EXAMPLE 1 Determination of Eigenvalues and Eigenvectors

We illustrate all the steps in terms of the matrix

Solution. (a) Eigenvalues. These must be determined first. Equation (1) is

Transferring the terms on the right to the left, we get

This can be written in matrix notation

because (1) is Ax − λx = Ax − λIx = (A − λI)x = 0, which gives (3*). We see that this is a homogeneous linear system. By Cramer's theorem in Sec. 7.7 it has a nontrivial solution x ≠ 0 (an eigenvector of A we are looking for) if and only if its coefficient determinant is zero, that is,

We call D(λ) the characteristic determinant or, if expanded, the characteristic polynomial, and D(λ) = 0 the characteristic equation of A. The solutions of this quadratic equation are λ₁ = −1 and λ₂ = −6. These are the eigenvalues of A.

(b₁) Eigenvector of A corresponding to λ₁. This vector is obtained from (2*) with λ = λ₁ = −1, that is,

A solution is x₂ = 2x₁, as we see from either of the two equations, so that we need only one of them. This determines an eigenvector corresponding to λ₁ = −1 up to a scalar multiple. If we choose x₁ = 1, we obtain the eigenvector

(b₂) Eigenvector of A corresponding to λ₂. For λ = λ₂ = −6, equation (2) becomes

A solution is x₂ = −x₁/2 with arbitrary x₁. If we choose x₁ = 2, we get x₂ = −1. Thus an eigenvector of A corresponding to λ₂ = −6 is

For the matrix in the intuitive opening example at the start of Sec. 8.1, the characteristic equation is λ² − 13λ + 30 = (λ − 10)(λ − 3) = 0. The eigenvalues are {10, 3}. Corresponding eigenvectors are [3 4]^T and [−1 1]^T, respectively. The reader may want to verify this.

THEOREM 1 Eigenvalues

The eigenvalues of a square matrix A are the roots of the characteristic equation (4) of A.

Hence an n × n matrix has at least one eigenvalue and at most n numerically different eigenvalues.

THEOREM 2 Eigenvectors, Eigenspace

If w and x are eigenvectors of a matrix A corresponding to the same eigenvalue λ, so are w + x (provided x ≠ −w) and kx for any k ≠ 0.

Hence the eigenvectors corresponding to one and the same eigenvalue λ of A, together with 0, form a vector space (cf. Sec. 7.4), called the eigenspace of A corresponding to that λ.

PROOF

Aw = λw and Ax = λx imply A(w + x) = Aw + Ax = λw + λx = λ(w + x) and A(kw) = k(Aw) = k(λw) = λ(kw); hence A(kw + x) = λ(kw + x).

EXAMPLE 2 Multiple Eigenvalues

Find the eigenvalues and eigenvectors of

Solution. For our matrix, the characteristic determinant gives the characteristic equation

The roots (eigenvalues of A) are λ₁ = 5, λ₂ = λ₃ = −3. (If you have trouble finding roots, you may want to use a root finding algorithm such as Newton's method (Sec. 19.2). Your CAS or scientific calculator can find roots. However, to really learn and remember this material, you have to do some exercises with paper and pencil.) To find eigenvectors, we apply the Gauss elimination (Sec. 7.3) to the system (A − λI)x = 0, first with λ = 5 and then with λ = −3. For λ = 5 the characteristic matrix is

Hence it has rank 2. Choosing x₃ = −1 we have x₂ = 2 from and then x₁ = 1 from −7x₁ + 2x₂ − 3x₃ = 0. Hence an eigenvector of A corresponding to λ = 5 is x₁ = [1 2 −1]^T.

For λ = −3 the characteristic matrix

Hence it has rank 1. From x₁ + 2x₂ − 3x₃ = 0 we have x₁ = −2x₂ + 3x₃. Choosing x₂ = 1, x₃ = 0 and x₂ = 0, x₃ = 1, we obtain two linearly independent eigenvectors of A corresponding to λ = −3 [as they must exist by (5), Sec. 7.5, with rank = 1 and n = 3],

and

EXAMPLE 3 Algebraic Multiplicity, Geometric Multiplicity. Positive Defect

The characteristic equation of the matrix

Hence λ = 0 is an eigenvalue of algebraic multiplicity M₀ = 2. But its geometric multiplicity is only m₀ = 1, since eigenvectors result from −0x₁ + x₂ = 0, hence x₂ = 0, in the form [x₁ 0]^T. Hence for λ = 0 the defect is Δ₀ = 1.

Similarly, the characteristic equation of the matrix

Hence λ = 3 is an eigenvalue of algebraic multiplicity M₃ = 2, but its geometric multiplicity is only m₃ = 1, since eigenvectors result from 0x₁ + 2x₂ = 0 in the form [x₁ 0]^T.

EXAMPLE 4 Real Matrices with Complex Eigenvalues and Eigenvectors

Since real polynomials may have complex roots (which then occur in conjugate pairs), a real matrix may have complex eigenvalues and eigenvectors. For instance, the characteristic equation of the skew-symmetric matrix

It gives the eigenvalues . Eigenvectors are obtained from −ix₁ + x₂ = 0 and ix₁ + x₂ = 0, respectively, and we can choose x₁ = 1 to get

THEOREM 3 Eigenvalues of the Transpose

The transpose A^T of a square matrix A has the same eigenvalues as A.

PROOF

Transposition does not change the value of the characteristic determinant, as follows from Theorem 2d in Sec. 7.7.

PROBLEM SET 8.1

1–16 EIGENVALUES, EIGENVECTORS

Find the eigenvalues. Find the corresponding eigenvectors. Use the given λ or factor in Probs. 11 and 15.

17–20 LINEAR TRANSFORMATIONS AND EIGENVALUES

Find the matrix A in the linear transformation y = Ax, where x = [x₁ x₂]^T (x = [x₁ x₂ x₃]^T) are Cartesian coordinates. Find the eigenvalues and eigenvectors and explain their geometric meaning.

• 17. Counterclockwise rotation through the angle π/2 about the origin in R².
• 18. Reflection about the x₁-axis in R².
• 19. Orthogonal projection (perpendicular projection) of R² onto the x₂-axis.
• 20. Orthogonal projection of R³ onto the plane x₂ = x₁.

21–25 GENERAL PROBLEMS

• 21. Nonzero defect. Find further 2 × 2 and 3 × 3 matrices with positive defect. See Example 3.
• 22. Multiple eigenvalues. Find further 2 × 2 and 3 × 3 matrices with multiple eigenvalues. See Example 2.
• 23. Complex eigenvalues. Show that the eigenvalues of a real matrix are real or complex conjugate in pairs.
• 24. Inverse matrix. Show that A⁻¹ exists if and only if the eigenvalues λ₁, …, λ_n are all nonzero, and then A⁻¹ has the eigenvalues 1/λ₁, …, 1/λ_n.
• 25. Transpose. Illustrate Theorem 3 with examples of your own.

EXAMPLE 1 Stretching of an Elastic Membrane

An elastic membrane in the x₁x₂-plane with boundary circle (Fig. 160) is stretched so that a point P: (x₁, x₂) goes over into the point Q: (y₁, y₂) given by

Find the principal directions, that is, the directions of the position vector x of P for which the direction of the position vector y of Q is the same or exactly opposite. What shape does the boundary circle take under this deformation?

Solution. We are looking for vectors x such that y = λx. Since y = Ax, this gives Ax = λx, the equation of an eigenvalue problem. In components, Ax = λx is

The characteristic equation is

Its solutions are λ₁ = 8 and λ₂ = 2. These are the eigenvalues of our problem. For λ = λ₁ = 8, our system (2) becomes

For λ₂ = 2, our system (2) becomes

We thus obtain as eigenvectors of A, for instance, [1 1]^T corresponding to λ₁ and [1 −1]^T corresponding to λ₂ (or a nonzero scalar multiple of these). These vectors make 45° and 135° angles with the positive x₁-direction. They give the principal directions, the answer to our problem. The eigenvalues show that in the principal directions the membrane is stretched by factors 8 and 2, respectively; see Fig. 160.

Accordingly, if we choose the principal directions as directions of a new Cartesian u₁u₂-coordinate system, say, with the positive u₁-semi-axis in the first quadrant and the positive u₂-semi-axis in the second quadrant of the x₁x₂-system, and if we set u₁ = r cos φ, u₂ = r sin φ, then a boundary point of the unstretched circular membrane has coordinates cos φ, sin φ. Hence, after the stretch we have

Since cos²φ + sin²φ = 1, this shows that the deformed boundary is an ellipse (Fig. 160)

EXAMPLE 2 Eigenvalue Problems Arising from Markov Processes

Markov processes as considered in Example 13 of Sec. 7.2 lead to eigenvalue problems if we ask for the limit state of the process in which the state vector x is reproduced under the multiplication by the stochastic matrix A governing the process, that is, Ax = x. Hence A should have the eigenvalue 1, and x should be a corresponding eigenvector. This is of practical interest because it shows the long-term tendency of the development modeled by the process.

In that example,

Hence A^T has the eigenvalue 1, and the same is true for A by Theorem 3 in Sec. 8.1. An eigenvector x of A for λ = 1 is obtained from

Taking x₃ = 1, we get x₂ = 6 from −x₂/30 + x₃/5 = 0 and then x₁ = 2 from −3x₁/10 + x₂/10 = 0. This gives x = [2 6 1]^T. It means that in the long run, the ratio Commercial:Industrial:Residential will approach 2:6:1, provided that the probabilities given by A remain (about) the same. (We switched to ordinary fractions to avoid rounding errors.)

EXAMPLE 3 Eigenvalue Problems Arising from Population Models. Leslie Model

The Leslie model describes age-specified population growth, as follows. Let the oldest age attained by the females in some animal population be 9 years. Divide the population into three age classes of 3 years each. Let the “Leslie matrix” be

where l_1k is the average number of daughters born to a single female during the time she is in age class k, and l_j,j−1(j = 2, 3) is the fraction of females in age class j − 1 that will survive and pass into class j. (a) What is the number of females in each class after 3, 6, 9 years if each class initially consists of 400 females? (b) For what initial distribution will the number of females in each class change by the same proportion? What is this rate of change?

Solution. (a) Initially, . After 3 years,

Similarly, after 6 years the number of females in each class is given by , and after 9 years we have .

(b) Proportional change means that we are looking for a distribution vector x such that Lx = λx, where λ is the rate of change (growth if λ > 1, decrease if λ < 1). The characteristic equation is (develop the characteristic determinant by the first column)

A positive root is found to be (for instance, by Newton's method, Sec. 19.2) λ = 1.2. A corresponding eigenvector x can be determined from the characteristic matrix

where x₃ = 0.125 is chosen, x₂ = 0.5 then follows from 0.3x₂ − 1.2x₃ = 0, and x₁ = 1 from −1.2x₁ + 2.3x₂ + 0.4x₃ = 0. To get an initial population of 1200 as before, we multiply x by 1200/(1 + 0.5 + 0.125) = 738. Answer: Proportional growth of the numbers of females in the three classes will occur if the initial values are 738, 369, 92 in classes 1, 2, 3, respectively. The growth rate will be 1.2 per 3 years.

EXAMPLE 4 Vibrating System of Two Masses on Two Springs (Fig. 161)

Mass–spring systems involving several masses and springs can be treated as eigenvalue problems. For instance, the mechanical system in Fig. 161 is governed by the system of ODEs

where y₁ and y₂ are the displacements of the masses from rest, as shown in the figure, and primes denote derivatives with respect to time t. In vector form, this becomes

Fig. 161. Masses on springs in Example 4

We try a vector solution of the form

This is suggested by a mechanical system of a single mass on a spring (Sec. 2.4), whose motion is given by exponential functions (and sines and cosines). Substitution into (7) gives

Dividing by e^ωt and writing ω² = λ, we see that our mechanical system leads to the eigenvalue problem

From Example 1 in Sec. 8.1 we see that A has the eigenvalues λ₁ = −1 and λ₂ = −6. Consequently, , respectively. Corresponding eigenvectors are

From (8) we thus obtain the four complex solutions [see (10), Sec. 2.2]

By addition and subtraction (see Sec. 2.2) we get the four real solutions

A general solution is obtained by taking a linear combination of these,

with arbitrary constants a₁, b₁, a₂, b₂ (to which values can be assigned by prescribing initial displacement and initial velocity of each of the two masses). By (10), the components of y are

These functions describe harmonic oscillations of the two masses. Physically, this had to be expected because we have neglected damping.

PROBLEM SET 8.2

1–6 ELASTIC DEFORMATIONS

Given A in a deformation y = Ax, find the principal directions and corresponding factors of extension or contraction. Show the details.

7–9 MARKOV PROCESSES

Find the limit state of the Markov process modeled by the given matrix. Show the details.

• 7.
• 8.
• 9.

10–12 AGE-SPECIFIC POPULATION

Find the growth rate in the Leslie model (see Example 3) with the matrix as given. Show the details.

• 10.
• 11.
• 12.

13–15 LEONTIEF MODELS¹

• 13. Leontief input–output model. Suppose that three industries are interrelated so that their outputs are used as inputs by themselves, according to the 3 × 3 consumption matrix

  

  where a_jk is the fraction of the output of industry k consumed (purchased) by industry j. Let p_j be the price charged by industry j for its total output. A problem is to find prices so that for each industry, total expenditures equal total income. Show that this leads to Ap = p, where p = [p₁ p₂ p₃]^T, and find a solution p with nonnegative p₁, p₂, p₃.

• 14. Show that a consumption matrix as considered in Prob. 13 must have column sums 1 and always has the eigenvalue 1.
• 15. Open Leontief input–output model. If not the whole output but only a portion of it is consumed by the industries themselves, then instead of Ax = x (as in Prob. 13), we have x − Ax = y, where x = [x₁ x₂ x₃]^T is produced, Ax is consumed by the industries, and, thus, y is the net production available for other consumers. Find for what production x a given demand vector y = [0.1 0.3 0.1]^T can be achieved if the consumption matrix is

  

16–20 GENERAL PROPERTIES OF EIGENVALUE PROBLEMS

Let A = [a_jk] be an n × n matrix with (not necessarily distinct) eigenvalues λ₁, …, λ_n Show.

• 16. Trace. The sum of the main diagonal entries, called the trace of A, equals the sum of the eigenvalues of A.
• 17. “Spectral shift.” A − kI has the eigenvalues λ₁ − k, …, λ_n − k and the same eigenvectors as A.
• 18. Scalar multiples, powers. kA has the eigenvalues kλ₁, …, kλ_n. A^m (m = 1, 2, …) has the eigenvalues . The eigenvectors are those of A.
• 19. Spectral mapping theorem. The “polynomial matrix”

  

  has the eigenvalues

  

  where j = 1, …, n, and the same eigenvectors as A.

• 20. Perron's theorem. A Leslie matrix L with positive l₁₂, l₁₃, l₂₁, l₃₂ has a positive eigenvalue. (This is a special case of the Perron–Frobenius theorem in Sec. 20.7, which is difficult to prove in its general form.)

DEFINITIONS Symmetric, Skew-Symmetric, and Orthogonal Matrices

A real square matrix A = [a_jk] is called

symmetric if transposition leaves it unchanged,

skew-symmetric if transposition gives the negative of A,

orthogonal if transposition gives the inverse of A,

EXAMPLE 1 Symmetric, Skew-Symmetric, and Orthogonal Matrices

The matrices

are symmetric, skew-symmetric, and orthogonal, respectively, as you should verify. Every skew-symmetric matrix has all main diagonal entries zero. (Can you prove this?)

EXAMPLE 2 Illustration of Formula (4)

THEOREM 1 Eigenvalues of Symmetric and Skew-Symmetric Matrices

1. The eigenvalues of a symmetric matrix are real.
2. The eigenvalues of a skew-symmetric matrix are pure imaginary or zero.

EXAMPLE 3 Eigenvalues of Symmetric and Skew-Symmetric Matrices

The matrices in (1) and (7) of Sec. 8.2 are symmetric and have real eigenvalues. The skew-symmetric matrix in Example 1 has the eigenvalues 0, −25i, and 25i. (Verify this.) The following matrix has the real eigenvalues 1 and 5 but is not symmetric. Does this contradict Theorem 1?

THEOREM 2 Invariance of Inner Product

An orthogonal transformation preserves the value of the inner product of vectors a and b in Rⁿ, defined by

That is, for any a and b in Rⁿ, orthogonal n × n matrix A, and u = Aa, v = Ab we have u • v = a • b.

Hence the transformation also preserves the length or norm of any vector a in given by Rⁿ given by

PROOF

Let A be orthogonal. Let u = Aa and v = Ab. We must show that u • v = a • b. Now (Aa)^T = a^TA^T by (10d) in Sec. 7.2 and A^TA = A⁻¹A = I by (3). Hence

From this the invariance of follows if we set b = a.

THEOREM 3 Orthonormality of Column and Row Vectors

A real square matrix is orthogonal if and only if its column vectors a₁, …, a_n (and also its row vectors) form an orthonormal system, that is,

PROOF

(a) Let A be orthogonal. Then A⁻¹A = A^TA = I. In terms of column vectors a₁, …, a_n,

The last equality implies (10), by the definition of the n × n unit matrix I. From (3) it follows that the inverse of an orthogonal matrix is orthogonal (see CAS Experiment 12). Now the column vectors of A⁻¹ (=A^T) are the row vectors of A. Hence the row vectors of A also form an orthonormal system.

(b) Conversely, if the column vectors of A satisfy (10), the off-diagonal entries in (11) must be 0 and the diagonal entries 1. Hence A^TA = I, as (11) shows. Similarly, AA^T = I. This implies A^T = A⁻¹ because also A⁻¹A = AA⁻¹ = I and the inverse is unique. Hence A is orthogonal. Similarly when the row vectors of A form an orthonormal system, by what has been said at the end of part (a).

THEOREM 4 Determinant of an Orthogonal Matrix

The determinant of an orthogonal matrix has the value +1 or −1.

PROOF

From det AB = det A det B (Sec. 7.8, Theorem 4) and det A^T = det A (Sec. 7.7, Theorem 2d), we get for an orthogonal matrix

EXAMPLE 4 Illustration of Theorems 3 and 4

The last matrix in Example 1 and the matrix in (6) illustrate Theorems 3 and 4 because their determinants are −1 and +1, as you should verify.

THEOREM 5 Eigenvalues of an Orthogonal Matrix

The eigenvalues of an orthogonal matrix A are real or complex conjugates in pairs and have absolute value 1.

PROOF

The first part of the statement holds for any real matrix A because its characteristic polynomial has real coefficients, so that its zeros (the eigenvalues of A) must be as indicated. The claim that |λ| = 1 will be proved in Sec. 8.5.

EXAMPLE 5 Eigenvalues of an Orthogonal Matrix

The orthogonal matrix in Example 1 has the characteristic equation

Now one of the eigenvalues must be real (why?), hence +1 or −1. Trying, we find −1. Division by λ + 1 gives −(λ² − 5λ/3 + 1) = 0 and the two eigenvalues , which have absolute value 1. Verify all of this.

PROBLEM SET 8.3

1–10 SPECTRUM

Are the following matrices symmetric, skew-symmetric, or orthogonal? Find the spectrum of each, thereby illustrating Theorems 1 and 5. Show your work in detail.

1. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 
9. 
10. 
11. WRITING PROJECT. Section Summary. Summarize the main concepts and facts in this section, giving illustrative examples of your own.
12. CAS EXPERIMENT. Orthogonal Matrices.

    (a) Products. Inverse. Prove that the product of two orthogonal matrices is orthogonal, and so is the inverse of an orthogonal matrix. What does this mean in terms of rotations?

    (b) Rotation. Show that (6) is an orthogonal transformation. Verify that it satisfies Theorem 3. Find the inverse transformation.

    (c) Powers. Write a program for computing powers A^m(m = 1, 2, …) of a 2 × 2 matrix A and their spectra. Apply it to the matrix in Prob. 1 (call it A). To what rotation does A correspond? Do the eigenvalues of A^m have a limit as m → ∞?

    (d) Compute the eigenvalues of (0.9A)^m, where A is the matrix in Prob. 1. Plot them as points. What is their limit? Along what kind of curve do these points approach the limit?

    (e) Find A such that y = Ax is a counterclockwise rotation through 30° in the plane.

13–20 GENERAL PROPERTIES

• 13. Verification. Verify the statements in Example 1.
• 14. Verify the statements in Examples 3 and 4.
• 15. Sum. Are the eigenvalues of A + B sums of the eigenvalues of A and of B?
• 16. Orthogonality. Prove that eigenvectors of a symmetric matrix corresponding to different eigenvalues are orthogonal. Give examples.
• 17. Skew-symmetric matrix. Show that the inverse of a skew-symmetric matrix is skew-symmetric.
• 18. Do there exist nonsingular skew-symmetric n × n matrices with odd n?
• 19. Orthogonal matrix. Do there exist skew-symmetric orthogonal 3 × 3 matrices?
• 20. Symmetric matrix. Do there exist nondiagonal symmetric 3 × 3 matrices that are orthogonal?

THEOREM 1 Basis of Eigenvectors

If an n × n matrix A has n distinct eigenvalues, then A has a basis of eigenvectors x₁, … x_n for Rⁿ.

PROOF

All we have to show is that x₁, …, x_n are linearly independent. Suppose they are not. Let r be the largest integer such that {x₁, …, x_r} is a linearly independent set. Then r < n and the set {x₁, …, x_r, x_r+1} is linearly dependent. Thus there are scalars c₁, …, c_r+1, not all zero, such that

(see Sec. 7.4). Multiplying both sides by A and using Ax_j = λ_jx_j, we obtain

To get rid of the last term, we subtract λ_r+1 times (2) from this, obtaining

Here c₁(λ₁ − λ_r+1) = 0, …, c_r(λ_r − λ_r+1) = 0 since {x₁, …, x_r} is linearly independent. Hence c₁ = … = c_r = 0, since all the eigenvalues are distinct. But with this, (2) reduces to c_r+1x_r+1 = 0, hence c_r+1 = 0, since x_r+1 ≠ 0 (an eigenvector!). This contradicts the fact that not all scalars in (2) are zero. Hence the conclusion of the theorem must hold.

EXAMPLE 1 Eigenbasis. Nondistinct Eigenvalues. Nonexistence

The matrix has a basis of eigenvectors corresponding to the eigenvalues λ₁ = 8, λ₂ = 2. (See Example 1 in Sec. 8.2.)

Even if not all n eigenvalues are different, a matrix A may still provide an eigenbasis for Rⁿ. See Example 2 in Sec. 8.1, where n = 3.

On the other hand, A may not have enough linearly independent eigenvectors to make up a basis. For instance, A in Example 3 of Sec. 8.1 is

THEOREM 2 Symmetric Matrices

A symmetric matrix has an orthonormal basis of eigenvectors for Rⁿ.

EXAMPLE 2 Orthonormal Basis of Eigenvectors

The first matrix in Example 1 is symmetric, and an orthonormal basis of eigenvectors is , .

DEFINITION Similar Matrices. Similarity Transformation

An n × n matrix is called similar to an n × n matrix A if

for some (nonsingular!) n × n matrix P. This transformation, which gives from A, is called a similarity transformation.

THEOREM 3 Eigenvalues and Eigenvectors of Similar Matrices

If is similar to A, then has the same eigenvalues as A.

Furthermore, if x is an eigenvector of A, then y = P⁻¹x is an eigenvector of corresponding to the same eigenvalue.

PROOF

From Ax = λx (λ an eigenvalue, x ≠ 0) we get P⁻¹Ax = λP⁻¹x. Now I = PP⁻¹. By this identity trick the equation P⁻¹Ax = λP⁻¹x gives

Hence λ is an eigenvalue of and P⁻¹x a corresponding eigenvector. Indeed, P⁻¹x ≠ 0 because P⁻¹x = 0 would give x = Ix = PP⁻¹x = P0 = 0, contradicting x ≠ 0.

EXAMPLE 3 Eigenvalues and Vectors of Similar Matrices

Here P⁻¹ was obtained from (4*) in Sec. 7.8 with det P = 1. We see that has the eigenvalues λ₁ = 3, λ₂ = 2. The characteristic equation of A is (6 − λ)(−1 − λ) + 12 = λ² − 5λ + 6 = 0. It has the roots (the eigenvalues of A) λ₁ = 3, λ₂ = 2, confirming the first part of Theorem 3.

We confirm the second part. From the first component of (A − λI)x = 0 we have (6 − λ)x₁ − 3x₂ = 0. For λ = 3 this gives 3x₁ − 3x₂ = 0, say, x₁ = [1 1]^T. For λ = 2 it gives 4x₁ − 3x₂ = 0, say, x₂ = [3 4]^T. In Theorem 3 we thus have

Indeed, these are eigenvectors of the diagonal matrix .

Perhaps we see that x₁ and x₂ are the columns of P. This suggests the general method of transforming a matrix A to diagonal form D by using P = X, the matrix with eigenvectors as columns.

THEOREM 4 Diagonalization of a Matrix

If an n × n matrix A has a basis of eigenvectors, then

is diagonal, with the eigenvalues of A as the entries on the main diagonal. Here X is the matrix with these eigenvectors as column vectors. Also,

PROOF

Let x₁, …, x_n be a basis of eigenvectors of A for Rⁿ. Let the corresponding eigenvalues of A be λ₁, …, λ_n, respectively, so that Ax₁ = λ₁x₁, …, Ax_n = λ_nx_n. Then X = [x₁ … x_n] has rank n, by Theorem 3 in Sec. 7.4. Hence X⁻¹ exists by Theorem 1 in Sec. 7.8. We claim that

where D is the diagonal matrix as in (5). The fourth equality in (6) follows by direct calculation. (Try it for n = 2 and then for general n.) The third equality uses Ax_k = λ_kx_k. The second equality results if we note that the first column of AX is A times the first column of X, which is x₁, and so on. For instance, when n = 2 and we write x₁ = [ x₁₁ x₂₁], x₂ = [x₁₂ x₂₂], we have

If we multiply (6) by X⁻¹ from the left, we obtain (5). Since (5) is a similarity transformation, Theorem 3 implies that D has the same eigenvalues as A. Equation (5*) follows if we note that

EXAMPLE 4 Diagonalization

Diagonalize

Solution. The characteristic determinant gives the characteristic equation −λ³ − λ² + 12λ = 0. The roots (eigenvalues of A) are λ₁ = 3, λ₂ = −4, λ₃ = 0. By the Gauss elimination applied to (A − λI)x = 0 with λ = λ₁, λ₂, λ₃ we find eigenvectors and then X⁻¹ by the Gauss–Jordan elimination (Sec. 7.8, Example 1). The results are

Calculating AX and multiplying by X⁻¹ from the left, we thus obtain

EXAMPLE 5 Quadratic Form. Symmetric Coefficient Matrix

Let

Here 4 + 6 = 10 = 5 + 5. From the corresponding symmetric matrix C = [c_jk], where , thus c₁₁ = 3, c₁₂ = c₂₁ = 5, c₂₂ = 2, we get the same result; indeed,

THEOREM 5 Principal Axes Theorem

The substitution (9) transforms a quadratic form

to the principal axes form or canonical form (10), where λ₁, …, λ_n are the (not necessarily distinct) eigenvalues of the (symmetric!) matrix A, and X is an orthogonal matrix with corresponding eigenvectors x₁, …, x_n, respectively, as column vectors.

EXAMPLE 6 Transformation to Principal Axes. Conic Sections

Find out what type of conic section the following quadratic form represents and transform it to principal axes:

Solution. We have Q = x^TAx, where

This gives the characteristic equation (17 − λ)² − 15² = 0. It has the roots λ₁ = 2, λ₂ = 32. Hence (10) becomes

We see that Q = 128 represents the ellipse , that is,

If we want to know the direction of the principal axes in the x₁x₂-coordinates, we have to determine normalized eigenvectors from (A − λI)x = 0 with λ = λ₁ = 2 and λ = λ₂ = 32 and then use (9). We get

hence

This is a 45° rotation. Our results agree with those in Sec. 8.2, Example 1, except for the notations. See also Fig. 160 in that example.

PROBLEM SET 8.4

1–5 SIMILAR MATRICES HAVE EQUAL EIGENVALUES

Verify this for A and A = P⁻¹AP. If y is an eigenvector of P, show that x = Py are eigenvectors of A. Show the details of your work.

1. 
2. 
3. 
4. 
5. 
6. PROJECT. Similarity of Matrices. Similarity is basic, for instance, in designing numeric methods.

   (a) Trace. By definition, the trace of an n × n matrix A = [a_jk] is the sum of the diagonal entries,

   

   Show that the trace equals the sum of the eigenvalues, each counted as often as its algebraic multiplicity indicates. Illustrate this with the matrices A in Probs. 1, 3, and 5.

   (b) Trace of product. Let B = [b_jk] be n × n. Show that similar matrices have equal traces, by first proving

   

   (c) Find a relationship between in (4) and .

   (d) Diagonalization. What can you do in (5) if you want to change the order of the eigenvalues in D, for instance, interchange d₁₁ = λ₁ and d₂₂ = λ₂?

7. No basis. Find further 2 × 2 and 3 × 3 matrices without eigenbasis.
8. Orthonormal basis. Illustrate Theorem 2 with further examples.

9–16 DIAGONALIZATION OF MATRICES

Find an eigenbasis (a basis of eigenvectors) and diagonalize. Show the details.

• 9.
• 10.
• 11.
• 12.
• 13.
• 14.
• 15.
• 16.

17–23 PRINCIPAL AXES. CONIC SECTIONS

What kind of conic section (or pair of straight lines) is given by the quadratic form? Transform it to principal axes. Express x^T = [x₁ x₂] in terms of the new coordinate vector y^T = [y₁ y₂], as in Example 6.

• 17.
• 18.
• 19.
• 20.
• 21.
• 22.
• 23.
• 24. Definiteness. A quadratic form Q(x) = x^TAx and its (symmetric!) matrix A are called (a) positive definite if Q(x) > 0 for all x ≠ 0, (b) negative definite if Q(x) < 0 for all x ≠ 0, (c) indefinite if Q(x) takes both positive and negative values. (See Fig. 162.) [Q(x) and A are called positive semidefinite (negative semidefinite) if Q(x) 0 (Q(x) 0) for all x.] Show that a necessary and sufficient condition for (a), (b), and (c) is that the eigenvalues of A are (a) all positive, (b) all negative, and (c) both positive and negative. Hint. Use Theorem 5.
• 25. Definiteness. A necessary and sufficient condition for positive definiteness of a quadratic form Q(x) = x^TAx with symmetric matrix A is that all the principal minors are positive (see Ref. [B3], vol. 1, p. 306), that is,

  

  Show that the form in Prob. 22 is positive definite, whereas that in Prob. 23 is indefinite.

  

  Fig. 162. Quadratic forms in two variables (Problem 24)

EXAMPLE 1 Notations

DEFINITION Hermitian, Skew-Hermitian, and Unitary Matrices

A square matrix A = [a_kj] is called

EXAMPLE 2 Hermitian, Skew-Hermitian, and Unitary Matrices

are Hermitian, skew-Hermitian, and unitary matrices, respectively, as you may verify by using the definitions.

THEOREM 1 Eigenvalues

(a) The eigenvalues of a Hermitian matrix (and thus of a symmetric matrix) are real.

(b) The eigenvalues of a skew-Hermitian matrix (and thus of a skew-symmetric matrix) are pure imaginary or zero.

(c) The eigenvalues of a unitary matrix (and thus of an orthogonal matrix) have absolute value 1.

EXAMPLE 3 Illustration of Theorem 1

For the matrices in Example 2 we find by direct calculation

PROOF

We prove Theorem 1. Let λ be an eigenvalue and x an eigenvector of A. Multiply Ax = λx from the left by , thus , and divide by , which is real and not 0 because x ≠ 0. This gives

(a) If A is Hermitian, ^T = A or A^T = and we show that then the numerator in (1) is real, which makes λ real. Ax is a scalar; hence taking the transpose has no effect. Thus

Hence, equals its complex conjugate, so that it must be real. (a + ib = a − ib implies b = 0.)

(b) If A is skew-Hermitian, A^T = − and instead of (2) we obtain

so that equals minus its complex conjugate and is pure imaginary or 0. (a + ib = −(a − ib) implies a = 0.)

(c) Let A be unitary. We take Ax = λx and its conjugate transpose

and multiply the two left sides and the two right sides,

But A is unitary, ^T = A⁻¹, so that on the left we obtain

Together, . We now divide by to get |λ|² = 1. Hence |λ| = 1.

This proves Theorem 1 as well as Theorems 1 and 5 in Sec. 8.3.

THEOREM 2 Invariance of Inner Product

A unitary transformation, that is, y = Ax with a unitary matrix A, preserves the value of the inner product (4), hence also the norm (5).

PROOF

The proof is the same as that of Theorem 2 in Sec. 8.3, which the theorem generalizes. In the analog of (9), Sec. 8.3, we now have bars,

DEFINITION Unitary System

A unitary system is a set of complex vectors satisfying the relationships

THEOREM 3 Unitary Systems of Column and Row Vectors

A complex square matrix is unitary if and only if its column vectors (and also its row vectors) form a unitary system.

PROOF

The proof is the same as that of Theorem 3 in Sec. 8.3, except for the bars required in ^T = A⁻¹ and in (4) and (6) of the present section.

THEOREM 4 Determinant of a Unitary Matrix

Let A be a unitary matrix. Then its determinant has absolute value one, that is, |det A| = 1.

PROOF

Similarly, as in Sec. 8.3, we obtain

Hence |det A| = 1 (where det A may now be complex).

EXAMPLE 4 Unitary Matrix Illustrating Theorems 1c and 2–4

For the vectors a^T = [2 −i] and b^T = [1 + i 4i] we get ^T = [2 i] and ^Tb = 2(1 + i) − 4 = −2 + 2i and with

as one can readily verify. This gives ()^TAb = −2 + 2i, illustrating Theorem 2. The matrix is unitary. Its columns form a unitary system,

and so do its rows. Also, det A = −1. The eigenvalues are 0.6 + 0.8i and −0.6 + 0.8i, with eigenvectors [1 1]^T and [1 −1]^T, respectively.

THEOREM 5 Basis of Eigenvectors

A Hermitian, skew-Hermitian, or unitary matrix has a basis of eigenvectors for Cⁿ that is a unitary system.

EXAMPLE 5 Unitary Eigenbases

The matrices A, B, C in Example 2 have the following unitary systems of eigenvectors, as you should verify.

EXAMPLE 6 Hermitian Form

For A in Example 2 and, say, x = [1 + i 5i]^T we get

PROBLEM SET 8.5

1–6 EIGENVALUES AND VECTORS

Is the given matrix Hermitian? Skew-Hermitian? Unitary? Find its eigenvalues and eigenvectors.

1. 
2. 
3. 
4. 
5. 
6. 
7. Pauli spin matrices. Find the eigenvalues and eigenvectors of the so-called Pauli spin matrices and show that , where

   

8. Eigenvectors. Find eigenvectors of A, B, C in Examples 2 and 3.

9–12 COMPLEX FORMS

Is the matrix A Hermitian or skew-Hermitian? Find . Show the details.

• 9.
• 10.
• 11.
• 12.

13–20 GENERAL PROBLEMS

• 13. Product. Show that for any n × n Hermitian A, skew-Hermitian B, and unitary C.
• 14. Product. Show for A and B in Example 2. For any n × n Hermitian A and skew-Hermitian B.
• 15. Decomposition. Show that any square matrix may be written as the sum of a Hermitian and a skew-Hermitian matrix. Give examples.
• 16. Unitary matrices. Prove that the product of two unitary n × n matrices and the inverse of a unitary matrix are unitary. Give examples.
• 17. Powers of unitary matrices in applications may sometimes be very simple. Show that C¹² = I in Example 2. Find further examples.
• 18. Normal matrix. This important concept denotes a matrix that commutes with its conjugate transpose, A^T = ^TA. Prove that Hermitian, skew-Hermitian, and unitary matrices are normal. Give corresponding examples of your own.
• 19. Normality criterion. Prove that A is normal if and only if the Hermitian and skew-Hermitian matrices in Prob. 18 commute.
• 20. Find a simple matrix that is not normal. Find a normal matrix that is not Hermitian, skew-Hermitian, or unitary.

The practical importance of matrix eigenvalue problems can hardly be overrated. The problems are defined by the vector equation

A is a given square matrix. All matrices in this chapter are square. λ is a scalar. To solve the problem (1) means to determine values of λ, called eigenvalues (or characteristic values) of A, such that (1) has a nontrivial solution x (that is, x ≠ 0), called an eigenvector of A corresponding to that λ. An n × n matrix has at least one and at most n numerically different eigenvalues. These are the solutions of the characteristic equation (Sec. 8.1)

D(λ) is called the characteristic determinant of A. By expanding it we get the characteristic polynomial of A, which is of degree n in λ. Some typical applications are shown in Sec. 8.2.

Section 8.3 is devoted to eigenvalue problems for symmetric (A^T = A), symmetric (A^T = −A), and orthogonal matrices (A^T = A⁻¹). Section 8.4 concerns the diagonalization of matrices and the transformation of quadratic forms to principal axes and its relation to eigenvalues.

Section 8.5 extends Sec. 8.3 to the complex analogs of those real matrices, called Hermitian (A^T = A), skew-Hermitian (A^T = −A), and unitary matrices (^T = A⁻¹). All the eigenvalues of a Hermitian matrix (and a symmetric one) are real. For a skew-Hermitian (and a skew-symmetric) matrix they are pure imaginary or zero. For a unitary (and an orthogonal) matrix they have absolute value 1.