Subsection 4.2.1 The characteristic polynomial
We will first see that the eigenvalues of a square matrix appear as the roots of a particular polynomial. To begin, notice that we originally defined an eigenvector as a nonzero vector that satisfies the equation We will rewrite this as
In other words, an eigenvector is a solution of the homogeneous equation This puts us in the familiar territory explored in the next activity.
Activity 4.2.2.
The eigenvalues of a square matrix are defined by the condition that there be a nonzero solution to the homogeneous equation
-
If there is a nonzero solution to the homogeneous equation
what can we conclude about the invertibility of the matrix
-
If there is a nonzero solution to the homogeneous equation
what can we conclude about the determinant
-
Let’s consider the matrix
from which we construct
Find the determinant What kind of equation do you obtain when we set this determinant to zero to obtain
-
Use the determinant you found in the previous part to find the eigenvalues
by solving the equation
We considered this matrix in
Activity 4.1.2 so we should find the same eigenvalues for
that we found by reasoning geometrically there.
-
Consider the matrix
and find its eigenvalues by solving the equation
-
Consider the matrix
and find its eigenvalues by solving the equation
-
Find the eigenvalues of the triangular matrix
What is generally true about the eigenvalues of a triangular matrix?
This activity demonstrates a technique that enables us to find the eigenvalues of a square matrix Since an eigenvalue is a scalar for which the equation has a nonzero solution, it must be the case that is not invertible. Therefore, its determinant is zero. This gives us the equation
whose solutions are the eigenvalues of This equation is called the characteristic equation of
Example 4.2.1.
If we write the characteristic equation for the matrix we see that
This shows us that the eigenvalues are and
In general, the expression
is a polynomial in
which is called the
characteristic polynomial of
If
is an
matrix, the degree of the characteristic polynomial is
For instance, if
is a
matrix, then
is a quadratic polynomial; if
is a
matrix, then
is a cubic polynomial.
The matrix in
Example 4.2.1 has a characteristic polynomial with two real and distinct roots. This will not always be the case, as demonstrated in the next two examples.
Example 4.2.2.
Consider the matrix whose characteristic equation is
In this case, the characteristic polynomial has one real root, which means that this matrix has a single real eigenvalue,
Example 4.2.3.
To find the eigenvalues of a triangular matrix, we remember that the determinant of a triangular matrix is the product of the entries on the diagonal. For instance, the following triangular matrix has the characteristic equation
showing that the eigenvalues are the diagonal entries
Subsection 4.2.2 Finding eigenvectors
Now that we can find the eigenvalues of a square matrix
by solving the characteristic equation
we will turn to the question of finding the eigenvectors associated to an eigenvalue
The key, as before, is to note that an eigenvector is a nonzero solution to the homogeneous equation
In other words, the eigenvectors associated to an eigenvalue
form the null space
This shows that the eigenvectors associated to an eigenvalue form a subspace of We will denote the subspace of eigenvectors of a matrix associated to the eigenvalue by and note that
We say that is the eigenspace of associated to the eigenvalue
Activity 4.2.3.
In this activity, we will find the eigenvectors of a matrix as the null space of the matrix
-
Let’s begin with the matrix
We have seen that
is an eigenvalue. Form the matrix
and find a basis for the eigenspace
What is the dimension of this eigenspace? For each of the basis vectors
verify that
-
We also saw that
is an eigenvalue. Form the matrix
and find a basis for the eigenspace
What is the dimension of this eigenspace? For each of the basis vectors
verify that
-
Is it possible to form a basis of
consisting of eigenvectors of
-
Now consider the matrix
Write the characteristic equation for
and use it to find the eigenvalues of
For each eigenvalue, find a basis for its eigenspace
Is it possible to form a basis of
consisting of eigenvectors of
-
Next, consider the matrix
Write the characteristic equation for
and use it to find the eigenvalues of
For each eigenvalue, find a basis for its eigenspace
Is it possible to form a basis of
consisting of eigenvectors of
-
Finally, find the eigenvalues and eigenvectors of the diagonal matrix
Explain your result by considering the geometric effect of the matrix transformation defined by
Once we find an eigenvalue of a matrix
describing the associated eigenspace
amounts to the familiar task of describing the null space
Example 4.2.4.
Revisiting the matrix
from
Example 4.2.1, we recall that we found eigenvalues
and
Considering the eigenvalue we have
Since the eigenvectors are the solutions of the equation we see that they are determined by the single equation or Therefore the eigenvectors in have the form
In other words, is a one-dimensional subspace of with basis vector or basis vector In the same way, we find that a basis for the eigenspace is
We note that, for this matrix, it is possible to construct a basis of consisting of eigenvectors, namely,
Example 4.2.5.
Consider the matrix whose characteristic equation is
There is a single eigenvalue and we find that
Therefore, the eigenspace is one-dimensional with a basis vector
Example 4.2.6.
If then
which implies that there is a single eigenvalue We find that
which says that every two-dimensional vector satisfies Therefore, every vector is an eigenvector and so This eigenspace is two-dimensional.
We can see this in another way. The matrix transformation defined by
rotates vectors by
which says that
for every vector
In other words, every two-dimensional vector is an eigenvector with associated eigenvalue
These last two examples illustrate two types of behavior when there is a single eigenvalue. In one case, we are able to construct a basis of
using eigenvectors; in the other, we are not. We will explore this behavior more in the next subsection.
A check on our work.
When finding eigenvalues and their associated eigenvectors in this way, we first find eigenvalues
by solving the characteristic equation. If
is a solution to the characteristic equation, then
is not invertible and, consequently,
must contain a row without a pivot position.
This serves as a check on our work. If we row reduce
and find the identity matrix, then we have made an error either in solving the characteristic equation or in finding
Subsection 4.2.3 The characteristic polynomial and the dimension of eigenspaces
Given a square
matrix
we saw in the previous section the value of being able to express any vector in
as a linear combination of eigenvectors of
For this reason,
Question 4.1.8 asks when we can construct a basis of
consisting of eigenvectors. We will explore this question more fully now.
As we saw above, the eigenvalues of are the solutions of the characteristic equation The examples we have considered demonstrate some different types of behavior. For instance, we have seen the characteristic equations
-
which has real and distinct roots,
-
which has repeated roots, and
-
If is an matrix, then the characteristic polynomial is a degree polynomial, and this means that it has roots. Therefore, the characteristic equation can be written as
giving eigenvalues As we have seen, some of the eigenvalues may be complex. Moreover, some of the eigenvalues may appear in this list more than once. However, we can always write the characteristic equation in the form
The number of times that appears as a factor in the characteristic polynomial, is called the multiplicity of the eigenvalue
Example 4.2.7.
We have seen that the matrix
has the characteristic equation
This matrix has a single eigenvalue
which has multiplicity
Example 4.2.8.
If a matrix has the characteristic equation
then that matrix has four eigenvalues: having multiplicity 2; having multiplicity 1; having multiplicity 7; and having multiplicity 2. The degree of the characteristic polynomial is the sum of the multiplicities so this matrix must be a matrix.
The multiplicities of the eigenvalues are important because they influence the dimension of the eigenspaces. We know that the dimension of an eigenspace must be at least one; the following proposition also tells us the dimension of an eigenspace can be no larger than the multiplicity of its associated eigenvalue.
Proposition 4.2.9.
If is a real eigenvalue of the matrix with multiplicity then
Example 4.2.10.
The diagonal matrix
has the characteristic equation
There is a single eigenvalue
having multiplicity
and we saw earlier that
Example 4.2.11.
The matrix
has the characteristic equation
This tells us that there is a single eigenvalue
having multiplicity
In contrast with the previous example, we have
Example 4.2.12.
We saw earlier that the matrix has the characteristic equation
There are three eigenvalues each having multiplicity By the proposition, we are guaranteed that the dimension of each eigenspace is that is,
It turns out that this is enough to guarantee that there is a basis of consisting of eigenvectors.
Example 4.2.13.
If a matrix has the characteristic equation
we know there are four eigenvalues Without more information, all we can say about the dimensions of the eigenspaces is
We can guarantee that but we cannot be more specific about the dimensions of the other eigenspaces.
Fortunately, if we have an matrix, it frequently happens that the characteristic equation has the form
where there are distinct real eigenvalues, each of which has multiplicity In this case, the dimension of each of the eigenspaces With a little work, it can be seen that choosing a basis vector for each of the eigenspaces produces a basis for We therefore have the following proposition.
Proposition 4.2.14.
If
is an
matrix having
distinct real eigenvalues, then there is a basis of
consisting of eigenvectors of
This proposition provides one answer to our
Question 4.1.8. The next activity explores this question further.
Activity 4.2.4.
-
Identify the eigenvalues, and their multiplicities, of an
matrix whose characteristic polynomial is
What can you conclude about the dimensions of the eigenspaces? What is the shape of the matrix? Do you have enough information to guarantee that there is a basis of
consisting of eigenvectors?
-
Find the eigenvalues of
and state their multiplicities. Can you find a basis of
consisting of eigenvectors of this matrix?
-
Consider the matrix whose characteristic equation is
-
Identify the eigenvalues and their multiplicities.
-
For each eigenvalue
find a basis of the eigenspace
and state its dimension.
-
Is there a basis of
consisting of eigenvectors of
-
Now consider the matrix whose characteristic equation is also
-
Identify the eigenvalues and their multiplicities.
-
For each eigenvalue
find a basis of the eigenspace
and state its dimension.
-
Is there a basis of
consisting of eigenvectors of
-
Consider the matrix whose characteristic equation is
-
Identify the eigenvalues and their multiplicities.
-
For each eigenvalue
find a basis of the eigenspace
and state its dimension.
-
Is there a basis of
consisting of eigenvectors of