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6 Eigenvalues and Eigenvectors

6.1 Introduction to Eigenvalues

Given a square matrix A, let us pose the following question: Does there exist a nonzero vector v such that the result Av of multiplying v by the matrix A is simply a scalar multiple of v? Thus we ask whether or not there exist a nonzero vector v and a scalar λ $λ$ such that

A v = λ v .

$Av = λ v .$ (1)

Section 6.3 and subsequent chapters include interesting applications in which this question arises. The following definition provides the terminology that we use in discussing Eq. (1).

The prefix eigen is a German word that (in some contexts) may be translated as proper, so eigenvalues are sometimes called proper values. Eigenvalues and eigenvectors are also called characteristic values and characteristic vectors in some books.

Example 1

Consider the 2×2 $2 \times 2$ matrix

A = [52 - 6 - 2] .

$A = [\begin{matrix} 5 & - 6 \\ 2 & - 2 \end{matrix}] .$

If v=(2,1)=[21]T, $v = (2, 1) = {[\begin{matrix} 2 & 1 \end{matrix}]}^{T},$ then

A v = [52 - 6 - 2] [21] = [42] = 2 v .

$Av = [\begin{matrix} 5 & - 6 \\ 2 & - 2 \end{matrix}] [\begin{matrix} 2 \\ 1 \end{matrix}] = [\begin{matrix} 4 \\ 2 \end{matrix}] = 2 v .$

Thus, v=[21]T $v = {[\begin{matrix} 2 & 1 \end{matrix}]}^{T}$ is an eigenvector of A that is associated with the eigenvalue λ=2 $λ = 2$ . If v=[32]T, $v = {[\begin{matrix} 3 & 2 \end{matrix}]}^{T},$ then

A v = [52 - 6 - 2] [32] = [32] = 1 v,

$Av = [\begin{matrix} 5 & - 6 \\ 2 & - 2 \end{matrix}] [\begin{matrix} 3 \\ 2 \end{matrix}] = [\begin{matrix} 3 \\ 2 \end{matrix}] = 1 v,$

so v=[32]T $v = {[\begin{matrix} 3 & 2 \end{matrix}]}^{T}$ is an eigenvector of A associated with the eigenvalue λ=1 $λ = 1$ . In summary, we see that the scalars λ1=2 $λ_{1} = 2$ and λ2=1 $λ_{2} = 1$ are both eigenvalues of the matrix A they correspond to the eigenvectors v1=[21]T $v_{1} = {[\begin{matrix} 2 & 1 \end{matrix}]}^{T}$ and v2=[32]T, $v_{2} = {[\begin{matrix} 3 & 2 \end{matrix}]}^{T},$ respectively.

Eigenvalues and eigenvectors have a simple geometric interpretation. Suppose that λ $λ$ is an eigenvalue of the matrix A and has the associated eigenvector v, so that Av=λv. $Av = λ v .$ Then the length |Av| $| Av |$ of the vector Av is ±λ|v|, $\pm λ | v |,$ depending on the sign of λ $λ$ . Thus, if λ>0 $λ > 0$ , then multiplication of v by the matrix A expands or contracts the vector v while preserving its direction; if λ<0 $λ < 0$ , then multiplication of v by A reverses the direction of v (see Fig. 6.1.1).

FIGURE 6.1.1.

(a) A positive eigenvalue; (b) a negative eigenvalue.

Remark 1

If v=0, $v = 0,$ then the equation Av=λv $Av = λ v$ holds for every scalar λ $λ$ and hence is of no significance. This is why only nonzero vectors qualify as eigenvectors in the definition.

Remark 2

Let λ $λ$ and v be an eigenvalue and associated eigenvector of the matrix A. If k is any nonzero scalar and u=kv, $u = k v,$ then

A u = A (k v) = k (A v) = k (λ v) = λ (k v) = λ u,

$A u = A (k v) = k (Av) = k (λ v) = λ (k v) = λ u,$

so u=kv $u = k v$ is also an eigenvector associated with λ $λ$ . Thus, any nonzero scalar multiple of an eigenvector is also an eigenvector and is associated with the same eigenvalue. In Example 1, for instance, u1=−3v1=[−6−3]T $u_{1} = - 3 v_{1} = {[\begin{matrix} - 6 & - 3 \end{matrix}]}^{T}$ is an eigenvector associated with λ1=2 $λ_{1} = 2$ , and u2=4v2=[128]T $u_{2} = 4 v_{2} = {[\begin{matrix} 12 & 8 \end{matrix}]}^{T}$ is an eigenvector associated with λ2=1 $λ_{2} = 1$ .

The Characteristic Equation

We now attack the problem of finding the eigenvalues and eigenvectors of a given n×n $n \times n$ square matrix A. According to the definition, the nonzero vector v is an eigenvector of A associated with the eigenvalue λ $λ$ exactly when

A v = λ v = λ I v;

$Av = λ v = λ I v;$

that is, when

(A - λ I) v = 0 .

$(A - λ I) v = 0 .$ (2)

For a fixed value of λ $λ$ , Eq. (2) is a homogeneous linear system of n equations in the n components of v. By Theorem 2 in Section 3.6 and Theorem 7 in Section 3.5, this system has a nontrivial solution v≠0 $v \neq 0$ if and only if the determinant

det (A - λ I) = | A - λ I |

$\det (A - λ I) = | A - λ I |$

of its coefficient matrix is zero. The equation |A−λI|=0 $| A - λ I | = 0$ is called the characteristic equation of the square matrix A, and we have proved that there exists an eigenvector v $v$ associated with λ $λ$ if and only if λ $λ$ satisfies this equation.

Now let us see just what sort of equation the characteristic equation in (3) is. Note that the matrix

A - λ I = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ a 11 - λ a 21 ⋮ a n 1 a 12 a 22 - λ ⋮ a n 2 \dots \dots ⋱ \dots a 1 n a 2 n ⋮ a n n - λ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥

$A - λ I = [\begin{matrix} a_{11} - λ & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} - λ & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} - λ \end{matrix}]$ (4)

is obtained simply by subtracting λ $λ$ from each diagonal element of A. If we think of expanding the determinant by cofactors, we see that |A−λI| $| A - λ I |$ is a polynomial in the variable λ $λ$ , and that the highest power of λ $λ$ comes from the product of the diagonal elements of the matrix in (4). Therefore, the characteristic equation of the n×n $n \times n$ matrix A takes the form

(- 1) n λ n + c n - 1 λ n - 1 + \dots + c 1 λ + c 0 = 0,

${(- 1)}^{n} λ^{n} + c_{n - 1} λ^{n - 1} + \dots + c_{1} λ + c_{0} = 0,$ (5)

an nth-degree polynomial equation in λ $λ$ .

According to the fundamental theorem of algebra, every nth-degree polynomial equation in one variable has n solutions (counting multiple solutions), but some of them can be complex. Hence, we can say that an n×n $n \times n$ matrix A always has n eigenvalues, though they might not be distinct and might not all be real. In this chapter we confine our attention mainly to real eigenvalues, but in Chapter 7 we will see important applications of complex eigenvalues to the solution of linear systems of differential equations.

Solving the characteristic equation is almost always easier said than done. In the examples that follow and in the problems, we have chosen matrices for which the characteristic polynomial |A−λI| $| A - λ I |$ readily factors to reveal the eigenvalues.

Example 2

Find the eigenvalues and associated eigenvectors of the matrix

A = [5 - 2 7 - 4] .

$A = [\begin{array}{r} 5 & 7 \\ - 2 & - 4 \end{array}] .$

Solution

Here

A - λ I = [5 - λ - 2 7 - 4 - λ],

$A - λ I = [\begin{array}{c} 5 - λ & 7 \\ - 2 & - 4 - λ \end{array}],$ (6)

so the characteristic equation of A is

0 = = = = | A - λ I | ∣ ∣ ∣ 5 - λ - 2 7 - 4 - λ ∣ ∣ ∣ (5 - λ) (- 4 - λ) - (- 2) (7) (λ - 5) (λ + 4) + 14 = λ 2 - λ - 6;

$\begin{array}{rcl} 0 & = & | A - λ I | \\ = & | \begin{array}{c} 5 - λ & 7 \\ - 2 & - 4 - λ \end{array} | \\ = & (5 - λ) (- 4 - λ) - (- 2) (7) \\ = & (λ - 5) (λ + 4) + 14 = λ^{2} - λ - 6; \end{array}$

that is, (λ+2)(λ−3)=0 $(λ + 2) (λ - 3) = 0$ . Thus, the matrix A has the two eigenvalues −2 $- 2$ and 3. To distinguish them, we write λ1=−2 $λ_{1} = - 2$ and λ2=3 $λ_{2} = 3$ . To find the associated eigenvectors, we must separately substitute each eigenvalue in (6) and then solve the resulting system (A−λI)v=0 $(A - λ I) v = 0$ .

Case 1: λ1=−2 $λ_{1} = - 2$ . With v=[xy]T, $v = {[\begin{array}{l} x & y \end{array}]}^{T},$ the system (A−λI)v=0 $(A - λ I) v = 0$ is

[7 - 2 7 - 2] [x y] = [00] .

$[\begin{array}{r} 7 & 7 \\ - 2 & - 2 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{l} 0 \\ 0 \end{array}] .$

Each of the two scalar equations here is a multiple of the equation x+y=0 $x + y = 0$ , and any nontrivial solution v=[xy]T $v = {[\begin{array}{l} x & y \end{array}]}^{T}$ of this equation is a nonzero multiple of [1−1]T. ${[\begin{array}{l} 1 & - 1 \end{array}]}^{T} .$ Hence, to within a constant multiple, the only eigenvector associated with λ1=−2 $λ_{1} = - 2$ is v1=[1−1]T $v_{1} = {[\begin{array}{l} 1 & - 1 \end{array}]}^{T}$ .

Case 2: λ2=3 $λ_{2} = 3$ . With v=[xy]T, $v = {[\begin{array}{l} x & y \end{array}]}^{T},$ the system (A−λI)v=0 $(A - λ I) v = 0$ is

[2 - 2 7 - 7] [x y] = [00] .

$[\begin{array}{r} 2 & 7 \\ - 2 & - 7 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{l} 0 \\ 0 \end{array}] .$

Again, we have only a single equation, 2x+7y=0 $2 x + 7 y = 0$ , and any nontrivial solution of this equation will suffice. The choice y=−2 $y = - 2$ yields x=7 $x = 7$ , so (to within a constant multiple) the only eigenvector associated with λ2=3 $λ_{2} = 3$ is v2=[7−2]T $v_{2} = {[\begin{array}{l} 7 & - 2 \end{array}]}^{T}$ .

Finally, note that it is not enough to say simply that the given matrix A has eigenvalues −2 $- 2$ and 3 and has eigenvectors [1−1]T ${[\begin{array}{l} 1 & - 1 \end{array}]}^{T}$ and [7−2]T. ${[\begin{array}{l} 7 & - 2 \end{array}]}^{T} .$ To give complete information, we must say which eigenvector is associated with each eigenvalue.

If the elements of the matrix A are all real, then so are the coefficients of its characteristic equation |A−λI|=0. $| A - λ I | = 0 .$ In this event, it follows that the complex eigenvalues (if any) of A occur only in conjugate pairs. The next example illustrates the possibility of imaginary or complex eigenvalues and eigenvectors.

Example 3

The characteristic equation of the matrix

A = [0 - 2 80]

$A = [\begin{aligned} 0 & 8 \\ - 2 & 0 \end{aligned}]$

| A - λ I | = [- λ - 2 8 - λ] = λ 2 + 16 = 0 .

$| A - λ I | = [\begin{array}{c} - λ & 8 \\ - 2 & - λ \end{array}] = λ^{2} + 16 = 0 .$

Hence the matrix A has the complex conjugate eigenvalues λ=±4i $λ = \pm 4 i$ .

Case 1: λ1=−4i $λ_{1} = - 4 i$ . With v=[xy]T, $v = {[\begin{array}{l} x & y \end{array}]}^{T},$ the system (A−λI)v=0 $(A - λ I) v = 0$ is

[4 i - 2 8 4 i] [x y] = [00] .

$[\begin{array}{lr} 4 i & 8 \\ - 2 & 4 i \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{l} 0 \\ 0 \end{array}] .$

The first equation here, 4ix+8y=0 $4 i x + 8 y = 0$ , is −2i $- 2 i$ times the second one and, obviously, is satisfied by x=2i, y=1 $x = 2 i, y = 1$ . Thus v1=[2i1]T $v_{1} = {[\begin{array}{l} 2 i & 1 \end{array}]}^{T}$ is a complex eigenvector associated with the complex eigenvalue λ1=−4i $λ_{1} = - 4 i$ .

Case 2: λ2=+4i $λ_{2} = + 4 i$ . In this case, you may verify similarly that the conjugate v2=[−2i1]T $v_{2} = {[- \begin{array}{l} 2 i & 1 \end{array}]}^{T}$ of v1 $v_{1}$ is an eigenvector associated with the conjugate λ2 $λ_{2}$ of the eigenvalue λ1 $λ_{1}$ . Thus the complex conjugate eigenvalues ∓4i $\mp 4 i$ of the matrix A correspond to the complex conjugate eigenvectors [±2i1]T ${[\begin{array}{l} \pm 2 i & 1 \end{array}]}^{T}$ (taking either both upper signs or both lower signs).

Example 4

The 2×2 $2 \times 2$ identity matrix I has characteristic equation

∣ ∣ ∣ 1 - λ 0 0 1 - λ ∣ ∣ ∣ = (1 - λ) 2 = 0,

$| \begin{array}{c} 1 - λ & 0 \\ 0 & 1 - λ \end{array} | = {(1 - λ)}^{2} = 0,$

so I has the single eigenvalue λ=1 $λ = 1$ . The equation (I−1λ)v=0 $(I - 1 λ) v = 0$ is

[0000] [x y] = [00],

$[\begin{array}{r} 0 & 0 \\ 0 & 0 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{l} 0 \\ 0 \end{array}],$

so every nonzero vector v=[xy]T $v = {[\begin{array}{r} x & y \end{array}]}^{T}$ is an eigenvector of I. In particular, the single eigenvalue λ=1 $λ = 1$ corresponds to the two linearly independent eigenvectors v1=[10]T $v_{1} = {[\begin{array}{l} 1 & 0 \end{array}]}^{T}$ and v2=[01]T. $v_{2} = {[\begin{array}{l} 0 & 1 \end{array}]}^{T} .$

Example 5

The characteristic equation of the matrix

A = [2032]

$A = [\begin{array}{r} 2 & 3 \\ 0 & 2 \end{array}]$

is (2−λ)2=0 ${(2 - λ)}^{2} = 0$ , so A has the single eigenvalue λ=2 $λ = 2$ . The equation (A−2I)v=0 $(A - 2 I) v = 0$ is

[0030] [x y] = [00] .

$[\begin{array}{r} 0 & 3 \\ 0 & 0 \end{array}] [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{l} 0 \\ 0 \end{array}] .$

Thus x is arbitrary, but $y = 0$ , so the eigenvalue $λ = 2$ corresponds (to within a constant multiple) to the single eigenvector $v = {[\begin{array}{l} 1 & 0 \end{array}]}^{T} .$

Examples 2–5 illustrate the four possibilities for a $2 \times 2$ matrix A. It can have either

two distinct real eigenvalues, each corresponding to a single eigenvector;
a single real eigenvalue corresponding to a single eigenvector;
a single real eigenvalue corresponding to two linearly independent eigenvectors; or
two complex conjugate eigenvalues corresponding to complex conjugate eigenvectors.

The characteristic equation of a $3 \times 3$ matrix is, by Eq. (5), of the form

$- λ^{3} + c_{2} λ^{2} + c_{1} λ + c_{0} = 0 .$ (7)

Now the value of the continuous function $- λ^{3} + \dots + c_{0}$ on the left-hand side here approaches $+ \infty$ as $λ \to - \infty$ and $- \infty$ as $λ \to + \infty$ . Hence every such cubic equation has at least one real solution, so every $3 \times 3$ matrix has (in contrast with $2 \times 2$ matrices) at least one real eigenvalue. A $3 \times 3$ matrix may have one, two, or three distinct eigenvalues, and a single eigenvalue of a $3 \times 3$ matrix may correspond to one, two, or three linearly independent eigenvectors. The remaining two examples of this section illustrate some of the possibilities. The next example shows also that, whereas the zero vector 0 cannot be an eigenvector of a matrix, there is nothing to prevent $λ = 0$ from being an eigenvalue.

Example 6

Find the eigenvalues and associated eigenvectors of the matrix

$A = [\begin{array}{r} 3 & 0 & 0 \\ - 4 & 6 & 2 \\ 16 & - 15 & - 5 \end{array}] .$

Solution

The matrix $A - λ I$ is

$A - λ I = [\begin{array}{c} 3 - λ & 0 & 0 \\ - 4 & 6 - λ & 2 \\ 16 & - 15 & - 5 - λ \end{array}] .$ (8)

Upon expansion of the determinant along its first row, we find that

$\begin{array}{rcl} | A - λ I | & = & (3 - λ) [(6 - λ) (- 5 - λ) + 30] \\ = & (3 - λ) (λ^{2} - λ) = λ (λ - 1) (3 - λ) . \end{array}$

Hence the characteristic equation $| A - λ I | = 0$ yields the three eigenvalues $λ_{1} = 0, λ_{2} = 1$ , and $λ_{3} = 3$ . To find the associated eigenvectors, we must solve the system $(A - λ I) v = 0$ separately for each of these three eigenvalues.

Case 1: $λ_{1} = 0$ . We write $v = {[\begin{array}{l} x & y & z \end{array}]}^{T}$ and substitute $λ = 0$ in the coefficient matrix in (8) to obtain the system

$[\begin{array}{r} 3 & 0 & 0 \\ - 4 & 6 & 2 \\ 16 & - 15 & - 5 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}] .$

From the first of the three equations here, $3 x = 0$ , we see that $x = 0$ . Then each of the remaining two equations is a multiple of the equation $3 y + z = 0$ . The choice $y = 1$ yields $z = - 3$ . Thus the eigenvector $v_{1} = {[\begin{array}{l} 0 & 1 & - 3 \end{array}]}^{T}$ is associated with $λ_{1} = 0$ .

Case 2: $λ_{2} = 1$ . Substitution of $λ = 1$ in the coefficient matrix in (8) yields the system

$[\begin{array}{r} 2 & 0 & 0 \\ - 4 & 5 & 2 \\ 16 & - 15 & - 6 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}]$

for $v = {[\begin{array}{l} x & y & z \end{array}]}^{T} .$ The first equation $2 x = 0$ implies that $x = 0$ . Then the third equation is a multiple of the second equation, $5 y + 2 z = 0$ . The choice of $y = 2$ yields $z = - 5$ , so the eigenvector $v_{2} = {[\begin{array}{l} 0 & 2 & - 5 \end{array}]}^{T}$ is associated with $λ_{2} = 1$ .

Case 3: $λ_{3} = 3$ . Substitution of $λ = 3$ in the coefficient matrix in (8) yields the system

$[\begin{array}{r} 0 & 0 & 0 \\ - 4 & 3 & 2 \\ 16 & - 15 & - 8 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}] .$

In this case, the first equation yields no information, but the result of adding 4 times the second equation to the third equation is $- 3 y = 0$ , so $y = 0$ . Consequently, the second and third equations are both multiples of the equation $2 x - z = 0$ . The choice $x = 1$ yields $z = 2$ , so the eigenvector $v_{3} = {[\begin{array}{l} 1 & 0 & 2 \end{array}]}^{T}$ is associated with $λ_{3} = 3$ .

In summary, we have found the eigenvectors $v_{1} = {[\begin{array}{l} 0 & 1 & - 3 \end{array}]}^{T},$ $v_{2} = {[\begin{array}{l} 0 & 2 & - 5 \end{array}]}^{T},$ and $v_{3} = {[\begin{array}{l} 1 & 0 & 2 \end{array}]}^{T}$ associated with the distinct eigenvalues $λ_{1} = 0, λ_{2} = 1$ , and $λ_{3} = 3$ , respectively.

Remark

Substitution of $λ = 0$ in the characteristic equation $| A - λ I | = 0$ yields $| A | = 0 .$ Therefore, $λ = 0$ is an eigenvalue of the matrix A if and only if A is singular: $| A | = 0 .$

Eigenspaces

Let $λ$ be a fixed eigenvalue of the $n \times n$ matrix A. Then the set of all eigenvectors associated with A is the set of all nonzero solution vectors of the system

$(A - λ I) v = 0 .$ (9)

The solution space of this system is called the eigenspace of A associated with the eigenvalue $λ$ . This subspace of $R^{n}$ consists of all eigenvectors associated with $λ$ together with the zero vector. In Example 6 we found (to within a constant multiple) only a single eigenvector associated with each eigenvalue $λ$ ; in this case, the eigenspace of $λ$ is 1-dimensional. In the case of an eigenspace of higher dimension, we generally want to find a basis for the solution space of Eq. (9).

Example 7

Find bases for the eigenspaces of the matrix

$A = [\begin{array}{r} 4 & - 2 & 1 \\ 2 & 0 & 1 \\ 2 & - 2 & 3 \end{array}] .$

Solution

Here we have

$A - λ I = [\begin{array}{c} 4 - λ & - 2 & 1 \\ 2 & - λ & 1 \\ 2 & - 2 & 3 - λ \end{array}] .$ (10)

We expand along the first row to obtain

$\begin{array}{rcl} | A - λ I | & = & (4 - λ) (λ^{2} - 3 λ + 2) - (- 2) (4 - 2 λ) + (1) (- 4 + 2 λ) \\ = & λ^{3} + 7 λ^{2} - 16 λ + 12 . \end{array}$

Thus, to find the eigenvalues we need to solve the cubic equation

$λ^{3} - 7 λ^{2} + 16 λ - 12 = 0 .$ (11)

We look (hopefully) for integer solutions. The factor theorem of algebra implies that if the polynomial equation

$λ^{n} + c_{n - 1} λ^{n - 1} + \dots + c_{1} λ + c_{0} = 0$

with integer coefficients and leading coefficient 1 has an integer solution, then that integer is a divisor of the constant $c_{0}$ . In the case of the cubic equation in (11), the possibilities for such a solution are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6$ , and $\pm 12$ . We substitute these numbers successively in (11) and thereby find that $+ 1$ and $- 1$ are not solutions but that $λ = + 2$ is a solution. Hence $λ - 2$ is a factor of the cubic polynomial in (11). Next, the long division

$\begin{array}{r} λ^{2} - 5 λ + 6 \\ λ - 2 λ^{3} - 7 λ^{2} + 16 λ - 12 \\ \underline{λ^{3} - 2 λ^{2}} \\ - 5 λ^{2} + 16 λ - 12 \\ \underline{- 5 λ^{2} + 10 λ} \\ 6 λ - 12 \\ \underline{6 λ - 12} \\ 0 \end{array}$

shows that

$\begin{array}{rcl} λ^{3} - 7 λ^{2} + 16 λ - 12 & = & (λ - 2) (λ^{2} - 5 λ + 6) \\ = & (λ - 2) (λ - 2) (λ - 3) \\ = & {(λ - 2)}^{2} (λ - 3) . \end{array}$

Thus we see finally that the given matrix A has the repeated eigenvalue $λ = 2$ and the eigenvalue $λ = 3$ .

Case 1: $λ = 2$ . The system $(A - λ I) v = 0$ is

$[\begin{array}{r} 2 & - 2 & 1 \\ 2 & - 2 & 1 \\ 2 & - 2 & 1 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}],$

which reduces to the single equation $2 x - 2 y + z = 0$ . This equation obviously has a 2-dimensional solution space. With $y = 1$ and $z = 0$ , we get $x = 1$ and, hence, obtain the basis eigenvector $v_{1} = {[\begin{array}{l} 1 & 1 & 0 \end{array}]}^{T} .$ With $y = 0$ and $z = 2$ , we get $x = - 1$ and, hence, the basis eigenvector $v_{2} = {[\begin{array}{l} - 1 & 0 & 2 \end{array}]}^{T} .$ The 2-dimensional eigenspace of A associated with the repeated eigenvalue $λ = 2$ has basis ${v_{1}, v_{2}}$ .

Case 2: $λ = 3$ . The system $(A - λ I) v = 0$ is

$[\begin{array}{r} 1 & - 2 & 1 \\ 2 & - 3 & 1 \\ 2 & - 2 & 0 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}] .$

The last equation here implies that $x = y$ , and then each of the first two equations yields $x = y = z$ . It follows that the eigenspace of A associated with $λ = 3$ is 1-dimensional and has $v_{3} = {[\begin{array}{l} 1 & 1 & 1 \end{array}]}^{T}$ as a basis eigenvector.

Remark

The typical higher-degree polynomial is not so easy to factor as the one in Example 7. Hence a numerical technique such as Newton’s method is often needed to solve the characteristic equation. Moreover, for an $n \times n$ matrix with n greater than about 4, the amount of labor required to find the characteristic equation by expanding the determinant $| A - λ I |$ is generally prohibitive; because of the presence of the variable $λ$ , row and column elimination methods do not work as they do with numerical determinants. Consequently, specialized techniques, beyond the scope of the present discussion, are often required to find the eigenvalues and eigenvectors of the large matrices that occur in many applications. Problems 40 and 41 at the end of this section outline a numerical technique that sometimes is useful with matrices of moderate size.

6.1 Problems

In Problems 1 through 26, find the (real) eigenvalues and associated eigenvectors of the given matrix A. Find a basis for each eigenspace of dimension 2 or larger.

$[\begin{array}{r} 4 & - 2 \\ 1 & 1 \end{array}]$
$[\begin{array}{r} 5 & - 6 \\ 3 & - 4 \end{array}]$
$[\begin{array}{r} 8 & - 6 \\ 3 & - 1 \end{array}]$
$[\begin{array}{r} 4 & - 3 \\ 2 & - 1 \end{array}]$
$[\begin{array}{r} 10 & - 9 \\ 6 & - 5 \end{array}]$
$[\begin{array}{r} 6 & - 4 \\ 3 & - 1 \end{array}]$
$[\begin{array}{r} 10 & - 8 \\ 6 & - 4 \end{array}]$
$[\begin{array}{r} 7 & - 6 \\ 12 & - 10 \end{array}]$
$[\begin{array}{r} 8 & - 10 \\ 2 & - 1 \end{array}]$
$[\begin{array}{r} 9 & - 10 \\ 2 & 0 \end{array}]$
$[\begin{array}{r} 19 & - 10 \\ 21 & - 10 \end{array}]$
$[\begin{array}{r} 13 & - 15 \\ 6 & - 6 \end{array}]$
$[\begin{array}{r} 2 & 0 & 0 \\ 2 & - 2 & - 1 \\ - 2 & 6 & 3 \end{array}]$
$[\begin{array}{r} 5 & 0 & 0 \\ 4 & - 4 & - 2 \\ - 2 & 12 & 6 \end{array}]$
$[\begin{array}{r} 2 & - 2 & 0 \\ 2 & - 2 & - 1 \\ - 2 & 2 & 3 \end{array}]$
$[\begin{array}{r} 1 & 0 & - 1 \\ - 2 & 3 & - 1 \\ - 6 & 6 & 0 \end{array}]$
$[\begin{array}{r} 3 & 5 & - 2 \\ 0 & 2 & 0 \\ 0 & 2 & 1 \end{array}]$
$[\begin{array}{r} 1 & 0 & 0 \\ - 6 & 8 & 2 \\ 12 & - 15 & - 3 \end{array}]$
$[\begin{array}{r} 3 & 6 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$
$[\begin{array}{r} 1 & 0 & 0 \\ - 4 & 7 & 2 \\ 10 & - 15 & - 4 \end{array}]$
$[\begin{array}{r} 4 & - 3 & 1 \\ 2 & - 1 & 1 \\ 0 & 0 & 2 \end{array}]$
$[\begin{array}{r} 5 & - 6 & 3 \\ 6 & - 7 & 3 \\ 6 & - 6 & 2 \end{array}]$
$[\begin{array}{r} 1 & 2 & 2 & 2 \\ 0 & 2 & 2 & 2 \\ 0 & 0 & 3 & 2 \\ 0 & 0 & 0 & 4 \end{array}]$
$[\begin{array}{r} 1 & 0 & 4 & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{array}]$
$[\begin{array}{r} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{array}]$
$[\begin{array}{r} 4 & 0 & 0 & - 3 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 6 & 0 & 0 & - 5 \end{array}]$

Find the complex conjugate eigenvalues and corresponding eigenvectors of the matrices given in Problems 27 through 32.

$A = [\begin{array}{r} 0 & 1 \\ - 1 & 0 \end{array}]$
$A = [\begin{array}{r} 0 & - 6 \\ 6 & 0 \end{array}]$
$A = [\begin{array}{r} 0 & - 3 \\ 12 & 0 \end{array}]$
$A = [\begin{array}{r} 0 & - 12 \\ 12 & 0 \end{array}]$
$A = [\begin{array}{r} 0 & 24 \\ - 6 & 0 \end{array}]$
$A = [\begin{array}{r} 0 & - 4 \\ 36 & 0 \end{array}]$
Suppose that $λ$ is an eigenvalue of the matrix A with associated eigenvector v and that n is a positive integer. Show that $λ^{n}$ is an eigenvalue of $A^{n}$ with associated eigen-vector v.
Show that $λ$ is an eigenvalue of the invertible matrix A if and only if $λ^{- 1}$ is an eigenvalue of $A^{- 1}$ . Are the associated eigenvectors the same?
1. Suppose that A is a square matrix. Use the characteristic equation to show that A and $A^{T}$ have the same eigenvalues.
2. Give an example of a $2 \times 2$ matrix A such that A and $A^{T}$ do not have the same eigenvectors.
Show that the eigenvalues of a triangular $n \times n$ matrix are its diagonal elements.
Suppose that the characteristic equation $| A - λ I | = 0$ is written as a polynomial equation [Eq. (5)]. Show that the constant term is $c_{0} = det A$ . Suggestion: Substitute an appropriate value for $λ$ .

Problems 38 through 42 introduce the trace of a square matrix and explore its connections with the determinant and the characteristic polynomial of the matrix.

If $A = [a_{i j}]$ is an $n \times n$ matrix, then the trace Tr A of A is defined to be

$Tr A = a_{11} + a_{22} + \dots + a_{n n},$

the sum of the diagonal elements of A. It can be proved that the coefficient of $λ^{n - 1}$ in Eq. (5) is $c_{n - 1} = {(- 1)}^{n - 1} (Tr A) .$ Show explicitly that this is true in the case of a $2 \times 2$ matrix.
Suppose that the $n \times n$ matrix A has n (real) eigenvalues $λ_{1}, λ_{2}, \dots, λ_{n}$ . Assuming the general result stated in Problem 38, prove that

$\begin{array}{rcl} λ_{1} + λ_{2} + \dots + λ_{n} & = & Tr A \\ = & a_{11} + a_{22} + \dots + a_{n n} . \end{array}$
According to the results stated in Problems 37 and 38, the characteristic polynomial

$p (λ) = | A - λ I |$

of a $3 \times 3$ matrix A is given by

$p (λ) = - λ^{3} + (Tr A) λ^{2} + c_{1} λ + (\det A) .$

The remaining coefficient $c_{1}$ can be found by substituting $λ = 1$ and then calculating the two determinants $| A |$ and $p (1) = | A - I |$ . Use this method to find the characteristic equation, eigenvalues, and associated eigenvectors of the matrix

$A = [\begin{array}{r} 32 & - 67 & 47 \\ 7 & - 14 & 13 \\ - 7 & 15 & - 6 \end{array}] .$
According to the results stated in Problems 37 and 38, the characteristic polynomial

$p (λ) = | A - λ I |$

of a $4 \times 4$ matrix A is given by

$p (λ) = λ^{4} - (Tr A) λ^{3} + c_{2} λ^{2} + c_{1} λ + (\det A) .$

The remaining coefficients $c_{1}$ and $c_{2}$ can be found by substituting $λ = \pm 1$ and calculating the three determinants $| A |, p (1) = | A - I |$ , and $p (- 1) = | A + I |$ . Use this method to find the characteristic equation, eigenvalues, and associated eigenvectors of the matrix

$A = [\begin{array}{r} 22 & - 9 & - 8 & - 8 \\ 10 & - 7 & - 14 & 2 \\ 10 & 0 & 8 & - 10 \\ 29 & - 9 & - 3 & - 15 \end{array}] .$
Combine ideas from Problems 37–40 to show explicitly that
1. If A is a $2 \times 2$ matrix then $\det (A - λ I) = {(- λ)}^{2} + Tr (A) (- λ) + \det (A) .$
2. If A is a $3 \times 3$ matrix then $\det (A - λ I) = {(- λ)}^{3} + Tr (A) {(- λ)}^{2} + c_{1} (- λ) \det (A),$ where $c_{1}$ is the sum of the minors of the diagonal elements of A. (Recall from Section 3.6 the distinction between minors and cofactors.)

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Table of Contents for 6 Eigenvalues and Eigenvectors

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Table of Contents for
6 Eigenvalues and Eigenvectors