INTRODUCTION

If A is an n × n matrix and K is an n × 1 matrix (column vector), then the product AK is defined and is another n × 1 matrix. It is important in many applications to determine whether there exist nonzero n × 1 matrices K such that the product vector AK is a constant multiple λ of K itself. The problem of solving AK = λK for nonzero vectors K is called the eigenvalue problem for the matrix A.

A Definition

The foregoing introductory remarks are summarized in the next definition.

The word eigenvalue is a combination of German and English terms adapted from the German word Eigenwert, which, translated literally, is proper value. Eigenvalues and eigenvectors are also called characteristic values and characteristic vectors, respectively.

Gauss–Jordan elimination introduced in Section 8.2 can be used to find the eigenvectors of a square matrix A.

EXAMPLE 1 Verification of an Eigenvector

Using properties of matrix algebra, we can write (1) in the alternative form

(2)

where I is the multiplicative identity. If we let

then (2) is the same as

(3)

Although an obvious solution of (3) is k₁ = 0, k₂ = 0, . . . , k_n = 0, we are seeking only nontrivial solutions. Now we know from Theorem 8.6.5 that a homogeneous system of n linear equations in n variables has a nontrivial solution if and only if the determinant of the coefficient matrix is equal to zero. Thus to find a nonzero solution K for (2), we must have

(4)

Inspection of (4) shows that expansion of det(A − λI) by cofactors results in an nth-degree polynomial in λ. The equation (4) is called the characteristic equation of A. Thus, the eigenvalues of A are the roots of the characteristic equation. To find an eigenvector corresponding to an eigenvalue λ, we simply solve the system of equations (A − λI)K = 0 by applying Gauss−Jordan elimination to the augmented matrix (A − λI | 0).

EXAMPLE 2 Finding Eigenvalues and Eigenvectors

Find the eigenvalues and eigenvectors of

(5)

SOLUTION

To expand the determinant in the characteristic equation

det(A − λI) = = 0,

we use the cofactors of the second row. It follows that the characteristic equation is

−λ³ − λ² + 12λ = 0 or λ(λ + 4)(λ − 3) = 0.

Hence the eigenvalues are λ₁ = 0, λ₂ = −4, λ₃ = 3. To find the eigenvectors, we must now reduce (A − λI | 0) three times corresponding to the three distinct eigenvalues.

For λ₁ = 0, we have

Thus we see that k₁ = −k₃ and k₂ = −k₃. Choosing k₃ = −13 gives the eigenvector

For λ₂ = −4,

implies k₁ = −k₃ and k₂ = 2k₃. Choosing k₃ = 1 then yields a second eigenvector

K₂ = .

Finally, for λ₃ = 3, Gauss–Jordan elimination gives

and so k₁ = −k₃ and k₂ = − k₃. The choice of k₃ = −2 leads to a third eigenvector,

K₃ = . ≡

It should be obvious from Example 2 that eigenvectors are not uniquely determined. For example, in obtaining, say, , had we not chosen a specific value for then a solution of the homogeneous system can be written

K₂ = = k₃ .

The point of showing in this form is:

For example, λ = 7 and K = are, in turn, an eigenvalue and corresponding eigenvector of the matrix A = . In practice, it may be more convenient to work with an eigenvector with integer entries:

K₁ = 6K = 6 = .

Observe that

AK = = = 7 = λK

and AK₁ = = = 7 = λK₁.

When an n × n matrix A possesses n distinct eigenvalues λ₁, λ₂, . . . , λ_n, it can be proved that a set of n linearly independent eigenvectors K₁, K₂, . . . , K_n can be found. However, when the characteristic equation has repeated roots, it may not be possible to find n linearly independent eigenvectors for A.

EXAMPLE 3 Finding Eigenvalues and Eigenvectors

Find the eigenvalues and eigenvectors of A = .

SOLUTION

From the characteristic equation

det(A − λI) = = (λ − 5)² = 0,

we see λ₁ = λ₂ = 5 is an eigenvalue of multiplicity 2. In the case of a 2 × 2 matrix, there is no need to use Gauss–Jordan elimination. To find the eigenvector(s) corresponding to λ₁ = 5, we resort to the system (A − 5I | 0) in its equivalent form

It is apparent from this system that k₁ = 2k₂. Thus, if we choose k₂ = 1, we find the single eigenvector K₁ = . ≡

EXAMPLE 4 Finding Eigenvalues and Eigenvectors

Find the eigenvalues and eigenvectors of A = .

SOLUTION

The characteristic equation

det(A − λI) = = −(λ − 11)(λ − 8)² = 0

shows that λ₁ = 11 and that λ₂ = λ₃ = 8 is an eigenvalue of multiplicity 2.

For λ₁ = 11, Gauss–Jordan elimination gives

Hence k₁ = k₃ and k₂ = k₃. If k₃ = 1, then

K₁ = .

Now for λ₂ = 8 we have

In the equation k₁ + k₂ + k₃ = 0 we are free to select two of the variables arbitrarily. Choosing, on the one hand, k₂ = 1, k₃ = 0, and on the other, k₂ = 0, k₃ = 1, we obtain two linearly independent eigenvectors:

corresponding to a single eigenvalue. ≡

Complex Eigenvalues

A matrix A may have complex eigenvalues.

PROOF:

Since A is a matrix with real entries, the characteristic equation det(A − λI) = 0 is a polynomial equation with real coefficients. From algebra we know that complex roots of such equations appear in conjugate pairs. In other words, if λ = α + iβ is a root, then = α − iβ is also a root. Now let K be an eigenvector of A corresponding to λ. By definition, AK = λK. Taking complex conjugates of the latter equation gives

since A is a real matrix. The last equation indicates that is an eigenvector corresponding to . ≡

EXAMPLE 5 Complex Eigenvalues and Eigenvectors

Find the eigenvalues and eigenvectors of A = .

SOLUTION

The characteristic equation is

det(A − λI) = = λ² − 10λ + 29 = 0.

From the quadratic formula, we find λ₁ = 5 + 2i and λ₂ = = 5 − 2i.

Now for λ₁ = 5 + 2i, we must solve

Since k₂ = (1 − 2i)k₁,^* it follows, after choosing k₁ = 1, that one eigenvector is

K₁ = .

From Theorem 8.8.1, we see that an eigenvector corresponding to λ₂ = = 5 − 2i is

≡

Eigenvalues and Singular Matrices

If the number 0 is an eigenvalue of an n × n matrix A, then by Definition 8.8.1 the homogeneous linear system

of n equations in n variables must possess a nontrivial solution vector K. But this fact tells us something important about the matrix A. By Theorem 8.6.6 a homogeneous system of n equations in n variables possesses a nontrivial solution if and only if the coefficient matrix A is singular. We summarize this result in the following theorem.

Put a different way:

A matrix A is nonsingular if and only if the number 0 is not an eigenvalue of A. (6)

Reexamination of Example 2 shows that is an eigenvalue of the 3 × 3 matrix (5). So we can conclude from Theorem 8.8.2 that the matrix (5) is singular, that is, does not possess an inverse. On the other hand, we conclude from (6) that the matrices in Examples 3 and 4 are nonsingular because none of the eigenvalues of the matrices are 0.

The eigenvalues of an n × n matrix A are related to det A. Because the characteristic equation det(A − λI) = 0 is an nth degree polynomial equation, it must, counting multiplicities and complex numbers, have n roots λ₁, λ₂, λ₃, … λ_n. By the Factor Theorem of algebra, the characteristic polynomial det(A − λI) can then be written

. (7)

By setting λ = 0 in (7) we see that

, (8)

that is:

The result in (8) provides an alternative proof of Theorem 8.8.2: If, say, then det A = 0. Conversely, if det A = 0, then implies that at least one of the eigenvalues of A is 0.

EXAMPLE 6 Examples 4 and 5 Revisited

Eigenvalues and Eigenvectors of A⁻¹

If an n × n matrix A is nonsingular, then A⁻¹ exists and we now know that none of the eigenvalues of A are 0. Suppose that λ is an eigenvalue of A with corresponding eigenvector K. By multiplying both sides of the equation AK = λK by A⁻¹, we get

or

(9)

Comparing the result in (9) with (1) leads us to the following conclusion.

EXAMPLE 7 Eigenvalues of an Inverse

Our last theorem follows immediately from the fact that the determinant of an upper triangular, lower triangular, and a diagonal matrix is the product of the main diagonal entries. See Theorem 8.5.8.

Notice in Example 7 that the matrix is a lower triangular matrix and its eigenvalues are the entries on the main diagonal.

8.8 Exercises Answers to selected odd-numbered problems begin on page ANS-19.

In Problems 1–6, determine which of the indicated column vectors are eigenvectors of the given matrix A. Give the corresponding eigenvalue.

In Problems 7–22, find the eigenvalues and eigenvectors of the given matrix. Using Theorem 8.8.2 or (6), state whether the matrix is singular or nonsingular.

In Problems 23–26, find the eigenvalues and eigenvectors of the given nonsingular matrix A. Then without finding A⁻¹, find its eigenvalues and corresponding eigenvectors.

Discussion Problems

27. Review the definitions of upper triangular, lower triangular, and diagonal matrices on page 379. Explain why the eigenvalues for those matrices are the main diagonal entries.
28. True or false: If λ is an eigenvalue of an n × n matrix A, then the matrix A − λI is singular. Justify your answer.
29. Suppose is an eigenvalue with corresponding eigenvector K of an matrix A.

    (a) If , then show that . Explain the meaning of the last equation.

    (b) Verify the result obtained in part (a) for the matrix

    (c) Generalize the result in part (a).

30. Let A and B be matrices. The matrix B is said to be similar to the matrix A if there exists a nonsingular matrix S such that . If B is similar to A, then show that A is similar to B.
31. Suppose A and B are similar matrices. See Problem 30. Show that A and B have the same eigenvalues. [Hint: Review Theorem 8.5.6 and Problem 37 in Exercises 8.6.]

Computer Lab Assignment

32. An n × n matrix A is said to be a stochastic matrix if all its entries are nonnegative and the sum of the entries in each row (or the sum of the entries in each column) add up to 1. Stochastic matrices are important in probability theory.
    1. Verify that

       

       and

       are stochastic matrices.

    2. Use a CAS or linear algebra software to find the eigenvalues and eigenvectors of the 3 × 3 matrix A in part (a). Make up at least six more stochastic matrices of various sizes, 2 × 2, 3 × 3, 4 × 4, and 5 × 5. Find the eigenvalues and eigenvectors of each matrix. If you discern a pattern, form a conjecture and then try to prove it.
    3. For the 3 × 3 matrix A in part (a), use the software to find A², A³, A⁴, . . . . Repeat for the matrices that you constructed in part (b). If you discern a pattern, form a conjecture and then try to prove it.

*Note that the second equation is simply 1 + 2i times the first.