Chapter 7 - Cayley-Hamilton Theorem PDF

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Chapter 7 Cayley-Hamilton Theorem   3 1 0 Example 7.1 Let A = 5 0 −1 . Does A satisfy any particular identity? 2 0 1

Solution. Consider A − λI. Clearly   3−λ 1 0 −λ −1  A − λI =  5 2 0 1−λ

and |A − λI| = −λ3 + 4λ2 + 2λ − 7.

Let B = A − λI and let C be the matrix of co-factors of B. Then   λ2 − λ λ−1 −1  B ∗ = C T =  5 λ − 7 3 − 4 λ + λ2 3−λ 2 2λ 2 λ − 3λ − 5  2      λ 0 0 −λ λ 0 0 −1 −1 3  =  0 λ2 0  +  5λ −4λ −λ  + −7 3 2 0 0 λ 2λ 0 −3λ 0 2 −5       0 −1 −1 −1 1 0 1 0 0 = λ2  0 1 0 + λ  5 −4 −1  +  −7 3 3 0 2 −5 2 0 −3 0 0 1

= λ2 M2 + λM1 + M0       1 0 0 −1 1 0 0 −1 −1 3 . where M2 = 0 1 0 , M1 =  5 −4 −1  and M0 = −7 3 0 0 1 2 0 −3 0 2 −5 Now • BB ∗ = |B| · I =⇒ BB ∗ = |B| · I = (−λ3 + 4λ2 + 2λ − 7)I and • BB ∗ = (A − λI)B ∗ = (A − λI )(λ2 M2 + λM1 + M0 ). 77

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CHAPTER 7. CAYLEY-HAMILTON THEOREM Therefore (−λ3 + 4λ2 + 2λ − 7)I = (A − λI )(λ2 M2 + λM1 + M0 ) −λ3 I + 4λ2 I + 2λI − 7I = A(λ2 M2 + λM1 + M0 ) − λ(λ2 M2 + λM1 + M0 ) −λ3 I + 4λ2 I + 2λI − 7I = λ2 AM2 + λAM1 + AM0 − λ3 M2 − λ2 M1 − λM0 Equating coefficients, we obtain −I = −M2 4I = AM2 − M1 2I = AM1 − M0 −7I = AM0 .

Now if we multiply the first equation by A3 , the second by A2 , the third by A1 and the last by A0 = I, we obtain −A3 = −A3 M2 4A2 = A3 M2 − A2 M1 2A = A2 M1 − AM0 −7I = AM0 . Adding these equations, we obtain the following identity: −A3 + 4A2 + 2A − 7I = 0. Clearly we can see that A satisfies it’s own characteristic equation.

♣

Theorem 7.2 (Cayley-Hamilton Theorem) Every square matrix satisfies it’s own characteristic equation. Let A be an n × n matrix and let p(λ) = |A − λI| =

n X

pi λi be the characteristic equation of A.

i=0

The Cayley-Hamilton Theorem states that p(A) = 0, i.e.

n X

pi Ai = 0

i=0

2

n−1

p0 + p1 A + p2 A + · · · + pn−1 A Proof. Let p(λ) =

n X

+ pn An = 0.

pi λi denote the characteristic polynomial of an n × n matrix A. Let

i=0

B = A − λI. Clearly

p(λ) = |A − λI | = |B|.

CHAPTER 7. CAYLEY-HAMILTON THEOREM

79

Clearly, B is an n × n matrix having first degree polynomial in λ for it’s diagonal elements and scalar elsewhere. It follows that every entry of B ∗ will contain polynomials (in λ) of degree less than or equal to n − 1. Therefore, B ∗ = λn−1 Mn−1 + λn−2 Mn−2 + · · · + λM1 + M0 where M0 , M1 , · · · , Mn−1 are all scalar matrices. Clearly BB ∗ = |B| · I. Therefore, BB ∗ = |B| · I =⇒ BB ∗ = p(λ) · I and BB ∗ = BB ∗ = (A − λI )B ∗ = (A − λI )(λn−1 Mn−1 + λn−2 Mn−2 + · · · + λM1 + M0 ). Thus,

n X i=0

p(λ) · I = (A − λI )(λn−1 Mn−1 + λn−2 Mn−2 + · · · + λM1 + M0 ) !

pi λi

· I = A(λn−1 Mn−1 + λn−2 Mn−2 + · · · + λM1 + M0 ) − λ(λn−1 Mn−1 + λn−2 Mn−2 + · · · + λM1 + M0 )

p0 I + p1 λI + p2 λ2 I + · · · + pn−1 λn−1 I + pn λn I = AMn−1 λn−1 + AMn−2 λn−2 + · · · + AM1 λ + AM0 − Mn−1 λn − Mn−2 λn−1 − · · · − M1 λ2 − λM0 . Equating coefficients, we obtain pn I = −Mn−1 pn−1 I = AMn−1 − Mn−2 pn−2 I = AMn−2 − Mn−3 .. . p1 I = AM1 − M0 p0 I = AM0 . If we multiply the first equation by An , the second equation by An−1 and so on, we obtain pn An = −An Mn−1 pn−1 An−1 = An Mn−1 − An−1 Mn−2 pn−2 An−2 = An−2 Mn−2 − An−2 Mn−3 .. . p1 A = A2 M1 − AM0 p0 = AM0 .

80

CHAPTER 7. CAYLEY-HAMILTON THEOREM If we add these equations, clearly p0 + p1 A + · · · + pn−2 An−2 + pn−1 An−1 + pn An = 0.



Example 7.3 Verify the Cayley-Hamilton theorem for A =



 3 5 . −2 −4

  3−λ 5 Solution. Now A − λ = and |A − λ| = λ2 + λ − 2. Therefore we need to show −2 −4 − λ that A2 + A − 2I = 0 to verify the Cayley-Hamilton theorem. Therefore 2    3 5 3 5 1 A + A − 2I = + −2 −2 −4 0 −2 −4      −1 −5 3 5 1 = + −2 2 6 0 −2 −4   −1 + 3 − 2 −5 + 5 + 0 = 2−2−0 6−4−2   0 0 . = 0 0 2



 0 1  0 1

♣

Example 7.4 Verify the Cayley-Hamilton theorem for A =



 3 4 . −5 −5

  3−λ 4 Solution. Now A − λ = and |A − λ| = λ2 + 2λ + 5. Therefore we need to show −5 −5 − λ that A2 + 2A + 5I = 0 to verify the Cayley-Hamilton theorem. Therefore 2      1 0 3 4 3 4 2 + 5 A + 2A + 5I = + −5 −5 −5 −5 0 1       6 8 5 0 −11 −8 + + = 10 5 −10 −10 0 5   0 0 = . 0 0 ♣

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CHAPTER 7. CAYLEY-HAMILTON THEOREM Corollary 7.5 Let A be an n × n matrix and let p(λ) = |A − λI| =

n X

pi λi be the characteristic

i=0

equation of A. The A

−1

" n # 1 X i−1 =− pi A p0 i=1

Proof. Let A be an n × n matrix and let p(λ) = |A − λI| =

n X

pi λi be the characteristic equation

i=0

of A. By the Cayley-Hamilton theorem,

n X

pi Ai = 0

i=0

p0 + p1 A + p2 A2 + · · · + pn−1 An−1 + pn An = 0. Now if we pre-multiply both sides by A−1 , then A−1 (p0 + p1 A + p2 A2 + · · · + pn−1 An−1 + pn An ) = 0 p0 A−1 + p1 I + p2 A + · · · + pn−1 An−2 + pn An−1 = 0 p0 A−1 = −1(p1 I + p2 A + · · · + pn−1 An−2 + pn An−1 ) 1 A−1 = − (p1 I + p2 A + · · · + pn−1 An−2 + pn An−1 ) p0 # " n 1 X i−1 =− pi A . p0 i=1    1 4 Example 7.6 Use the Cayley-Hamilton theorem to calculate the inverse of A = . 2 3   1−λ 4 Solution. A − λ = and 2 3−λ |A − λ| = (1 − λ)(3 − λ) − (4)(2) = λ2 − 4λ − 5. By the Cayley-Hamilton theorem, A satisfies: A2 − 4A − 5I = 0. Therefore A−1 (A2 − 4A − 5I) = 0 A − 4I − 5A−1 = 0

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CHAPTER 7. CAYLEY-HAMILTON THEOREM 5A−1 = A − 4I 1 A−1 = {A − 4I } 5     1 4 0 1 4 = − 2 3 0 4 5   1 −3 4 = . 5 2 −1

♣



 3 −2 −1 Example 7.7 Let A =  2 3 4 . Use the Cayley-Hamilton theorem to calculate A−1 and −2 0 5 A4 .   3 − λ −2 −1 Solution. Clearly A − λI =  2 3−λ 4  and −2 0 5−λ        2 3 − λ 2 3 − λ 4  4      − 1 − (−2)  |A − λI| = (3 − λ)  −2 0  −2 5 − λ 0 5 − λ = (3 − λ)[(3 − λ)(5 − λ) − 0] + 2[10 − 2λ + 8] − 1[0 + 2(3 − λ)] = (3 − λ)(3 − λ)(5 − λ) + 2[18 − 2λ] − 2(3 − λ) = (9 − 6λ + λ2 )(5 − λ) + 2[18 − 2λ] − 2(3 − λ) = 45 − 9λ − 30λ + 6λ2 + 5λ2 − λ3 + 36 − 4λ − 6 + 2λ = −λ3 + 11λ2 − 41λ + 75. By the Cayley-Hamilton theorem, A satisfies: −A3 + 11A2 − 41A + 75I = 0. Therefore A−1 (−A3 + 11A2 − 41A + 75I) = 0 −A2 + 11A − 41I + 75A−1 = 0 75A−1 = A2 − 11A + 41I  1  2 A − 11A + 41I 75   2 3 −2 −1 3  1  2 3 4  − 11  2 = 75  −2 0 5 −2    7 −12 −16 −33 1    4 5 30 + −22 = 75  −16 −4 −27 22   15 10 −5 1  = −18 13 −14  . 75 6 4 13

A−1 =

  −2 −1 41 0 3 4  +  0 41 0 5 0 0   22 11 41 −33 −44 +  0 0 −55 0

 0  0  41  0 0  41 0   0 41

CHAPTER 7. CAYLEY-HAMILTON THEOREM

83

Also A(−A3 + 11A2 − 41A + 75I) = 0 −A4 + 11A3 − 41A2 + 75A = 0. Thus A4 = 11A3 − 41A2 + 75A       319 −550 −1485 −287 492 656 225 −150 −75 =  −418 77 1826  +  −164 −205 −1230  +  150 225 300  −1034 484 1837 656 −164 −1107 −150 0 375   257 −208 −904 896  =  −432 97 −528 320 1105

♣

  3 1 Example 7.8 Let A = . Use the Cayley-Hamilton theorem to find constants a and b 5 −2 such that: A3 = aA + bI. Solution.   3−λ 1 A−λ= and 5 −2 − λ |A − λ| = (3 − λ)(−2 − λ) − (5)(1) = λ2 − λ − 6 − 5 = λ2 − λ − 11. By the Cayley-Hamilton theorem, A satisfies: A2 − A − 11I = 0. Therefore A(A2 − A − 11I) = 0 A3 − A2 − 11A = 0 A3 = A2 + 11A A3 = A + 11I + 11A A3 = 12A + 11I. Therefore a = 12 and b = 11. ♣

CHAPTER 7. CAYLEY-HAMILTON THEOREM 3  2 Example 7.9 Let A = 

−2

and c such that:



84

−2 −1 3 4 . Use the Cayley-Hamilton theorem to find constants a, b 0

5

A4 = aA2 + bA + cI. 

3 −2 −1 3 4 . Recall that the characteristic equation of A is p(λ) = Solution. Let A =  2 −2 0 5 −λ3 + 11λ2 − 41λ + 75. Therefore A3 = 11A2 − 41A + 75I and A4 = 11A3 − 41A2 + 75A = 11(11A2 − 41A + 75I) − 41A2 + 75A = 121A2 − 451A + 825I − 41A2 + 75A = 80A2 − 376A + 825I. Therefore a = 80, b = −376 and c = 825.

♣

  2 4 . Use the Cayley-Hamilton Theorem to formulate An+2 . Example 7.10 Let A = 3 1   2 4 Solution. Let A = . |A − λI| = λ2 − 3λ − 10, By CH-Theorem A2 − 3A − 10I = 0. 3 1 Therefore A2 = 3A + 10 I A3 = 19A + 30 I A4 = 87A + 190 I A5 = 451A + 870 I .. . First, we solve yn+2 = 3yn+1 + 10yn for yn where y0 = 3 and y1 = 19. Recall that the solution of the difference equation ayn+2 + byn+1 + cyn = 0 is yn = A(m1 )n + B (m2 )n where m1 , m2 (m1 6= m2 ) are the solutions of the equation am2 + bm + c = 0. Now yn+2 = 3yn+1 + 10yn =⇒ yn+2 − 3yn+1 − 10yn = 0. Now m2 − 3m − 10 = 0 =⇒ (m − 5)(m + 2) = 0 =⇒ m1 = 5, m2 = −2. Therefore yn = A5n + B(−2)n . Now, we need to use the fact that y0 = 3 and y1 = 19 to calculate A and B. Clearly 4 25 and B = − and y0 = 3 =⇒ A + B = 3 and y1 = 19 =⇒ 5A − 2B = 19, therefore A = 7 7     25 n 4 yn = 5 − (−2)n . 7 7

85

CHAPTER 7. CAYLEY-HAMILTON THEOREM

Next solve gn+2 = gyn+1 + 10gn for gn where g0 = 10 and g1 = 30. Clearly, gn = C5n + D(−2)n . 50 20 Now, g0 = 10 =⇒ C + D = 10 and g1 = 30 =⇒ 5C − 2D = 30, therefore C = and and B = 7 7     50 n 20 gn = 5 + (−2)n . 7 7 Therefore An+2 = yn · A + gn · I n ≥ 0.           4 20 50 n 25 n n n n+2 5 − (−2) A + 5 + (−2) I A = 7 7 7 7

n ≥ 0. ♣...