EIGENVALUES AND EIGENVECTORS


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1 Chapter 6 EIGENVALUES AND EIGENVECTORS 61 Motivation We motivate the chapter on eigenvalues b discussing the equation ax + hx + b = c, where not all of a, h, b are zero The expression ax + hx + b is called a quadratic form in x and and we have the identit where X = form ax + hx + b = x a h h b x and A = a h h b x = X t AX, A is called the matrix of the quadratic We now rotate the x, axes anticlockwise through θ radians to new x 1, 1 axes The equations describing the rotation of axes are derived as follows: Let P have coordinates (x, ) relative to the x, axes and coordinates (x 1, 1 ) relative to the x 1, 1 axes Then referring to Figure 61: 115
2 116 CHAPTER 6 EIGENVALUES AND EIGENVECTORS 1 P x 1 R α θ O Q x Figure 61: Rotating the axes x = OQ = OP cos (θ + α) = OP(cos θ cos α sinθ sinα) = (OP cos α)cos θ (OP sinα)sin θ = OR cos θ PR sinθ = x 1 cos θ 1 sin θ Similarl = x 1 sinθ + 1 cos θ We can combine these transformation equations into the single matrix equation: x cos θ sinθ x1 =, sinθ cos θ 1 x x1 cos θ sinθ or X = PY, where X =, Y = and P = 1 sinθ cos θ We note that the columns of P give the directions of the positive x 1 and 1 axes Also P is an orthogonal matrix we have PP t = I and so P 1 = P t The matrix P has the special propert that detp = 1 cos θ sinθ A matrix of the tpe P = is called a rotation matrix sinθ cos θ We shall show soon that an real orthogonal matrix with determinant
3 61 MOTIVATION 117 equal to 1 is a rotation matrix We can also solve for the new coordinates in terms of the old ones: x1 = Y = P t cos θ sinθ x X =, sinθ cos θ 1 so x 1 = xcos θ + sinθ and 1 = xsinθ + cos θ Then X t AX = (PY ) t A(PY ) = Y t (P t AP)Y Now suppose, as we later show, that it is possible to choose an angle θ so that P t AP is a diagonal matrix, sa diag(λ 1, λ ) Then X t AX = λ x x1 1 = λ 0 λ 1 x 1 + λ 1 (61) and relative to the new axes, the equation ax + hx + b = c becomes λ 1 x 1 + λ 1 = c, which is quite eas to sketch This curve is smmetrical about the x 1 and 1 axes, with P 1 and P, the respective columns of P, giving the directions of the axes of smmetr Also it can be verified that P 1 and P satisf the equations 1 AP 1 = λ 1 P 1 and AP = λ P These equations force a restriction on λ 1 and λ For if P 1 = first equation becomes a h u1 u1 = λ h b 1 v 1 v 1 or a λ1 h h b λ 1 u1 v 1 = u1 v 1 0 0, the Hence we are dealing with a homogeneous sstem of two linear equations in two unknowns, having a non trivial solution (u 1, v 1 ) Hence a λ 1 h h b λ 1 = 0 Similarl, λ satisfies the same equation In expanded form, λ 1 and λ satisf λ (a + b)λ + ab h = 0 This equation has real roots λ = a + b ± (a + b) 4(ab h ) = a + b ± (a b) + 4h (6) (The roots are distinct if a b or h 0 The case a = b and h = 0 needs no investigation, as it gives an equation of a circle) The equation λ (a+b)λ+ab h = 0 is called the eigenvalue equation of the matrix A
4 118 CHAPTER 6 EIGENVALUES AND EIGENVECTORS 6 Definitions and examples DEFINITION 61 (Eigenvalue, eigenvector) Let A be a complex square matrix Then if λ is a complex number and X a non zero complex column vector satisfing AX = λx, we call X an eigenvector of A, while λ is called an eigenvalue of A We also sa that X is an eigenvector corresponding to the eigenvalue λ So in the above example P 1 and P are eigenvectors corresponding to λ 1 and λ, respectivel We shall give an algorithm which starts from the a h eigenvalues of A = and constructs a rotation matrix P such that h b P t AP is diagonal As noted above, if λ is an eigenvalue of an n n matrix A, with corresponding eigenvector X, then (A λi n )X = 0, with X 0, so det(a λi n ) = 0 and there are at most n distinct eigenvalues of A Conversel if det (A λi n ) = 0, then (A λi n )X = 0 has a non trivial solution X and so λ is an eigenvalue of A with X a corresponding eigenvector DEFINITION 6 (Characteristic polnomial, equation) The polnomial det (A λi n ) is called the characteristic polnomial of A and is often denoted b ch A (λ) The equation det(a λi n ) = 0 is called the characteristic equation of A Hence the eigenvalues of A are the roots of the characteristic polnomial of A a b For a matrix A =, it is easil verified that the characteristic polnomial is λ (tracea)λ+det A, where trace A = a+d is the sum c d of the diagonal elements of A 1 EXAMPLE 61 Find the eigenvalues of A = and find all eigenvectors 1 Solution The characteristic equation of A is λ 4λ + 3 = 0, or (λ 1)(λ 3) = 0 Hence λ = 1 or 3 The eigenvector equation (A λi n )X = 0 reduces to λ 1 1 λ x = 0 0,
5 6 DEFINITIONS AND EXAMPLES 119 or Taking λ = 1 gives ( λ)x + = 0 x + ( λ) = 0 x + = 0 x + = 0, which has solution x =, arbitrar Consequentl the eigenvectors corresponding to λ = 1 are the vectors, with 0 Taking λ = 3 gives x + = 0 x = 0, which has solution x =, arbitrar Consequentl the eigenvectors corresponding to λ = 3 are the vectors, with 0 Our next result has wide applicabilit: THEOREM 61 Let A be a matrix having distinct eigenvalues λ 1 and λ and corresponding eigenvectors X 1 and X Let P be the matrix whose columns are X 1 and X, respectivel Then P is non singular and P 1 λ1 0 AP = 0 λ Proof Suppose AX 1 = λ 1 X 1 and AX = λ X We show that the sstem of homogeneous equations xx 1 + X = 0 has onl the trivial solution Then b theorem 510 the matrix P = X 1 X is non singular So assume xx 1 + X = 0 (63) Then A(xX 1 + X ) = A0 = 0, so x(ax 1 ) + (AX ) = 0 Hence xλ 1 X 1 + λ X = 0 (64)
6 10 CHAPTER 6 EIGENVALUES AND EIGENVECTORS Multipling equation 63 b λ 1 and subtracting from equation 64 gives (λ λ 1 )X = 0 Hence = 0, as (λ λ 1 ) 0 and X 0 Then from equation 63, xx 1 = 0 and hence x = 0 Then the equations AX 1 = λ 1 X 1 and AX = λ X give AP = AX 1 X = AX 1 AX = λ 1 X 1 λ X λ1 0 λ1 0 = X 1 X = P 0 λ 0 λ so P 1 λ1 0 AP = 0 λ 1 EXAMPLE 6 Let A = be the matrix of example 61 Then X 1 = and X 1 = are eigenvectors corresponding to eigenvalues and 3, respectivel Hence if P =, we have 1 1 P AP = 0 3 There are two immediate applications of theorem 61 The first is to the calculation of A n : If P 1 AP = diag (λ 1, λ ), then A = Pdiag (λ 1, λ )P 1 and ( ) n n A n λ1 0 = P P 1 λ1 0 = P P 1 λ n = P λ 0 λ 0 λ n P 1 The second application is to solving a sstem of linear differential equations = ax + b d = cx + d, a b where A = is a matrix of real or complex numbers and x and c d are functions of t The sstem can be written in matrix form as Ẋ = AX, where x ẋ X = and Ẋ = = ẏ d,
7 6 DEFINITIONS AND EXAMPLES 11 We make the substitution X = PY, where Y = x1 are also functions of t and Ẋ = PẎ = AX = A(PY ), so Ẏ = (P 1 λ1 0 AP)Y = 0 λ 1 Then x 1 and 1 Y Hence x 1 = λ 1 x 1 and 1 = λ 1 These differential equations are well known to have the solutions x 1 = x 1 (0)e λ1t and 1 = 1 (0)e λt, where x 1 (0) is the value of x 1 when t = 0 If = kx, where k is a constant, then d ( ) e kt x = ke kt x + e kt = ke kt x + e kt kx = 0 Hence e kt x is constant, so e kt x = e k0 x(0) = x(0) Hence x = x(0)e kt x1 (0) x(0) However = P 1 (0) 1, so this determines x (0) 1 (0) and 1 (0) in terms of x(0) and (0) Hence ultimatel x and are determined as explicit functions of t, using the equation X = PY 3 EXAMPLE 63 Let A = Use the eigenvalue method to 4 5 derive an explicit formula for A n and also solve the sstem of differential equations d given x = 7 and = 13 when t = 0 = x 3 = 4x 5, Solution The characteristic polnomial of A is λ +3λ+ which has distinct 1 roots λ 1 = 1 and λ = We find corresponding eigenvectors X 1 = and X = Hence if P =, we have P AP = diag ( 1, ) Hence A n = ( Pdiag ( 1, )P 1) n = Pdiag (( 1) n, ( ) n )P ( 1) n = ( ) n 1 1
8 1 CHAPTER 6 EIGENVALUES AND EIGENVECTORS = ( 1) n = ( 1) n 1 3 n 1 4 n n = ( 1) n 4 3 n n 4 4 n n To solve the differential equation sstem, make the substitution X = PY Then x = x , = x The sstem then becomes ẋ 1 = x 1 ẏ 1 = 1, so x 1 = x 1 (0)e t, 1 = 1 (0)e t Now x1 (0) 1 (0) = P 1 x(0) (0) = = 11 6 so x 1 = 11e t and 1 = 6e t Hence x = 11e t + 3(6e t ) = 11e t + 18e t, = 11e t + 4(6e t ) = 11e t + 4e t For a more complicated example we solve a sstem of inhomogeneous recurrence relations EXAMPLE 64 Solve the sstem of recurrence relations given that x 0 = 0 and 0 = 1 x n+1 = x n n 1 n+1 = x n + n +, Solution The sstem can be written in matrix form as where A = X n+1 = AX n + B, 1 1 It is then an eas induction to prove that and B = 1 X n = A n X 0 + (A n A + I )B (65),
9 6 DEFINITIONS AND EXAMPLES 13 Also it is eas to verif b the eigenvalue method that A n = n 1 3 n 1 3 n n = 1 U + 3n V, where U = and V = Hence A n A + I = n U + (3n ) V = n U + (3n 1) V 4 Then equation 65 gives ( ) 1 X n = U + 3n V which simplifies to xn n = 0 1 ( n + U + (3n 1) V 4 (n n )/4 (n n )/4 Hence x n = (n n )/4 and n = (n n )/4 ) 1 REMARK 61 If (A I ) 1 existed (that is, if det(a I ) 0, or equivalentl, if 1 is not an eigenvalue of A), then we could have used the formula A n A + I = (A n I )(A I ) 1 (66) However the eigenvalues of A are 1 and 3 in the above problem, so formula 66 cannot be used there Our discussion of eigenvalues and eigenvectors has been limited to matrices The discussion is more complicated for matrices of size greater than two and is best left to a second course in linear algebra Nevertheless the following result is a useful generalization of theorem 61 The reader is referred to 8, page 350 for a proof THEOREM 6 Let A be an n n matrix having distinct eigenvalues λ 1,,λ n and corresponding eigenvectors X 1,,X n Let P be the matrix whose columns are respectivel X 1,,X n Then P is non singular and P 1 AP = λ λ λ n,
10 14 CHAPTER 6 EIGENVALUES AND EIGENVECTORS Another useful result which covers the case where there are multiple eigenvalues is the following (The reader is referred to 8, pages for a proof): THEOREM 63 Suppose the characteristic polnomial of A has the factorization det(λi n A) = (λ c 1 ) n1 (λ c t ) nt, where c 1,,c t are the distinct eigenvalues of A Suppose that for i = 1,,t, we have nullit (c i I n A) = n i For each such i, choose a basis X i1,,x ini for the eigenspace N(c i I n A) Then the matrix P = X 11 X 1n1 X t1 X tnt is non singular and P 1 AP is the following diagonal matrix P 1 AP = c 1 I n c I n c t I nt (The notation means that on the diagonal there are n 1 elements c 1, followed b n elements c,, n t elements c t ) 63 PROBLEMS Let A = Find an invertible matrix P such that P AP = diag (1, 3) and hence prove that If A = as n A n = 3n 1 A + 3 3n I, prove that A n tends to a limiting matrix /3 /3 1/3 1/3
11 63 PROBLEMS 15 3 Solve the sstem of differential equations d given x = 13 and = when t = 0 = 3x = 5x 4, Answer: x = 7e t + 6e t, = 7e t + 15e t 4 Solve the sstem of recurrence relations x n+1 = 3x n n n+1 = x n + 3 n, given that x 0 = 1 and 0 = Answer: x n = n 1 (3 n ), n = n 1 (3 + n ) a b 5 Let A = be a real or complex matrix with distinct eigenvalues c d λ 1, λ and corresponding eigenvectors X 1, X Also let P = X 1 X (a) Prove that the sstem of recurrence relations x n+1 n+1 = ax n + b n = cx n + d n has the solution xn n = αλ n 1X 1 + βλ n X, where α and β are determined b the equation α = P 1 x0 β 0 (b) Prove that the sstem of differential equations has the solution x d = ax + b = cx + d = αe λ1t X 1 + βe λt X,
12 16 CHAPTER 6 EIGENVALUES AND EIGENVECTORS where α and β are determined b the equation α = P 1 x(0) β (0) a11 a 6 Let A = 1 be a real matrix with non real eigenvalues λ = a 1 a a + ib and λ = a ib, with corresponding eigenvectors X = U + iv and X = U iv, where U and V are real vectors Also let P be the real matrix defined b P = U V Finall let a + ib = re iθ, where r > 0 and θ is real (a) Prove that AU AV = au bv = bu + av (b) Deduce that P 1 AP = a b b a (c) Prove that the sstem of recurrence relations x n+1 n+1 = a 11 x n + a 1 n = a 1 x n + a n has the solution xn = r n {(αu + βv )cos nθ + (βu αv )sinnθ}, n where α and β are determined b the equation α = P 1 x0 β 0 (d) Prove that the sstem of differential equations d = ax + b = cx + d
13 63 PROBLEMS 17 has the solution x = e at {(αu + βv )cos bt + (βu αv )sin bt}, where α and β are determined b the equation α = P 1 x(0) β (0) x x1 Hint: Let = P Also let z = x 1 + i 1 Prove that and deduce that 1 ż = (a ib)z x 1 + i 1 = e at (α + iβ)(cos bt + i sinbt) Then equate real and imaginar parts to solve for x 1, 1 and hence x, a b 7 (The case of repeated eigenvalues) Let A = and suppose c d that the characteristic polnomial of A, λ (a + d)λ + (ad bc), has a repeated root α Also assume that A αi Let B = A αi (i) Prove that (a d) + 4bc = 0 (ii) Prove that B = 0 (iii) Prove that BX 0 for some vector X ; indeed, show that X 1 0 can be taken to be or 0 1 (iv) Let X 1 = BX Prove that P = X 1 X is non singular, and deduce that AX 1 = αx 1 and AX = αx + X 1 P 1 AP = α 1 0 α 8 Use the previous result to solve sstem of the differential equations d = 4x = 4x + 8,
14 18 CHAPTER 6 EIGENVALUES AND EIGENVECTORS given that x = 1 = when t = 0 To solve the differential equation kx = f(t), k a constant, multipl throughout b e kt, thereb converting the left hand side to (e kt x) Answer: x = (1 3t)e 6t, = (1 + 6t)e 6t 9 Let A = 1/ 1/ 0 1/4 1/4 1/ 1/4 1/4 1/ (a) Verif that det(λi 3 A), the characteristic polnomial of A, is given b (λ 1)λ(λ 1 4 ) (b) Find a non singular matrix P such that P 1 AP = diag (1, 0, 1 4 ) (c) Prove that if n 1 A n = n Let A = (a) Verif that det(λi 3 A), the characteristic polnomial of A, is given b (λ 3) (λ 9) (b) Find a non singular matrix P such that P 1 AP = diag (3, 3, 9)
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