get specific eigenvectors for each eigenvalue the basis
vector for each eigenspace would work.
So, if we do that we could use the following eigenvectors (and their
corresponding eigenvalues) if we’d like.
Note as well
that these eigenvectors are linearly independent vectors.
[Return to Problems]
(b)
From Example 2
we know that is a double eigenvalue and so there will be
a single eigenspace to compute for this matrix. Using the formula for from Example 2 and plugging into this gives the following system that
we’ll need to solve for the eigenvector and eigenspace.
The solution to
this system is,
The general
eigenvector and a basis for the eigenspace corresponding is then,
In this case we
get only a single eigenvector and so a good eigenvalue/eigenvector pair is,
[Return to Problems]
|
We didn’t look at the second matrix from Example 2 in the
previous example. You should try and
determine the eigenspace(s) for that matrix.
The work will follow what we did in the first part of the previous
example since there are two simple eigenvalues.
Now, let’s determine the eigenspaces for the matrices in
Example 3
|
Example 6 Determine
the eigenvectors corresponding to each eigenvalue and a basis for the
eigenspace corresponding to each eigenvalue for each of the matrices from
Example 3 above.
Solution
The work finding the eigenvalues for each of these is
shown in Example 3 above. Also, we’ll
be doing this with less detail than those in the previous example. In each part we’ll use the formula for found in Example 3 and plug in each
eigenvalue to find the system that we need to solve for the eigenvector(s)
and eigenspace.
(a)
The eigenvalues for this matrix are ,
and so we’ll have three eigenspaces to find.
Starting with we’ll need to find the solution to the
following system,
The general
eigenvector and a basis for the eigenspace corresponding to is then,
Now, let’s take
a look at . Here is the system we need to solve,
Here is the
general eigenvector and a basis for the eigenspace corresponding to .
Finally, here
is the system for .
The general eigenvector and a basis for the eigenspace
corresponding to is then,
Now, we as with
the previous example, let’s write down a specific set of
eigenvalue/eigenvector pairs for this matrix just in case we happen to need
them for some reason. We can get
specific eigenvectors using the basis vectors for the eigenspace as we did in
the previous example.
You might want
to verify that these three vectors are linearly independent vectors.
(b)
The eigenvalues for this matrix are and so it looks like we’ll have two eigenspaces
to find for this matrix.
We’ll start with . Here is the system that we need to solve
and its solution.
The general
eigenvector and a basis for the eigenspace corresponding is then,
Note that even
though we have a double eigenvalue we get a single basis vector here.
Next, the
system for that we need to solve and its solution is,
The general
eigenvector and a basis for the eigenspace corresponding to is,
A set of
eigenvalue/eigenvector pairs for this matrix is,
Unlike the
previous part we only have two eigenvectors here even though we have three
eigenvalues (if you include repeats anyway).
(c)
As with the
previous part we’ve got two eigenvalues, and and so we’ll again have two eigenspaces to
find here.
We’ll start
with . Here is the system we need to solve,
The general
eigenvector corresponding to is then,
Now, the
eigenspace is spanned by the two vectors above and since they are linearly
independent we can see that a basis for the eigenspace corresponding to is,
Here is the
system for that we need to solve,
The general
eigenvector and a basis for the eigenspace corresponding to is,
Okay, in this
case if we want to write down a set of eigenvalue/eigenvector pairs we’ve got
a slightly different situation than we’ve seen to this point. In the previous example we had an
eigenvalue of multiplicity two but only got a single eigenvector for it. In this case because the eigenspace for our
multiplicity two eigenvalue has a dimension of two we can use each basis vector
as a separate eigenvector and so the eigenvalue/eigenvector pairs this time
are,
Note that we
listed the eigenvalue of “-1” twice, once for each eigenvector. You should verify that these are all
linearly independent.
(d)
In this case we
had a single eigenvalue, so we’ll have a single eigenspace to
find. Here is the system and its
solution for this eigenvalue.
The general
eigenvector corresponding to is then,
As with the
previous example we can see that the eigenspace is spanned by the two vectors
above and since they are linearly independent we can see that a basis for the
eigenspace corresponding to is,
Note that the
two vectors above would also make a nice pair of eigenvectors for the single
eigenvalue in this case.
|
Okay, let’s go back and take a look at the
eigenvector/eigenvalue pairs for a second.
First, there are reasons for wanting these as we’ll see in the next
section. On occasion we really do want
specific eigenvectors and we generally want them to be linearly independent as
well as we’ll see. Also, we saw two
examples of eigenvalues of multiplicity 2.
In one case we got a single eigenvector and in the other we got two
linearly independent eigenvectors. This
will always be the case, if is an eigenvalue of multiplicity k then there will be anywhere from 1 to k linearly independent eigenvectors
depending upon the dimension of the eigenspace.
Now there is one type of eigenvalue that we’ve been avoiding
to this point and you may have noticed that already. Let’s take a look at the following example to
see what we’ve been avoiding to this point.
|
Example 7 Find
all the eigenvalues of
Solution
First we’ll need the matrix and then we’ll use that to find the
characteristic equation.
From this we
can see that the eigenvalues will be complex.
In fact they will be,
|
So we got a complex eigenvalue. We’ve not seen any of these to this point and
this will be the only one we’ll look at here.
We have avoided complex eigenvalues to this point for a very important reason. Let’s recall just what an eigenvalue is. An eigenvalue is a scalar such that,
Can you see the problem?
In order to talk about the scalar multiplication of the right we need to
be in a complex vector space! Up to this
point we’ve been working exclusively with real vector spaces and remember that
in a real vector space the scalars are all real numbers. So, in order to work with complex eigenvalues
we would need to be working in a complex vector space and we haven’t looked at
those yet. So, since we haven’t looked
at complex vector spaces yet we will be working only with matrices that have
real eigenvalues.
Note that this doesn’t mean that complex eigenvalues are not
important. There are some very important
applications to complex eigenvalues in various areas of math, engineering and
the sciences. Of course, there are also
times where we would ignore complex eigenvalues.
We will leave finish this section with a couple of nice
theorems.
Proof : The proof
here is pretty simple.
So, from this we can see that is an eigenvalue of with corresponding eigenvector x.
We’ll prove part (a)
here. The proof of part (b) involves some fairly messy algebra
and so we won’t show it here.
Proof of (a) : First,
recall that eigenvalues of A are
roots of the characteristic polynomial and hence we can write the
characteristic polynomial as,
Now, plug in into this and we get,
Finally, from Theorem 1 of the Properties
of Determinant section we know that
So, plugging this in gives,