###
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 \(\vvec\) that satisfies the equation \(A\vvec =
\lambda\vvec\text{.}\) We will rewrite this as

\begin{equation*}
\begin{aligned}
A\vvec \amp {}={} \lambda\vvec \\
A\vvec - \lambda\vvec \amp {}={} \zerovec \\
A\vvec - \lambda I\vvec \amp {}={} \zerovec \\
(A-\lambda I)\vvec \amp {}={} \zerovec\text{.} \\
\end{aligned}
\end{equation*}

In other words, an eigenvector \(\vvec\) is a solution of the homogeneous equation \((A-\lambda I)\vvec=\zerovec\text{.}\) 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 \((A-\lambda I)\vvec=\zerovec\text{.}\)

If there is a nonzero solution to the homogeneous equation \((A-\lambda I)\vvec = \zerovec\text{,}\) what can we conclude about the invertibility of the matrix \(A-\lambda
I\text{?}\)

If there is a nonzero solution to the homogeneous equation \((A-\lambda I)\vvec = \zerovec\text{,}\) what can we conclude about the determinant \(\det(A-\lambda I)\text{?}\)

Let’s consider the matrix

\begin{equation*}
A = \left[\begin{array}{rr}
1 \amp 2 \\
2 \amp 1 \\
\end{array}\right]
\end{equation*}

from which we construct

\begin{equation*}
A-\lambda I =
\left[\begin{array}{rr}
1 \amp 2 \\
2 \amp 1 \\
\end{array}\right]
- \lambda
\left[\begin{array}{rr}
1 \amp 0 \\
0 \amp 1 \\
\end{array}\right]
=
\left[\begin{array}{rr}
1-\lambda \amp 2 \\
2 \amp 1-\lambda \\
\end{array}\right]\text{.}
\end{equation*}

Find the determinant \(\det(A-\lambda I)\text{.}\) What kind of equation do you obtain when we set this determinant to zero to obtain \(\det(A-\lambda I) = 0\text{?}\)

Use the determinant you found in the previous part to find the eigenvalues

\(\lambda\) by solving the equation

\(\det(A-\lambda I) = 0\text{.}\) We considered this matrix in

Activity 4.1.2 so we should find the same eigenvalues for

\(A\) that we found by reasoning geometrically there.

Consider the matrix \(A = \left[\begin{array}{rr}
2 \amp 1 \\
0 \amp 2 \\
\end{array}\right]\) and find its eigenvalues by solving the equation \(\det(A-\lambda I) = 0\text{.}\)

Consider the matrix \(A = \left[\begin{array}{rr}
0 \amp -1 \\
1 \amp 0 \\
\end{array}\right]\) and find its eigenvalues by solving the equation \(\det(A-\lambda I) = 0\text{.}\)

Find the eigenvalues of the triangular matrix \(\left[\begin{array}{rrr}
3 \amp -1 \amp 4 \\
0 \amp -2 \amp 3 \\
0 \amp 0 \amp 1 \\
\end{array}\right]
\text{.}\) 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 \(A\text{.}\) Since an eigenvalue \(\lambda\) is a scalar for which the equation \((A-\lambda
I)\vvec = \zerovec\) has a nonzero solution, it must be the case that \(A-\lambda I\) is not invertible. Therefore, its determinant is zero. This gives us the equation

\begin{equation*}
\det(A-\lambda I) = 0
\end{equation*}

whose solutions are the eigenvalues of \(A\text{.}\) This equation is called the *characteristic equation* of \(A\text{.}\)

####
Example 4.2.1.

If we write the characteristic equation for the matrix \(A = \left[\begin{array}{rr}
-4 \amp 4 \\
-12 \amp 10 \\
\end{array}\right]
\text{,}\) we see that

\begin{equation*}
\begin{aligned}
\det(A-\lambda I) \amp {}={} 0 \\
\\
\det \left[\begin{array}{rr}
-4 - \lambda \amp 4 \\
-12 \amp 10 - \lambda \\
\end{array}\right] \amp {}={} 0 \\
\\
(-4-\lambda)(10-\lambda) + 48 \amp {}={} 0 \\
\lambda^2-6\lambda + 8 {}={} 0 \\
(\lambda-4)(\lambda-2) {}={} 0\text{.} \\
\end{aligned}
\end{equation*}

This shows us that the eigenvalues are \(\lambda = 4\) and \(\lambda=2\text{.}\)

In general, the expression \(\det(A-\lambda I)\) is a polynomial in \(\lambda\text{,}\) which is called the *characteristic polynomial* of \(A\text{.}\) If \(A\) is an \(n\times n\) matrix, the degree of the characteristic polynomial is \(n\text{.}\) For instance, if \(A\) is a \(2\times2\) matrix, then \(\det(A-\lambda I)\) is a quadratic polynomial; if \(A\) is a \(3\times3\) matrix, then \(\det(A-\lambda I)\) 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 \(A=\begin{bmatrix}
5 \amp -1\\
4 \amp 1 \\
\end{bmatrix}
\text{,}\) whose characteristic equation is

\begin{equation*}
\lambda^2 - 6\lambda + 9 = (\lambda-3)^2 = 0.
\end{equation*}

In this case, the characteristic polynomial has one real root, which means that this matrix has a single real eigenvalue, \(\lambda = 3\text{.}\)

####
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

\begin{equation*}
\begin{aligned}
\det\left(
\left[\begin{array}{rrr}
4 \amp 2 \amp 3 \\
0 \amp -2 \amp -1 \\
0 \amp 0 \amp 3 \\
\end{array}\right]
-\lambda I\right) \amp {}={}
\det
\left[\begin{array}{rrr}
4-\lambda \amp 2 \amp 3 \\
0 \amp -2-\lambda \amp -1 \\
0 \amp 0 \amp 3-\lambda \\
\end{array}\right] \\
\\
\amp {}={}(4-\lambda)(-2-\lambda)(3-\lambda) = 0\text{,}
\end{aligned}
\end{equation*}

showing that the eigenvalues are the diagonal entries \(\lambda =
4,-2,3\text{.}\)

###
Subsection 4.2.2 Finding eigenvectors

Now that we can find the eigenvalues of a square matrix \(A\) by solving the characteristic equation \(\det(A-\lambda
I) = 0\text{,}\) we will turn to the question of finding the eigenvectors associated to an eigenvalue \(\lambda\text{.}\) The key, as before, is to note that an eigenvector is a nonzero solution to the homogeneous equation \((A-\lambda I)\vvec =
\zerovec\text{.}\) In other words, the eigenvectors associated to an eigenvalue \(\lambda\) form the null space \(\nul(A-\lambda
I)\text{.}\)

This shows that the eigenvectors associated to an eigenvalue form a subspace of \(\real^n\text{.}\) We will denote the subspace of eigenvectors of a matrix \(A\) associated to the eigenvalue \(\lambda\) by \(E_\lambda\) and note that

\begin{equation*}
E_\lambda = \nul(A-\lambda I)\text{.}
\end{equation*}

We say that \(E_\lambda\) is the *eigenspace* of \(A\) associated to the eigenvalue \(\lambda\text{.}\)

####
Activity 4.2.3.

In this activity, we will find the eigenvectors of a matrix as the null space of the matrix \(A-\lambda I\text{.}\)

Let’s begin with the matrix \(A = \left[\begin{array}{rr}
1 \amp 2 \\
2 \amp 1 \\
\end{array}\right]
\text{.}\) We have seen that \(\lambda = 3\) is an eigenvalue. Form the matrix \(A-3I\) and find a basis for the eigenspace \(E_3 = \nul(A-3I)\text{.}\) What is the dimension of this eigenspace? For each of the basis vectors \(\vvec\text{,}\) verify that \(A\vvec = 3\vvec\text{.}\)

We also saw that \(\lambda = -1\) is an eigenvalue. Form the matrix \(A-(-1)I\) and find a basis for the eigenspace \(E_{-1}\text{.}\) What is the dimension of this eigenspace? For each of the basis vectors \(\vvec\text{,}\) verify that \(A\vvec = -\vvec\text{.}\)

Is it possible to form a basis of \(\real^2\) consisting of eigenvectors of \(A\text{?}\)

Now consider the matrix \(A = \left[\begin{array}{rr}
3 \amp 0 \\
0 \amp 3 \\
\end{array}\right]
\text{.}\) Write the characteristic equation for \(A\) and use it to find the eigenvalues of \(A\text{.}\) For each eigenvalue, find a basis for its eigenspace \(E_\lambda\text{.}\) Is it possible to form a basis of \(\real^2\) consisting of eigenvectors of \(A\text{?}\)

Next, consider the matrix \(A = \left[\begin{array}{rr}
2 \amp 1 \\
0 \amp 2 \\
\end{array}\right]
\text{.}\) Write the characteristic equation for \(A\) and use it to find the eigenvalues of \(A\text{.}\) For each eigenvalue, find a basis for its eigenspace \(E_\lambda\text{.}\) Is it possible to form a basis of \(\real^2\) consisting of eigenvectors of \(A\text{?}\)

Finally, find the eigenvalues and eigenvectors of the diagonal matrix \(A = \left[\begin{array}{rr}
4 \amp 0 \\
0 \amp -1 \\
\end{array}\right]
\text{.}\) Explain your result by considering the geometric effect of the matrix transformation defined by \(A\text{.}\)

Once we find an eigenvalue of a matrix \(A\text{,}\) describing the associated eigenspace \(E_\lambda\) amounts to the familiar task of describing the null space \(\nul(A-\lambda I)\text{.}\)

####
Example 4.2.4.

Revisiting the matrix

\(A = \left[\begin{array}{rr}
-4 \amp 4 \\
-12 \amp 10 \\
\end{array}\right]\) from

Example 4.2.1, we recall that we found eigenvalues

\(\lambda = 4\) and

\(\lambda =
2\text{.}\)
Considering the eigenvalue \(\lambda = 4\text{,}\) we have

\begin{equation*}
A -4I =
\begin{bmatrix}
-8 \amp 4 \\
-12 \amp 6 \\
\end{bmatrix}
\sim
\begin{bmatrix}
1 \amp -1/2 \\
0 \amp 0 \\
\end{bmatrix}.
\end{equation*}

Since the eigenvectors \(\vvec = \twovec{v_1}{v_2}\) are the solutions of the equation \((A-4I)\vvec=\zerovec\text{,}\) we see that they are determined by the single equation \(v_1-\frac12 v_2 = 0\) or \(v_1 =
\frac12v_2\text{.}\) Therefore the eigenvectors in \(E_4\) have the form

\begin{equation*}
\vvec=\twovec{v_1}{v_2} = \ctwovec{\frac12 v_2}{v_2} =
v_2\ctwovec{1/2}{1}.
\end{equation*}

In other words, \(E_4\) is a one-dimensional subspace of \(\real^2\) with basis vector \(\ctwovec{1/2}{1}\) or basis vector \(\twovec12\text{.}\) In the same way, we find that a basis for the eigenspace \(E_2\) is \(\twovec23\text{.}\)

We note that, for this matrix, it is possible to construct a basis of \(\real^2\) consisting of eigenvectors, namely,

\begin{equation*}
\bcal = \left\{\twovec12, \twovec23\right\}.
\end{equation*}

####
Example 4.2.5.

Consider the matrix \(A=\begin{bmatrix}
1 \amp 1 \\
-1 \amp 3 \\
\end{bmatrix}\) whose characteristic equation is

\begin{equation*}
\det(A-\lambda I) = \lambda^2 - 4\lambda + 4 = (\lambda-2)^2
=0.
\end{equation*}

There is a single eigenvalue \(\lambda= 2\text{,}\) and we find that

\begin{equation*}
A - 2\lambda = \begin{bmatrix}
-1 \amp 1 \\
-1 \amp 1 \\
\end{bmatrix}
\sim
\begin{bmatrix}
1 \amp -1 \\
0 \amp 0 \\
\end{bmatrix}.
\end{equation*}

Therefore, the eigenspace \(E_2 = \nul(A-2I)\) is one-dimensional with a basis vector \(\twovec11\text{.}\)

####
Example 4.2.6.

If \(A=\begin{bmatrix}
-1 \amp 0 \\
0 \amp -1 \\
\end{bmatrix}
\text{,}\) then

\begin{equation*}
\det(A - \lambda I) = (\lambda+1)^2 = 0,
\end{equation*}

which implies that there is a single eigenvalue \(\lambda=-1\text{.}\) We find that

\begin{equation*}
A-(-1)I = \begin{bmatrix}
0 \amp 0 \\
0 \amp 0 \\
\end{bmatrix},
\end{equation*}

which says that every two-dimensional vector \(\vvec\) satisfies \((A-(-1)I)\vvec = \zerovec\text{.}\) Therefore, every vector is an eigenvector and so \(E_{-1} =
\real^2\text{.}\) This eigenspace is two-dimensional.

We can see this in another way. The matrix transformation defined by \(A\) rotates vectors by \(180^\circ\text{,}\) which says that \(A\xvec = -\xvec\) for every vector \(\xvec\text{.}\) In other words, every two-dimensional vector is an eigenvector with associated eigenvalue \(\lambda=-1\text{.}\)

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 \(\real^2\) 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 \(\lambda\) by solving the characteristic equation. If \(\lambda\) is a solution to the characteristic equation, then \(A-\lambda I\) is not invertible and, consequently, \(A-\lambda I\) must contain a row without a pivot position.

This serves as a check on our work. If we row reduce \(A-\lambda I\) and find the identity matrix, then we have made an error either in solving the characteristic equation or in finding \(\nul(A-\lambda I)\text{.}\)

###
Subsection 4.2.3 The characteristic polynomial and the dimension of eigenspaces

Given a square

\(n\times n\) matrix

\(A\text{,}\) we saw in the previous section the value of being able to express any vector in

\(\real^n\) as a linear combination of eigenvectors of

\(A\text{.}\) For this reason,

Question 4.1.8 asks when we can construct a basis of

\(\real^n\) consisting of eigenvectors. We will explore this question more fully now.

As we saw above, the eigenvalues of \(A\) are the solutions of the characteristic equation \(\det(A-\lambda I) = 0\text{.}\) The examples we have considered demonstrate some different types of behavior. For instance, we have seen the characteristic equations

\((4-\lambda)(-2-\lambda)(3-\lambda) = 0\text{,}\) which has real and distinct roots,

\((2-\lambda)^2 = 0\text{,}\) which has repeated roots, and

\(\lambda^2 +1 = (i-\lambda)(-i-\lambda) = 0\text{,}\) which has complex roots.

If \(A\) is an \(n\times n\) matrix, then the characteristic polynomial is a degree \(n\) polynomial, and this means that it has \(n\) roots. Therefore, the characteristic equation can be written as

\begin{equation*}
\det(A-\lambda I) = (\lambda_1 -
\lambda)(\lambda_2-\lambda)\ldots(\lambda_n-\lambda) = 0
\end{equation*}

giving eigenvalues \(\lambda_1,
\lambda_2,\ldots,\lambda_n\text{.}\) 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

\begin{equation*}
(\lambda_1-\lambda)^{m_1}
(\lambda_2-\lambda)^{m_2}
\ldots
(\lambda_p-\lambda)^{m_p}
= 0.
\end{equation*}

The number of times that \(\lambda_j - \lambda\) appears as a factor in the characteristic polynomial, is called the *multiplicity* of the eigenvalue \(\lambda_j\text{.}\)

####
Example 4.2.7.

We have seen that the matrix \(A = \left[\begin{array}{rr}
1 \amp 1 \\
-1 \amp 3 \\
\end{array}\right]\) has the characteristic equation \((2-\lambda)^2 =
0\text{.}\) This matrix has a single eigenvalue \(\lambda = 2\text{,}\) which has multiplicity \(2\text{.}\)

####
Example 4.2.8.

If a matrix has the characteristic equation

\begin{equation*}
(4-\lambda)^2(-5-\lambda)(1-\lambda)^7(3-\lambda)^2 = 0\text{,}
\end{equation*}

then that matrix has four eigenvalues: \(\lambda=4\) having multiplicity 2; \(\lambda=-5\) having multiplicity 1; \(\lambda=1\) having multiplicity 7; and \(\lambda=3\) having multiplicity 2. The degree of the characteristic polynomial is the sum of the multiplicities \(2+1+7+2 =
12\) so this matrix must be a \(12\times12\) 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 \(\lambda\) is a real eigenvalue of the matrix \(A\) with multiplicity \(m\text{,}\) then

\begin{equation*}
1 \leq \dim~E_\lambda \leq m\text{.}
\end{equation*}

####
Example 4.2.10.

The diagonal matrix \(\left[\begin{array}{rr}
-1 \amp 0 \\
0 \amp -1 \\
\end{array}\right]\) has the characteristic equation \((-1-\lambda)^2 =
0\text{.}\) There is a single eigenvalue \(\lambda = -1\) having multiplicity \(m = 2\text{,}\) and we saw earlier that \(\dim E_{-1} = 2 \leq m = 2\text{.}\)

####
Example 4.2.11.

The matrix \(\left[\begin{array}{rr}
1 \amp 1 \\
-1 \amp 3 \\
\end{array}\right]\) has the characteristic equation \((2-\lambda)^2 =
0\text{.}\) This tells us that there is a single eigenvalue \(\lambda =
2\) having multiplicity \(m = 2\text{.}\) In contrast with the previous example, we have \(\dim~ E_2 = 1 \leq m = 2\text{.}\)

####
Example 4.2.12.

We saw earlier that the matrix \(\left[\begin{array}{rrr}
4 \amp 2 \amp 3 \\
0 \amp -2 \amp -1 \\
0 \amp 0 \amp 3 \\
\end{array}\right]\) has the characteristic equation

\begin{equation*}
(4-\lambda)(-2-\lambda)(3-\lambda)=0\text{.}
\end{equation*}

There are three eigenvalues \(\lambda=3,-2,1\) each having multiplicity \(1\text{.}\) By the proposition, we are guaranteed that the dimension of each eigenspace is \(1\text{;}\) that is,

\begin{equation*}
\dim E_3 = \dim E_{-2} = \dim E_1 = 1\text{.}
\end{equation*}

It turns out that this is enough to guarantee that there is a basis of \(\real^3\) consisting of eigenvectors.

####
Example 4.2.13.

If a \(12\times12\) matrix has the characteristic equation

\begin{equation*}
(4-\lambda)^2(-5-\lambda)(1-\lambda)^7(3-\lambda)^2 = 0\text{,}
\end{equation*}

we know there are four eigenvalues \(\lambda=4,-5,1,3\text{.}\) Without more information, all we can say about the dimensions of the eigenspaces is

\begin{equation*}
\begin{aligned}
1 \leq \dim E_4 \amp {}\leq{} 2 \\
1 \leq \dim E_{-5} \amp {}\leq{} 1 \\
1 \leq \dim E_1 \amp {}\leq{} 7 \\
1 \leq \dim E_3 \amp {}\leq{} 2\text{.} \\
\end{aligned}
\end{equation*}

We can guarantee that \(\dim E_{-5} = 1\text{,}\) but we cannot be more specific about the dimensions of the other eigenspaces.

Fortunately, if we have an \(n\times n\) matrix, it frequently happens that the characteristic equation has the form

\begin{equation*}
(\lambda_1-\lambda)
(\lambda_2-\lambda)
\ldots
(\lambda_n-\lambda) = 0
\end{equation*}

where there are \(n\) distinct real eigenvalues, each of which has multiplicity \(1\text{.}\) In this case, the dimension of each of the eigenspaces \(\dim E_{\lambda_j} = 1\text{.}\) With a little work, it can be seen that choosing a basis vector \(\vvec_j\) for each of the eigenspaces produces a basis for \(\real^n\text{.}\) We therefore have the following proposition.

####
Proposition 4.2.14.

If \(A\) is an \(n\times n\) matrix having \(n\) distinct real eigenvalues, then there is a basis of \(\real^n\) consisting of eigenvectors of \(A\text{.}\)

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 \(n\times n\) matrix whose characteristic polynomial is \((2-\lambda)^3(-3-\lambda)^{10}(5-\lambda)\text{.}\) 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 \(\real^n\) consisting of eigenvectors?

Find the eigenvalues of \(\left[\begin{array}{rr}
0 \amp -1 \\
4 \amp -4 \\
\end{array}\right]\) and state their multiplicities. Can you find a basis of \(\real^2\) consisting of eigenvectors of this matrix?

Consider the matrix \(A =
\left[\begin{array}{rrr}
-1 \amp 0 \amp 2 \\
-2 \amp -2 \amp -4 \\
0 \amp 0 \amp -2 \\
\end{array}\right]\) whose characteristic equation is

\begin{equation*}
(-2-\lambda)^2(-1-\lambda) = 0\text{.}
\end{equation*}

Identify the eigenvalues and their multiplicities.

For each eigenvalue \(\lambda\text{,}\) find a basis of the eigenspace \(E_\lambda\) and state its dimension.

Is there a basis of \(\real^3\) consisting of eigenvectors of \(A\text{?}\)

Now consider the matrix \(A =
\left[\begin{array}{rrr}
-5 \amp -2 \amp -6 \\
-2 \amp -2 \amp -4 \\
2 \amp 1 \amp 2 \\
\end{array}\right]\) whose characteristic equation is also

\begin{equation*}
(-2-\lambda)^2(-1-\lambda) = 0\text{.}
\end{equation*}

Identify the eigenvalues and their multiplicities.

For each eigenvalue \(\lambda\text{,}\) find a basis of the eigenspace \(E_\lambda\) and state its dimension.

Is there a basis of \(\real^3\) consisting of eigenvectors of \(A\text{?}\)

Consider the matrix \(A =
\left[\begin{array}{rrr}
-5 \amp -2 \amp -6 \\
4 \amp 1 \amp 8 \\
2 \amp 1 \amp 2 \\
\end{array}\right]\) whose characteristic equation is

\begin{equation*}
(-2-\lambda)(1-\lambda)(-1-\lambda) = 0\text{.}
\end{equation*}

Identify the eigenvalues and their multiplicities.

For each eigenvalue \(\lambda\text{,}\) find a basis of the eigenspace \(E_\lambda\) and state its dimension.

Is there a basis of \(\real^3\) consisting of eigenvectors of \(A\text{?}\)