Cayley–Hamilton theorem

Every matrix satisfies its own characteristic equation — and how to use that to find inverses and reduce high powers without heavy computation.

Module contents

Statement

Every square matrix satisfies its own characteristic equation.

If the characteristic equation of $A$ is

$λ^{n} + c_{n - 1} λ^{n - 1} + \dots + c_{1} λ + c_{0} = 0$

then replacing $λ$ by $A$ and the constant term by $c_{0} I$ gives a true matrix identity:

$A^{n} + c_{n - 1} A^{n - 1} + \dots + c_{1} A + c_{0} I = O$

Note carefully: the right-hand side is the null matrix $O$ , not the scalar $0$ , and the constant term carries an $I$ . Writing $c_{0}$ alone is dimensionally meaningless.

Order 2

$A^{2} - (tr A) A + ∣ A ∣ I = O$

Order 3

$A^{3} - S_{1} A^{2} + S_{2} A - S_{3} I = O$

with $S_{1} = tr A$ , $S_{2} =$ sum of principal $2 \times 2$ minors, $S_{3} = ∣ A ∣$ .

Use 1 — Finding the inverse

This is the most common exam application. Start from the theorem for order $3$ :

$A^{3} - S_{1} A^{2} + S_{2} A - S_{3} I = O$

Multiply throughout by $A^{- 1}$ (legal when $∣ A ∣ \neq = 0$ ), using $A^{- 1} A^{k} = A^{k - 1}$ :

$A^{2} - S_{1} A + S_{2} I - S_{3} A^{- 1} = O$

Rearrange:

$A^{- 1} = \frac{1}{S _{3}} (A^{2} - S_{1} A + S_{2} I)$

For order $2$ the same procedure gives

$A^{- 1} = \frac{1}{∣ A ∣} (tr A \cdot I - A)$

No cofactors, no adjoint — just one matrix multiplication and some scalar arithmetic.

Why this fails for singular matrices. The constant term $S_{3}$ is $∣ A ∣$ . Dividing by it is exactly the step that requires $∣ A ∣ \neq = 0$ .

Use 2 — Reducing high powers

The theorem expresses $A^{n}$ in terms of lower powers:

$A^{3} = S_{1} A^{2} - S_{2} A + S_{3} I$

Multiply by $A$ to get $A^{4}$ in terms of $A^{3}, A^{2}, A$ — then substitute the line above to bring it back down. Repeating this, every power of $A$ reduces to a combination of $I, A, \dots, A^{n - 1}$ .

For large exponents there is a cleaner route: divide the polynomial $λ^{k}$ by the characteristic polynomial $p (λ)$ ,

$λ^{k} = p (λ) q (λ) + r (λ), de g r < n$

Since $p (A) = O$ , substituting $A$ collapses the first term:

$A^{k} = r (A)$

So computing $A^{50}$ needs only a polynomial remainder and then a matrix of degree at most $n - 1$ .

Verifying the theorem

When a question says "verify the Cayley–Hamilton theorem for $A$ ", it wants:

The characteristic equation of $A$ .
The powers $A^{2}$ (and $A^{3}$ if order $3$ ), computed explicitly.
Substitution into the equation, position by position, ending at $O$ .

Show the arithmetic. The mark scheme is for the working, not the conclusion.

Worked examples

2 solved

Every step is shown, in the order you would write it in an answer book.

Example 1

Verifying the theorem for a 2 × 2 and finding A⁻¹

Question

Verify the Cayley–Hamilton theorem for $A = [1423]$ and use it to find $A^{- 1}$ .

Characteristic equation
$tr A = 1 + 3 = 4, ∣ A ∣ = 3 - 8 = - 5$

$λ^{2} - 4 λ - 5 = 0$
Compute $A^{2}$
$A^{2} = [1423] [1423] = [1 + 8 4 + 12 2 + 6 8 + 9] = [916817]$
Substitute into $A^{2} - 4 A - 5 I$
$4 A = [416812], 5 I = [5005]$

$A^{2} - 4 A - 5 I = [9 - 4 - 5 16 - 16 - 0 8 - 8 - 0 17 - 12 - 5] = [0000] = O ✓$

The theorem is verified.
Multiply the identity by $A^{- 1}$
$A^{2} - 4 A - 5 I = O$

$\times A^{- 1} A - 4 I - 5 A^{- 1} = O$
Solve for $A^{- 1}$
$5 A^{- 1} = A - 4 I ⟹ A^{- 1} = \frac{1}{5} (A - 4 I)$

$A - 4 I = [1 - 4 4 2 3 - 4] = [- 3 4 2 - 1]$

$A^{- 1} = \frac{1}{5} [- 3 4 2 - 1]$
Cross-check with the adjoint formula
$A^{- 1} = \frac{1}{∣ A ∣} adj A = \frac{1}{- 5} [3 - 4 - 2 1] = \frac{1}{5} [- 3 4 2 - 1] ✓$

Answer

$A^{2} - 4 A - 5 I = O$ , and $A^{- 1} = \frac{1}{5} [- 3 4 2 - 1]$ .

Example 2

The theorem for a 3 × 3, and the inverse from it

Question

Verify the Cayley–Hamilton theorem for $A = 123 2 - 1 1 34 - 1$ and hence find $A^{- 1}$ .

Compute $S_{1}, S_{2}, S_{3}$
$S_{1} = 1 + (- 1) + (- 1) = - 1$

Principal $2 \times 2$ minors:
$- 1 1 4 - 1 = 1 - 4 = - 3, 13 3 - 1 = - 1 - 9 = - 10, 12 2 - 1 = - 1 - 4 = - 5$

$S_{2} = - 3 - 10 - 5 = - 18$

$S_{3} = ∣ A ∣ = 1 (1 - 4) - 2 (- 2 - 12) + 3 (2 + 3) = - 3 + 28 + 15 = 40$
Write the characteristic equation
$λ^{3} - S_{1} λ^{2} + S_{2} λ - S_{3} = 0$

$λ^{3} + λ^{2} - 18 λ - 40 = 0$

By the theorem: $A^{3} + A^{2} - 18 A - 40 I = O$ .
Compute $A^{2}$
Row 1 $(1, 2, 3)$ : $1 + 4 + 9 = 14$ , $2 - 2 + 3 = 3$ , $3 + 8 - 3 = 8$
Row 2 $(2, - 1, 4)$ : $2 - 2 + 12 = 12$ , $4 + 1 + 4 = 9$ , $6 - 4 - 4 = - 2$
Row 3 $(3, 1, - 1)$ : $3 + 2 - 3 = 2$ , $6 - 1 - 1 = 4$ , $9 + 4 + 1 = 14$

$A^{2} = 14122394 8 - 2 14$
Compute $A^{3} = A^{2} A$
Row 1 $(14, 3, 8)$ : $14 + 6 + 24 = 44$ , $28 - 3 + 8 = 33$ , $42 + 12 - 8 = 46$
Row 2 $(12, 9, - 2)$ : $12 + 18 - 6 = 24$ , $24 - 9 - 2 = 13$ , $36 + 36 + 2 = 74$
Row 3 $(2, 4, 14)$ : $2 + 8 + 42 = 52$ , $4 - 4 + 14 = 14$ , $6 + 16 - 14 = 8$

$A^{3} = 44245233131446748$
Substitute and verify
$18 A = 183654 36 - 18 18 5472 - 18, 40 I = 400004000040$

Position $(1, 1)$ : $44 + 14 - 18 - 40 = 0$ ✓
Position $(1, 2)$ : $33 + 3 - 36 - 0 = 0$ ✓
Position $(1, 3)$ : $46 + 8 - 54 - 0 = 0$ ✓
Position $(2, 1)$ : $24 + 12 - 36 - 0 = 0$ ✓
Position $(2, 2)$ : $13 + 9 + 18 - 40 = 0$ ✓
Position $(2, 3)$ : $74 - 2 - 72 - 0 = 0$ ✓
Position $(3, 1)$ : $52 + 2 - 54 - 0 = 0$ ✓
Position $(3, 2)$ : $14 + 4 - 18 - 0 = 0$ ✓
Position $(3, 3)$ : $8 + 14 + 18 - 40 = 0$ ✓

$A^{3} + A^{2} - 18 A - 40 I = O ✓$
Derive $A^{- 1}$
Multiply through by $A^{- 1}$ :

$A^{2} + A - 18 I - 40 A^{- 1} = O$

$A^{- 1} = \frac{1}{40} (A^{2} + A - 18 I)$
Evaluate
$A^{2} + A = 1514558511213$

$A^{2} + A - 18 I = - 3 145 5 - 10 5 112 - 5$

$A^{- 1} = \frac{1}{40} - 3 145 5 - 10 5 112 - 5$

Check (row 1 of $A$ × column 1 of the bracket): $1 (- 3) + 2 (14) + 3 (5) = - 3 + 28 + 15 = 40$ , and $40/40 = 1$ ✓

Answer

$A^{3} + A^{2} - 18 A - 40 I = O$ ; $A^{- 1} = \frac{1}{40} - 3 145 5 - 10 5 112 - 5$

NoteCompute S₁, S₂, S₃ first and write the equation before touching A² or A³. It tells you exactly which powers you need.

Practice problems

4 with solutions

Work each one on paper first. The full solution — not just the answer — is one click away.

Problem 1Basic

Verify the Cayley–Hamilton theorem for $A = [23 - 1 1]$ .

Show solution

$tr A = 3, ∣ A ∣ = 2 + 3 = 5$

Characteristic equation: $λ^{2} - 3 λ + 5 = 0$

$A^{2} = [23 - 1 1] [23 - 1 1] = [4 - 3 6 + 3 - 2 - 1 - 3 + 1] = [19 - 3 - 2]$

$A^{2} - 3 A + 5 I = [19 - 3 - 2] - [69 - 3 3] + [5005]$

$= [1 - 6 + 5 9 - 9 + 0 - 3 + 3 + 0 - 2 - 3 + 5] = [0000] = O ✓$

Answer

A^{2} - 3 A + 5 I = O

✓

Problem 2Exam level

Use the Cayley–Hamilton theorem to find $A^{- 1}$ for $A = [3 - 1 12]$ .

Show solution

$tr A = 5, ∣ A ∣ = 6 + 1 = 7$

$λ^{2} - 5 λ + 7 = 0 ⟹ A^{2} - 5 A + 7 I = O$

Multiply by $A^{- 1}$ :

$A - 5 I + 7 A^{- 1} = O ⟹ A^{- 1} = \frac{1}{7} (5 I - A)$

$5 I - A = [5 - 3 1 - 1 5 - 2] = [21 - 1 3]$

$A^{- 1} = \frac{1}{7} [21 - 1 3]$

Check: $\frac{1}{7} [3 - 1 12] [21 - 1 3] = \frac{1}{7} [7007] = I$ ✓

Answer

A^{- 1} = \frac{1}{7} [21 - 1 3]

Problem 3Exam level

Verify the Cayley–Hamilton theorem for $A = 102020213$ and find $A^{- 1}$ .

Show solution

Coefficients.

$S_{1} = 1 + 2 + 3 = 6$

$S_{2} = 2013 + 1223 + 1002 = 6 + (- 1) + 2 = 7$

$S_{3} = ∣ A ∣ = 1 (6 - 0) - 0 + 2 (0 - 4) = 6 - 8 = - 2$

Characteristic equation: $λ^{3} - 6 λ^{2} + 7 λ + 2 = 0$

Compute $A^{2}$ .

Row 1 $(1, 0, 2)$ : $1 + 0 + 4 = 5$ , $0 + 0 + 0 = 0$ , $2 + 0 + 6 = 8$
Row 2 $(0, 2, 1)$ : $0 + 0 + 2 = 2$ , $0 + 4 + 0 = 4$ , $0 + 2 + 3 = 5$
Row 3 $(2, 0, 3)$ : $2 + 0 + 6 = 8$ , $0$ , $4 + 0 + 9 = 13$

$A^{2} = 5280408513$

Compute $A^{3} = A^{2} A$ .

Row 1 $(5, 0, 8)$ : $5 + 0 + 16 = 21$ , $0$ , $10 + 0 + 24 = 34$
Row 2 $(2, 4, 5)$ : $2 + 0 + 10 = 12$ , $0 + 8 + 0 = 8$ , $4 + 4 + 15 = 23$
Row 3 $(8, 0, 13)$ : $8 + 0 + 26 = 34$ , $0$ , $16 + 0 + 39 = 55$

$A^{3} = 211234080342355$

Verify $A^{3} - 6 A^{2} + 7 A + 2 I$ :

$(1, 1)$ : $21 - 30 + 7 + 2 = 0$ $(1, 2)$ : $0$ $(1, 3)$ : $34 - 48 + 14 = 0$
$(2, 1)$ : $12 - 12 + 0 = 0$ $(2, 2)$ : $8 - 24 + 14 + 2 = 0$ $(2, 3)$ : $23 - 30 + 7 = 0$
$(3, 1)$ : $34 - 48 + 14 = 0$ $(3, 2)$ : $0$ $(3, 3)$ : $55 - 78 + 21 + 2 = 0$

$A^{3} - 6 A^{2} + 7 A + 2 I = O ✓$

Inverse. Multiply by $A^{- 1}$ :

$A^{2} - 6 A + 7 I + 2 A^{- 1} = O ⟹ A^{- 1} = - \frac{1}{2} (A^{2} - 6 A + 7 I)$

$6 A = 6012012012618$

$A^{2} - 6 A + 7 I = 5 - 6 + 7 2 - 0 8 - 12 0 4 - 12 + 7 0 8 - 12 5 - 6 13 - 18 + 7 = 62 - 4 0 - 1 0 - 4 - 1 2$

$A^{- 1} = - \frac{1}{2} 62 - 4 0 - 1 0 - 4 - 1 2 = - 3 - 1 2 0 \frac{1}{2} 0 2 \frac{1}{2} - 1$

Check (row 1 of $A$ × column 1 of $A^{- 1}$ ): $1 (- 3) + 0 (- 1) + 2 (2) = 1$ ✓

Answer

A^{3} - 6 A^{2} + 7 A + 2 I = O

;

A^{- 1} = - 3 - 1 2 0 1/2 0 2 1/2 - 1

Problem 4Challenge

If $A = [1221]$ , use the Cayley–Hamilton theorem to express $A^{5}$ in the form $p A + q I$ .

Show solution

$tr A = 2, ∣ A ∣ = 1 - 4 = - 3$

$λ^{2} - 2 λ - 3 = 0 ⟹ A^{2} = 2 A + 3 I$

Now climb the powers, reducing at each stage.

$A^{3} = A \cdot A^{2} = A (2 A + 3 I) = 2 A^{2} + 3 A = 2 (2 A + 3 I) + 3 A = 7 A + 6 I$

$A^{4} = A \cdot A^{3} = A (7 A + 6 I) = 7 A^{2} + 6 A = 7 (2 A + 3 I) + 6 A = 20 A + 21 I$

$A^{5} = A \cdot A^{4} = A (20 A + 21 I) = 20 A^{2} + 21 A = 20 (2 A + 3 I) + 21 A = 61 A + 60 I$

$A^{5} = 61 A + 60 I$

Verification. $61 A + 60 I = [61 + 60 122 122 61 + 60] = [121122122121]$

Directly: the eigenvalues of $A$ are $3$ and $- 1$ , so those of $A^{5}$ are $243$ and $- 1$ . Then $tr (A^{5}) = 242 = 121 + 121$ ✓ and $∣ A^{5} ∣ = - 243 = 12 1^{2} - 12 2^{2} = (121 - 122) (121 + 122) = - 243$ ✓

Answer

A^{5} = 61 A + 60 I = [121122122121]