Cayley-Hamilton Theorem

1. Statement

Every square matrix satisfies its own characteristic equation.

If the characteristic equation of an n × n matrix A is

det(A - λI) = 0, say λ^n + a_1 λ^{n-1} + a_2 λ^{n-2} + ... + a_{n-1} λ + a_n = 0

then A satisfies exactly the same polynomial, with λ replaced by A and the constant term multiplied by I:

A^n + a_1 A^{n-1} + a_2 A^{n-2} + ... + a_{n-1} A + a_n I = 0

The identity matrix appears in the constant term because a matrix polynomial must be a matrix. Equivalently, Aⁿ can always be written as a linear combination of the lower powers I, A, A², …, Aⁿ⁻¹.

Example (verification on a 2 × 2):

        [ 1  2 ]
   A =  [ 2  1 ]        ⇒  det(A - λI) = (1-λ)² - 4 = λ² - 2λ - 3 = 0
        ⇒  A² - 2A - 3I = 0

        A² = [ 1  2 ][ 1  2 ] = [ 5  4 ]  ,  2A = [ 2  4 ]  ,  3I = [ 3  0 ]
             [ 2  1 ][ 2  1 ]   [ 4  5 ]        [ 4  2 ]        [ 0  3 ]
        A² - 2A - 3I = [ 5-2-3    4-4-0 ] = [ 0  0 ]  =  O   ✓
                       [ 4-4-0    5-2-3 ]   [ 0  0 ]

2. Three Standard Uses

(a) Finding the inverse A⁻¹ — from Aⁿ + a₁Aⁿ⁻¹ + … + aₙI = 0, multiply throughout by A⁻¹ and solve (legitimate only if aₙ = det A ≠ 0, i.e. A is non-singular):

A^{-1} = -(A^{n-1} + a_1 A^{n-2} + ... + a_{n-1} I) / a_n

(b) Computing high powers of A — repeatedly replace the highest power using the characteristic equation, so A⁴, A⁵ … stay in terms of I, A, A².

(c) Evaluating a polynomial in A — an expression such as A⁵ − 4A⁴ + 7A³ − 11A² + 5A − 2I can be reduced by dividing by the characteristic polynomial, or by repeated substitution.

        [ 1  2 ]
   A =  [ 2  1 ]        characteristic equation:  λ² - 2λ - 3 = 0
   ⇒  A² - 2A - 3I = 0        ⇒   A² - 2A = 3I
   Multiply by A⁻¹ :            A - 2I = 3 A⁻¹
   ⇒  A⁻¹ = (A - 2I)/3

            A - 2I = [ 1-2   2  ] = [ -1   2 ]
                     [  2   1-2 ]   [  2  -1 ]
   ⇒  A⁻¹ = (1/3)[ -1   2 ]
                 [  2  -1 ]

   Check:  A · A⁻¹ = (1/3)[ 1  2 ][ -1   2 ] = (1/3)[ -1+4    2-2  ] = (1/3)[ 3  0 ] = I ✓
                          [ 2  1 ][  2  -1 ]        [ -2+2    4-1  ]        [ 0  3 ]

6. Why It Works (short justification for theory questions)

Write the characteristic polynomial as p(λ) = det(λI − A) = (λ − λ₁)(λ − λ₂)…(λ − λₙ). The adjoint identity (λI − A)·adj(λI − A) = det(λI − A)·I shows that the entries of the adjoint are polynomials in λ of degree ≤ n − 1; replacing λ by A and simplifying the resulting matrix series gives p(A) = 0. For a diagonalizable A the result is immediate: A = PDP⁻¹, so p(A) = P·p(D)·P⁻¹, and p evaluated at each eigenvalue on the diagonal is zero.

7. Exam Points

State the Cayley–Hamilton theorem — 3 marks, almost certain.
Verify the theorem for a given 2 × 2 or 3 × 3 matrix — 5 marks, very frequent.
Find A⁻¹ using the Cayley–Hamilton theorem — 5-7 marks, the most frequent use.
Find A⁴ or A⁵ using the theorem (express in terms of A and I) — 5-7 marks, frequent.
Evaluate a polynomial in A, e.g. A⁴ − 5A³ + 7A² − 3A + 8I — 5 marks.
Compute Aⁿ for large n once the power pattern becomes clear — 5 marks.
Prove the theorem for a diagonalizable matrix A = PDP⁻¹ — 3-5 marks.

Cayley-Hamilton Theorem