Cayley-Hamilton Theorem

CAYLEY-HAMILTON THEOREM

Theorem 1.3.1 Every square matrix satisfies its characteristic equation

Cayley-Hamilton theorem has two important uses (1) to find the inverse of a non-singular matrix A and (2) to find higher integral powers of A.

WORKED EXAMPLE

Example 1

Verify that A= satisfies its own characteristic equation and hence find A4.

Solution

Given A =

The characteristic equation of A is |A - λI| = 0

⇒ λ2 − S1λ + S2 = 0

where S1 = 1 + (-1) = 0, S2 = |A| = -1 - 4 = -5 

⸫ the characteristic equation is λ2 - 5 = 0           (1)

By Cayley-Hamilton theorem, A satisfies (1) 

i.e.      A2 – 5I = 0                                               (2)

We shall now verify this by direct computation.

⇒       A2 – 5I = 0

Hence A satisfies its characteristic equation.

To find A4: We have A2 = 5I                 [from (2)]

Example 2

Verify Cayley-Hamilton theorem for the matrix A = and find its inverse. Also express A5 - 4A4 - 7A3 + 11A2 – A – 10I as a linear polynomial in A.

Solution

Given A =

The characteristic equation of A is |A - λI| = 0

⇒  λ2 − S1λ + S2 = 0

where S1 = 1 + 3 = 4, S2 = |A| = 3 – 8 = -5

⸫ the characteristic equation is λ2 - 4λ - 5 = 0         (1)

50 By Cayley-Hamilton theorem, A satisfies (1)

⸫ A2- 4A - 5I = 0                                                       (2)

We shall now verify this by direct computations.

⇒ A2 - 4A - 51 = 0. Hence the theorem is verified.

To find A-1: We have 5I = A2 - 4A

Multiply by A-1, we get 5A-1 = A-1A2 - 4A-1 A

= A – 4I

Finally, to find A5 - 4A4 -7A3 + 11A2 - A – 10I:

Consider the polynomial λ5 - 4 λ4 - 7 λ3 + 11 λ2 - λ - 10

Divide it by λ2 - 4 λ - 5. We get the quotient λ3 - 2 λ + 3 and remainder λ + 5

⸫ λ5 - 4 λ4 - 7 λ3 + 11 λ2 - λ - 10 = (λ2 - 4 λ — 5)( λ3 - 2 λ + 3) + λ + 5

Replace λ by A, we get 

A5 - 4 A4 - 7 A3 + 11 A2 - A – 10I = (A2 - 4 A — 5I)( A3 - 2 A + 3) + A + 5I

= 0 + A + 5I      [using (2)]

= A + 5I,

which is a linear polynomial in A.

Example 3

Find the characteristic equation of the matrix A given Hence find A-1 and A4.

Solution

where S1 = sum of the diagonal elements of A

= 2 + 2 + 2 = 6

S2 = sum of minors of the diagonal elements of A

= 4 - 1 + 4 - 1 + 4 - 1 = 9

S3 = |A| = 2(4 − 1) + (−2 + 1) + (1 − 2) = 6 – 1 – 1 = 4

⸫ the characteristic equation is λ3 − 6λ2 + 9λ − 4 = 0

By Cayley-Hamilton theorem, A satisfies its characteristic equation

Example 4

Use Cayley-Hamilton theorem to find the matrix

A8 - 5A7 + 7A6 - 3A5 + 8A4 - 3A5 + 8A4 - 5A3+ 8A2 – 2A + I if the matrix

Solution

The characteristic equation is |A –λI| = 0

⸫ the characteristic equation is λ3 — 5λ2 + 7λ − 3 = 0

By Cayley-Hamilton theorem, we get A3 - 5A2 + 7A – 3I = 0         (1)

We have to find the matrix

A8 — 5A2 + 7A6 – 3A5 + 8A4 - 5A3 + 8A2 - 2A + I = ƒ(A), say We shall rewrite this matrix polynomial in terms of

A3 - 5A2 + 7A – 3I

⸫ the polynomial

ƒ(A) = A5(A3 - 5A2 + 7A – 3I) + 8A4 - 5A3 + 8A2 − 2A + I

= 8A4 - 5A3 + 8A2 - 2A+ I          [Using (1)]

= 8A(A3 - 5A2 + 7A – 3I) + 35A3 - 48A2 + 22A + I

= 35A3 - 48A2 + 22A + I            [Using (1)]

= 35(A3 - 5A2 + 7A – 3I) + 127A2 – 223A + 106I

= 127A2 - 223A + 106I              [Using (1)]

Example 5

Verify Cayley-Hamilton theorem and find the inverse of

Solution

The characteristic equation of A is │A - λI│ = 0

⸫ the characteristic equation is λ3 - 10 λ2 + 8 λ + 64 = 0       (1)

By Cayley-Hamilton theorem A statisfies (1)

⸫ A3 - 10A2 + 8A + 64I = 0       (2)

We shall now verify this by direct computations.

Now A3 - 10A2 + 8A + 64I

⸫ A3 - 10A2 + 8A + 64I = 0

Hence the theorem is verified.

To find A-1: We have      64I = -A3 +10A2 – 8A

Multiplying by A-1, we get

Example 6

If A = then show that An = An¬2 + A2 - I for n ≥ 3. Hence find A50.

Solution

⸫ the characteristic equation is λ3 - λ2 + λ + 1 = 0     (1)

By Cayley-Hamilton theorem A statisfies (1)

⸫     A3 - A2 + A + I = 0

⇒ A3 - A2 = A - I                                                       (2)

Multiplying (2) by A, A2, ..., An-3, we get the equations

A4 - A3 = A2 - A

A5 - A4 = A3 - A2

Adding (2) and all these equations we get

= An−4 + 2(A2 –I)                   (ii)

= An-6 + A2 – 1 + 2(A2 –I)                   

= An-6 + 3(A2 –I)                      (iii)             

Example 7

If A = find An in terms of A and I.

Solution

Given

The characteristic equation is λ2 - 5 = 0     [see example 17]

By Cayley-Hamiltons theorem A2 – 5I = 0                       (1)

To find An, consider the polynomial λn

Dividing λn by λ2 – 5 we get λn = (λ2 - 5) φ (λ) + aλ + b    (2)

where φ (λ) is the quotient and aλ + b is the remainder. 

We shall now find the values of a and b.

Example 8

Using Cayley-Hamilton theorem find A-1, where A =

Solution

= (1 - 4) + (-7 + 12) + (−7 + 12)

= - 3 + 5 + 5 = 7

S3 = |A| = 7(1 - 4) -2(6 - 12) + (-2) (− 12 + 6) = −21 + 12 + 12 = 3

⸫ the characteristic equation is λ3 − 5 λ2 + 7λ − 3 = 0

By Cayley --Hamilton theorem, A satisfies this equation.

⸫  A3 - 5A2 + 7A – 3I = 0

⇒ 3I = A3 - 5A2 + 7A

Multiply by A-1, 3A-1 = A2 - 5A + 7I 

Example 9

Verify Cayley-Hamilton theorem for A = Hence using it find A-1.

Solution

= 3 + 28 + 15 = 40

⸫ The characteristic equation is λ3 + λ2 – 18 λ – 40 = 0.

By Cayley-Hamilton theorem A satisfies this equation

⸫ A3 + A2 – 18A – 40I = 0                 (1)

We shall verify (1) by actual computation.

⸫ A3 + A2 – 18A – 40I = 0.

Hence Cayley-Hamilton theorem is verified.

We have 40I = A3 + A2 – 18A

Multiply by A-1.

⸫ 40A-1 = A2 + A – 18I

Example 10

If verify Cayley-Hamilton theorem and hence find A-1

Solution

where S1 = sum of the diagonal elements of A

= 3 +5 + 3 = 11

 S1 = sum of the minors of diagonal elements of A 

= 15 -1 + 9 – 1 +15  = 36

= 3 (15 − 1) − (−1)(−3 + 1) + 1.(1 − 5)

= 42 - 2 - 4 = 36

⸫  the characteristic equation is λ3 − 11 λ2 + 36λ – 36 = 0       (1)

A satisfies (1)

⸫  A3 – 11A2 + 36A – 36I = 0                                                      (2)

We now verify the equation (2)

We find A2, A3

Equation (2) is satisfied

Hence the Cayley-Hamilton is verified.

Row to find A-1

Pneumatic applying (2) A-1, we get

Example 11

Using Cayley Hamilton Theorem find A-1 and A4 if A =

Solution

= 1(3-0) -2(-1 - 0) + (-2) (2 - 0)

= 3 + 2 - 4 = 1

The characteristic equation is λ3 − 4 λ2 + 9 λ − 1 = 0

By Cayley Hamilton Theorem, it satisfies the characteristic equation

⸫  A3 - 4А2 + 9A − 1 = 0

Post multiplying by A-1,

⇒ A3A-1 - 4A2A-1 + 9AA-1 - IA-1 = 0

⇒ A2 - 4A + 9I - A-1 = 0

A-1 = А2 + 4A + 9I

and pre multiplying by A,

A4 - 4A3 + 9A2 - A = 0

A4 = 4A3 + 9A2 + A 

EXERCISE

Verify Cayley-Hamilton theorem for the following matrices and hence find their inverses.

5. Verify that the matrix A = satisfies its characteristic equation and hence find A4.

6. find An in terms of A and I using Cayley-Hamilton theorem and hence find A3.

7. Find A4 using Cayley-Hamilton theorem for the matrix

Find A4 + A3 – 18A2 – 39A + 2I

8. Find the eigen values and eigen vectors of the system of equations 10x1 + 2x2 + x3 = λx1, 2x1 + 10x2 + x3 = λx2, 2x1 + x2 + 10x3 = λx3 

[Hint: Equations can be rewritten as (10 - λ)x1 + 2x2 + x3 = 0, 2x1 + (10 − λ)x2 + x3 = 0, 2x1 + x2 + (10 - λ)x3 = 0

then these equations in matrix form is (A – λI)X = 0 and so | A – λI | = 0 is the characteristic equation of A and λ = 8, 9, 13. Eigen vectors are given by (I)]

9. If λ is an eigen value of a non-singular matrix A, show that |A| / λ is an eigen value of the matrix adj A.

[Hint: AX = λ X ⇒ (adj A) (AX) = (adj A) (λX)

10. Verify Cayley-Hamilton theorem for the matrix and find its inverse and also find A6 – 4A5 + 8A4 – 12A3 + 14A2.

ANSWERS TO EXERCISE 1.2