Linear Algebra: Matrix Eigenvalue Problems

Chapter 8

Linear Algebra:

Matrix Eigenvalue Problems

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Contents

 8.1 Eigenvalues, Eigenvectors

 8.2 Some Applications of Eigenvalue Problems

 8.3 Symmetric, Skew-Symmetric, and Orthogonal Matrices

 8.4 Eigenbases. Diagonalization. Quadratic Forms

 8.5 Complex Matrices and Forms. Optional

 Summary of Chapter 8

8.1 Eigenvalues, Eigenvectors



_{Let A = [ajk}] be a given n × n matrix and consider the ve ctor equation

(1) Ax = λx.

Here x is an unknown vector and λ an unknown scalar.

Our task is to determine x’s and λ’s that satisfy (1). Ge ometrically, we are looking for vectors x for which the multiplication by A has the same effect as the multiplic ation by a scalar λ; in other words, Ax should be propor tional to x.

continued

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

Clearly, the zero vector x = 0 is a solution of (1) for any value of λ, because A0 = 0. This is of no interest. A val ue of for which (1) has a solution x ≠ 0 is called an eig envalue or characteristic value (or latent root) of the m atrix A. (“Eigen” is German and means “proper” or “cha racteristic.”) The corresponding solutions x ≠ 0 of (1) ar e called the eigenvectors or characteristic vectors of A corresponding to that eigenvalue λ. The set of all the ei genvalues of A is called the spectrum of A. We shall s ee that the spectrum consists of at least one eigenvalu e and at most of n numerically different eigenvalues. T he largest of the absolute values of the eigenvalues of A is called the spectral radius of A, a name to be motiv ated later.

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How to Find Eigenvalues and Eigenvectors

E X A M P L E 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

continued

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Transferring the terms on the right to the left, we get (2*)

This can be written in matrix notation (3*) (A – λI) x = 0

because (1) is Ax – λx = Ax – λIx = (A – λI)x = 0, which gives (3*). We see that this is a homogeneous linear sy stem. 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,

continued

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(4*)



We call D(λ) the characteristic determinant or, if expa nded, the characteristic polynomial, and D(λ) = 0 the characteristic equation of A. The solutions of this qua dratic equation are λ1 = –1 and λ2 = –6. These are the e igenvalues of A.

continued

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

_(b1) Eigenvector of A corresponding to λ1. This vect or 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 dete rmines an eigenvector corresponding to λ₁ = –1 up to a scalar multiple. If we choose x₁ = 1, we obtain the eige nvector

continued

335



_(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 correspo nding to λ₂ = –6 is

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

This example illustrates the general case as follows. E quation (1) written in components is

continued

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

Transferring the terms on the right side to the left side, we have

(2)

In matrix notation,

(3) (A – λI)x = 0.

continued

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

By Cramer’s theorem in Sec. 7.7, this homogeneous lin ear system of equations has a nontrivial solution if and only if the corresponding determinant of the coefficients is zero:

(4)

continued

336



A – λI is called the characteristic matrix and D(λ) the characteristic determinant of A. Equation (4) is called the characteristic equation of A. By developing D(λ) we obtain a polynomial of nth degree in . This is called the characteristic polynomial of A.

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Eigenvalues

THEOREM 1

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.

336



The eigenvalues must be determined first. Once the se are known, corresponding eigenvectors are obtaine d from the system (2), for instance, by the Gauss elimin ation, where λ is the eigenvalue for which an eigenvect or is wanted. This is what we did in Example 1 and shal l do again in the examples below.

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Eigenvectors, Eigenspace

THEOREM 2

If w and x are eigenvectors of a matrix A correspondin g 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 s ame eigenvalue λ of A, together with 0, form a vector s pace (cf. Sec. 7.4), called the eigenspace of A corresp onding to that λ.

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E X A M P L E 2

Multiple Eigenvalues



Find the eigenvalues and eigenvectors of



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

–λ³ – λ² + 21λ + 45 = 0.

continued

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The roots (eigenvalues of A) are λ1 = 5, λ2 = λ3 = –3. To find eigenvectors, we apply the Gauss elimination (Sec . 7.3) to the system (A – λI)x = 0, first with λ = 5 and th en with λ = –3. For λ = 5 the characteristic matrix is

Hence it has rank 2. Choosing x3 = –1 we have x2 = 2 fr om –(24/7)x₂ – (48/7)x₃ = 0 and then x₁ = 1 from –7x₁ + 2x₂ – 3x₃ = 0. Hence an eigenvector of A coresponding to λ = 5 is x₁ = [1 2 –1]^T.

continued

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

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 c orresponding to λ = –3 [as they must exist by (5), Sec.

7.5, with rank = 1 and n = 3], and

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E X A M P L E 3

Algebraic Multiplicity,

Geometric Multiplicity. Positive Defect



The characteristic equation of the matrix

Hence λ = 0 is an eigenvalue of algebraic multiplicity M

0 = 2. But its geometric multiplicity is only m0 = 1, since eigenvectors result from –0x₁ + x₂ = 0, hence x₂ = 0, in t he form [x₁ 0]^T. Hence for λ = 0 the defect is ∆₀ = 1.

continued

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

Similarly, the characteristic equation of the matrix

Hence λ = 3 is an eigenvalue of algebraic multiplicity M

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

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E X A M P L E 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, t he characteristic equation of the skew-symmetric matri x

It gives the eigenvalues λ1 = i , λ2 = –i. Eigenvectors are obtained from –ix₁ + x₂ = 0 and ix₁ + x₂ = 0, respecti vely, and we can choose x₁ = 1 to get

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Eigenvalues of the Transpose

THEOREM 3

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

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8.2 Some Applications of Eigenvalue Problems

E X A M P L E 1

Stretching of an Elastic Membrane

An elastic membrane in the x₁x₂-plane with boundary circle x₁₂ + x₂₂ = 1 (Fig. 158) is stretched so that a point P:

(x₁, x₂) goes over into the point Q: (y₁, y₂) given by (1)

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?

continued

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Fig. 158.

Undeformed and deformed membrane in Example 1

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

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

(2)

The characteristic equation is (3)

continued

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Its solutions are λ1 = 8 and λ2 = 2. These are the eigen values of our problem. For λ = λ1 = 8, our system (2) be comes

For λ₂ = 2, our system (2) becomes

continued

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

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 p roblem. The eigenvalues show that in the principal dire ctions the membrane is stretched by factors 8 and 2, re spectively; see Fig. 158.

continued

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

Accordingly, if we choose the principal directions as dir ections of a new Cartesian u₁u₂-coordinate system, say, with the positive u₁-semi-axis in the first quadrant and th e positive u₂-semi-axis in the second quadrant of the x₁ x₂-system, and if we set u₁ = r cos φ, u₂ = r sin φ, then a boundary point of the unstretched circular membrane h as coordinates cos φ, sin φ. Hence, after the stretch we have

Since cos² φ + sin² φ = 1, this shows that the deformed boundary is an ellipse (Fig. 158)

(4)

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E X A M P L E 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 repro duced under the multiplication by the stochastic matrix A governing the process, that is, Ax = x. Hence A shou ld have the eigenvalue 1, and x should be a correspon ding eigenvector. This is of practical interest because it shows the long-term tendency of the development mod eled by the process.

continued

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

In that example,

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

continued

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

_{Taking x3} = 1, we get x2 = 6 from – x2 / 30 + x3 / 5 = 0 a nd 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 Co mmercial : Industrial : Residential will approach 2 : 6 : 1 , provided that the probabilities given by A remain (abo ut) the same. (We switched to ordinary fractions to avoi d rounding errors.)

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E X A M P L E 3

Eigenvalue Problems Arising from Population Models. Leslie Model



The Leslie model describes age-specified population gr owth, as follows. Let the oldest age attained by the fem ales in some animal population be 9 years. Divide the p opulation into three age classes of 3 years each. Let th e “Leslie matrix” be

(5)

continued

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where l1k is the average number of daughters born to a single female during the time she is in age class k, and lj, 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 nu mber of females in each class after 3, 6, 9 years if each class initially consists of 400 females? (b) For what initi al distribution will the number of females in each class change by the same proportion? What is this rate of ch ange?

continued

341



Solution. (a) Initially, = [400 400 400]. After 3 yea rs,

Similarly, after 6 years the number of females in each c lass is given by = (Lx(3))^T = [600 648 72], and after 9 years we have = (Lx(6))^T = [1519.2 360 194.4].

continued

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

(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 de terminant 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 eigenvect or x can be determined from the characteristic matrix

continued

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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: Pro portional 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.

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E X A M P L E 4

Vibrating System of Two Masses on Two

Springs



Mass–spring systems involving several masses and sp rings can be treated as eigenvalue problems. For insta nce, the mechanical system in Fig. 159 is governed by the system of ODEs

(6)

where y₁ and y₂ are the displacements of the masses fr om rest, as shown in the figure, and primes denote deri vatives with respect to time t. In vector form, this beco mes

(7)

continued

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Fig. 159.

Masses on springs in Example 4

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

We try a vector solution of the form (8)

This is suggested by a mechanical system of a single m ass on a spring (Sec. 2.4), whose motion is given by ex ponential functions (and sines and cosines). Substitutio n into (7) gives

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

(9)

continued

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

From Example 1 in Sec. 8.1 we see that A has the eige nvalues λ1 = –1 and λ2 = –6. Consequently, ω = = ±i and = , respectively. Corresponding eigenvectors are

(10)

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

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

continued

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

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.

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8.3 Symmetric, Skew-Symmetric, and Orthogonal Matrices

Symmetric, Skew-Symmetric, and Orthogonal Matrices

DEFINITION

A real square matrix A = [a_jk] is called

symmetric if transposition leaves it unchanged, (1) A^T = A, thus a_kj = a_jk,

skew-symmetric if transposition gives the negative of A ,

(2) A^T = –A, thus a_kj = –a_jk,

orthogonal if transposition gives the inverse of A,

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E X A M P L E 1

Symmetric, Skew-Symmetric,

and Orthogonal Matrices



The matrices

are symmetric, skew-symmetric, and orthogonal, respe ctively, as you should verify. Every skew-symmetric ma trix has all main diagonal entries zero. (Can you prove t his?)

345



Any real square matrix A may be written as the sum of a symmetric matrix R and a skew-symmetric matrix S, where

(4)

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E X A M P L E 2

Illustration of Formula (4)

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Eigenvalues of Symmetric and Skew-Symmetric Matrices

THEOREM 1

(a) The eigenvalues of a symmetric matrix are real.

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

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E X A M P L E 3

Eigenvalues of Symmetric and

Skew-Symmetric Matrices



The matrices in (1) and (7) of Sec. 8.2 are symmetric a nd have real eigenvalues. The skew-symmetric matrix i n Example 1 has the eigenvalues 0, –25i, and 25i. (Veri fy this.) The following matrix has the real eigenvalues 1 and 5 but is not symmetric. Does this contradict Theore m 1?

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Orthogonal Transformations and Orthogonal Matrices



Orthogonal transformations are transformations

(5) y = Ax where A is an orthogonal matrix .

With each vector x in Rⁿ such a transformation assigns a vector y in Rⁿ. For instance, the plane rotation throug h an angle θ

(6)

continued

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is an orthogonal transformation. It can be shown that a ny orthogonal transformation in the plane or in three-di mensional space is a rotation (possibly combined with a reflection in a straight line or a plane, respectively).

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Invariance of Inner Product (1)

THEOREM 2

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

(7)

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

continued

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Invariance of Inner Product (2)

THEOREM 2

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

(8)

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Orthonormality of Column and Row Vectors

THEOREM 3

A real square matrix is orthogonal if and only if its col umn vectors a₁, ‥‥, a_n (and also its row vectors) for m an orthonormal system, that is,

(10)

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Determinant of an Orthogonal Matrix

THEOREM 4

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

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E X A M P L E 4

Illustration of Theorems 3 and 4



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

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Eigenvalues of an Orthogonal Matrix

THEOREM 5

The eigenvalues of an orthogonal matrix A are real o r complex conjugates in pairs and have absolute valu e 1.

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E X A M P L E 5

Eigenvalues of an Orthogonal Matrix



The orthogonal matrix in Example 1 has the characteris tic equation

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

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8.4 Eigenbases. Diagonalization.

Quadratic Forms



Eigenvectors of an n × n matrix A may (or may not!) for m a basis for Rⁿ. If we are interested in a transformatio n y = Ax, such an “eigenbasis” (basis of eigenvector s)—if it exists—is of great advantage because then we can represent any x in Rⁿ uniquely as a linear combinat ion of the eigenvectors x₁, ‥‥, x_n, say,

continued

349

And, denoting the corresponding (not necessarily distin ct) eigenvalues of the matrix A by λ₁, ‥‥, λ_n, we have Axj = λjxj, so that we simply obtain

(1)

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Basis of Eigenvectors

THEOREM 1

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

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E X A M P L E 1

Eigenbasis. Nondistinct Eigenvalues.

Nonexistence



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



Even if not all n eigenvalues are different, a matrix A m ay 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 i ndependent eigenvectors to make up a basis. For in stance, A in Example 3 of Sec. 8.1 is

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Symmetric Matrices

THEOREM 2

A symmetric matrix has an orthonormal basis of eige nvectors for Rⁿ.

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E X A M P L E 2

Orthonormal Basis of Eigenvectors



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

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Diagonalization of Matrices

Similar Matrices. Similarity Transformation

DEFINITION

An n × n matrix  is called similar to an n × n matrix A if (4)  = P^-1AP

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

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Eigenvalues and Eigenvectors of Similar Matrices

THEOREM 3

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

Furthermore, if x is an eigenvector of A, then y = P^-1x is an eigenvector of  corresponding to the same eige nvalue.

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E X A M P L E 3

Eigenvalues and Vectors of Similar Matrices



_Let

Then

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. T he characteristic equation of A is (6 – λ)(–1 – λ) + 12 = λ² – 5λ + 6 = 0. It has the roots (the eigenvalues of A) λ

1 = 3, λ2 = 2, confirming the first part of Theorem 3.

continued

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

We confirm the second part. From the first component o f (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 4 x₁ – 3x₂ = 0, say, x₂ = [3 4]^T. In Theorem 3 we thus hav e

Indeed, these are eigenvectors of the diagonal matrix  .



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

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Diagonalization of a Matrix

THEOREM 4

If an n × n matrix A has a basis of eigenvectors, then (5) D = X^-1AX

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

(5*) D^m = X^-1A^mX (m = 2, 3, ‥‥).

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E X A M P L E 4

Diagonalization



Diagonalize



Solution. The characteristic determinant gives the char acteristic equation –λ³ – λ² + 12λ = 0. The roots (eigenv alues of A) are λ1 = 3, λ2 = 4, λ3 = 0. By the Gauss elimi nation applied to (A – λI)x = 0 with λ = λ1, λ2, λ3 we find eigenvectors and then X^-1 by the Gauss–Jordan elimina tion (Sec. 7.8, Example 1). The results are

continued

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

Calculating AX and multiplying by X^-1 from the left, we t hus obtain

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Quadratic Forms. Transformation to Principal Axes



By definition, a quadratic form Q in the components x1

, ‥‥, xn of a vector x is a sum of n² terms, namely,

(7)

continued

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

_{A = [ajk}] is called the coefficient matrix of the form. W e may assume that A is symmetric, because we can t ake off-diagonal terms together in pairs and write the re sult as a sum of two equal terms; see the following exa mple.

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E X A M P L E 5

Quadratic Form.

Symmetric Coefficient Matrix



_Let

Here 4 + 6 = 10 = 5 + 5. From the corresponding sym metric matrix C = [c_jk], where c_jk = 1/2 (a_jk + a_kj), thus c₁₁

= 3, c12 = c21 = 5, c22 = 2, we get the same result; indee d,

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

By Theorem 2 the symmetric coefficient matrix A of (7) has an orthonormal basis of eigenvectors. Hence if we take these as column vectors, we obtain a matrix X that is orthogonal, so that X^-1 = X^T. From (5) we thus have A

= XDX^-1 = XDX^T. Substitution into (7) gives (8) Q = x^TXDX^Tx.

continued

353

If we set X^Tx = y, then, since X^T = X^-1, we get (9) x = Xy.

Furthermore, in (8) we have x^TX = (X^Tx)^T = y^T and X^Tx

= y, so that Q becomes simply (10)

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Principal Axes Theorem

THEOREM 5

The substitution (9) transforms a quadratic form

to the principal axes form or canonical form (10), whe re λ₁, ‥‥, λ_n are the (not necessarily distinct) eigenva lues of the (symmetric!) matrix A, and X is an orthogon al matrix with corresponding eigenvectors x₁, ‥‥, x_n, respectively, as column vectors.

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E X A M P L E 6

Transformation to Principal Axes.

Conic Sections



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

Q = 17x₁₂ – 30x₁x₂ + 17x₂₂ = 128.



Solution. We have Q = x^TAx, where

continued

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

This gives the characteristic equation (17 – λ)² – 15² = 0. It has the roots λ₁ = 2, λ₂ = 32. Hence (10) becomes Q = 2y12 + 32y22.

We see that Q = 128 represents the ellipse 2y₁₂ + 32y₂₂

= 128, that is,

continued

354



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 S ec. 8.2, Example 1, except for the notations. See also Fig. 158 in that example.

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

A quadratic form Q(x) = xTAx and its matrix A are calle d (a) positive definite if Q(x) > 0 for all x ≠ 0, (b) nega tive definite if Q(x) < 0 for all x ≠ 0, (c) indefinite if Q (x) takes both positive and negative values. A necessar y and sufficient condition for positive definiteness is tha t all the “principal minors” are positive.

continued

356

Fig. 160.

Quadratic forms in two variables

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8.5 Complex Matrices and Forms. Optional



_Notations

Ā = [ājk] is obtained from A = [ajk] by replacing each ent ry ajk = α + iβ (α, β real) with its complex conjugate ājk = α – iβ. Also, Ā^T = [ākj] is the transpose of Ā, hence the c onjugate transpose of A.

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E X A M P L E 1

Notations

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Hermitian, Skew-Hermitian, and Unitary Matrices

DEFINITION

A square matrix A = [akj] is called

Hermitian if Ā^T = A, that is, ā_kj = a_jk skew-Hermitian if Ā^T = –A, that is, ā_kj = –a_jk unitary if Ā^T = A^-1,

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E X A M P L E 2

Hermitian, Skew-Hermitian, and Unitary Matrices

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

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Eigenvalues

Eigenvalues

THEOREM 1

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

(b) The eigenvalues of a skew-Hermitian matrix (an d 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.

continued

358

Fig. 161.

Location of the eigenvalues of Hermitian,

skew-Hermitian, and unitary matrices in the complex λ-plane

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E X A M P L E 3

Illustration of Theorem 1

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For the matrices in Example 2 we find by direct calculat ion

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Invariance of Inner Product

THEOREM 2

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).

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Unitary System

DEFINITION

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

(6)

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Unitary Systems of Column and Row Vectors

THEOREM 3

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

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Determinant of a Unitary Matrix

THEOREM 4

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

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E X A M P L E 4

Unitary Matrix Illustrating Theorems 1c and 2–4

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For the vectors a^T = [2 –i] and b^T = [1 + i 4i] we get ā^T

= [2 i]^T and ā^Tb = 2(1 + i) – 4 = –2 + 2i and with

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

continued

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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.

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Basis of Eigenvectors

THEOREM 5

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

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E X A M P L E 5

Unitary Eigenbases

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

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Hermitian and Skew-Hermitian Forms

E X A M P L E 6

Hermitian Form

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

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SUMMARY OF CHAPTER 8

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The practical importance of matrix eigenvalue problem s can hardly be overrated. The problems are defined b y the vector equation

(1) Ax = λx.

A is a given square matrix. All matrices in this chapter are square. λ is a scalar. To solve the problem (1) mea ns to determine values of λ, called eigenvalues (or ch aracteristic values) of A, such that (1) has a nontrivial solution x (that is, x ≠ 0), called an eigenvector of A co rresponding to that λ. An n × n matrix has at least one and at most n numerically different eigenvalues.

continued

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These are the solutions of the characteristic equation (Sec. 8.1)

(2)

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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.

continued

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

Section 8.3 is devoted to eigenvalue problems for sym metric (A^T = A), skew-symmetric (A^T = –A), and orth ogonal matrices (A^T = A^-1). Section 8.4 concerns the d iagonalization of matrices and the transformation of qu adratic forms to principal axes and its relation to eigenv alues.

continued

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

Section 8.5 extends Sec. 8.3 to the complex analogs of those real matrices, called Hermitian (Ā^T = A), skew-H ermitian (Ā^T = –A), and unitary matrices (Ā^T = A^-1). Al l the eigenvalues of a Hermitian matrix (and a symmetri c one) are real. For a skew-Hermitian (and a skew-sym metric) matrix they are pure imaginary or zero. For a un itary (and an orthogonal) matrix they have absolute val ue 1.