The Cayley-Hamilton Theorem

There is a surprising, almost magical, fact that ties together everything in this course: every square matrix satisfies its own characteristic equation. You compute the characteristic polynomial $p(\lambda) = \det(A-\lambda I)$p(λ)=det(A−λI) — the very polynomial whose roots are the eigenvalues — and then, astonishingly, if you substitute the matrix $A$A in place of the scalar $\lambda$λ, you get the zero matrix: $p(A) = O$p(A)=O. This is the Cayley–Hamilton theorem, and it is far more than a curiosity. It gives a slick way to find the inverse of a matrix, a powerful method for reducing high powers $A^n$An to low-degree expressions in $A$A, and a route to $A^n$An that works even for matrices that cannot be diagonalised (Lesson 9's defective case). The theorem is the natural capstone of the matrices strand: it links the determinant, the characteristic polynomial, eigenvalues, inverses and powers into a single elegant identity. This lesson states it carefully, derives the indispensable $2\times2$2×2 form, and shows its three big applications — inverses, powers, and polynomial reduction — with fully verified worked examples and the exam technique examiners reward.

---

1. Where this sits in AQA 7367

Cayley–Hamilton is compulsory pure content for Papers 1 and 2 and the final topic of the matrices strand. Stating and verifying the theorem for a given matrix is AO1/AO2; using it to find $A^{-1}$A−1 or to reduce a power is AO1; and the multi-step "express $A^n$An in the form $\alpha A + \beta I$αA+βI" or "use Cayley–Hamilton to evaluate a matrix polynomial" problems are AO3. The topic rests on the characteristic polynomial (Lesson 8), determinants (Lesson 3), inverses (Lesson 4) and powers/diagonalisation (Lesson 9) — it is genuinely synoptic within the strand.

---

2. Statement of the theorem

Let $A$A be an $n\times n$n×n matrix with characteristic polynomial

$$p(\lambda) = \det(A - \lambda I) = (-1)^n\lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_1\lambda + c_0.$$p(λ)=det(A−λI)=(−1)nλn+cn−1​λn−1+⋯+c1​λ+c0​.

The Cayley–Hamilton theorem states:

$$\boxed{\;p(A) = (-1)^n A^n + c_{n-1}A^{n-1} + \cdots + c_1 A + c_0 I = O.\;}$$p(A)=(−1)nAn+cn−1​An−1+⋯+c1​A+c0​I=O.​

In words: substitute the matrix $A$A for $\lambda$λ throughout the characteristic polynomial — interpreting the constant term $c_0$c0​ as $c_0 I$c0​I — and the result is the zero matrix. The single most common error is to forget that the constant becomes $c_0 I$c0​I (not the scalar $c_0$c0​): you are building a matrix equation, so every term must be a matrix.

A warning about the "obvious" wrong proof. It is tempting to argue "$p(\lambda) = \det(A-\lambda I)$p(λ)=det(A−λI), so $p(A) = \det(A - A) = \det(O) = 0$p(A)=det(A−A)=det(O)=0." This is nonsense: $p(A)$p(A) is an $n\times n$n×n matrix, not the scalar $\det(A-AI)$det(A−AI), and the substitution of a matrix into a polynomial is a completely different operation from evaluating a determinant. The genuine theorem is much deeper than this fake one-liner — see §12.

---

3. The 2×2 form (derived) — the one to know cold

For a $2\times2$2×2 matrix $A$A, the characteristic polynomial is (Lesson 8)

$$p(\lambda) = \lambda^2 - (\operatorname{tr}A)\lambda + \det A.$$p(λ)=λ2−(trA)λ+detA.

Cayley–Hamilton therefore gives the identity every Further-Maths student should have at their fingertips:

$$\boxed{\;A^2 - (\operatorname{tr}A)\,A + (\det A)\,I = O,\qquad\text{equivalently}\qquad A^2 = (\operatorname{tr}A)\,A - (\det A)\,I.\;}$$A2−(trA)A+(detA)I=O,equivalentlyA2=(trA)A−(detA)I.​

The rearranged form is the workhorse: it expresses $A^2$A2 as a linear combination of $A$A and $I$I — and, by repeated substitution, so is every higher power. That single fact powers both the inverse trick and the power-reduction method below.

Worked Example 1 — verify Cayley–Hamilton for a 2×2 (with mark scheme)

Verify the theorem for $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$A=(13​24​).

$\operatorname{tr}A = 5$trA=5, $\det A = 4 - 6 = -2$detA=4−6=−2, so the characteristic equation is $\lambda^2 - 5\lambda - 2 = 0$λ2−5λ−2=0 and Cayley–Hamilton claims $A^2 - 5A - 2I = O$A2−5A−2I=O. Compute:

$$A^2 = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix},\quad 5A = \begin{pmatrix} 5 & 10 \\ 15 & 20 \end{pmatrix},\quad 2I = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}.$$A2=(13​24​)(13​24​)=(715​1022​),5A=(515​1020​),2I=(20​02​). $$A^2 - 5A - 2I = \begin{pmatrix} 7-5-2 & 10-10-0 \\ 15-15-0 & 22-20-2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O.\ ✓$$A2−5A−2I=(7−5−215−15−0​10−10−022−20−2​)=(00​00​)=O. ✓

Mark scheme: M1 find $\operatorname{tr}A$trA and $\det A$detA / state characteristic equation; M1 compute $A^2$A2; M1 form $A^2 - 5A - 2I$A2−5A−2I (constant term as $-2I$−2I); A1 obtain $O$O. Treating the constant as the scalar $-2$−2 instead of $-2I$−2I is the standard error and loses the structural mark.

---

4. Application 1 — finding the inverse

Start from the $2\times2$2×2 identity and isolate $I$I. From $A^2 - (\operatorname{tr}A)A + (\det A)I = O$A2−(trA)A+(detA)I=O,

$$(\det A)\,I = (\operatorname{tr}A)\,A - A^2 = A\bigl((\operatorname{tr}A)\,I - A\bigr).$$(detA)I=(trA)A−A2=A((trA)I−A).

If $\det A \neq 0$detA=0, divide by $\det A$detA and read off the inverse:

$$\boxed{\;A^{-1} = \frac{(\operatorname{tr}A)\,I - A}{\det A}.\;}$$A−1=detA(trA)I−A​.​

This is a genuinely useful alternative to the cofactor method, and it generalises: for an $n\times n$n×n matrix, Cayley–Hamilton expresses $A^{-1}$A−1 as a polynomial in $A$A of degree $n-1$n−1.

Worked Example 2 — inverse via Cayley–Hamilton (with mark scheme)

Use Cayley–Hamilton to find $A^{-1}$A−1 for $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$A=(13​24​).

$\operatorname{tr}A = 5$trA=5, $\det A = -2$detA=−2, so

$$A^{-1} = \frac{5I - A}{-2} = \frac{1}{-2}\begin{pmatrix} 5-1 & 0-2 \\ 0-3 & 5-4 \end{pmatrix} = \frac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}.$$A−1=−25I−A​=−21​(5−10−3​0−25−4​)=−21​(4−3​−21​)=(−23/2​1−1/2​).

Check: $AA^{-1} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix} = \begin{pmatrix} -2+3 & 1-1 \\ -6+6 & 3-2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.\ ✓$AA−1=(13​24​)(−23/2​1−1/2​)=(−2+3−6+6​1−13−2​)=(10​01​). ✓

Mark scheme: M1 quote/derive $A^{-1} = ((\operatorname{tr}A)I - A)/\det A$A−1=((trA)I−A)/detA; M1 substitute $\operatorname{tr}A = 5,\det A = -2$trA=5,detA=−2; A1 form $5I - A$5I−A; A1 $A^{-1} = \binom{-2\ \ 1}{3/2\,-1/2}$A−1=(3/2−1/2−2  1​). The $AA^{-1}=I$AA−1=I check secures the accuracy mark.

---

5. Application 2 — computing higher powers

Because $A^2 = (\operatorname{tr}A)A - (\det A)I$A2=(trA)A−(detA)I is a linear combination of $A$A and $I$I, so is $A^3 = A\cdot A^2$A3=A⋅A2, and so on: every power of a $2\times2$2×2 matrix can be written as $\alpha A + \beta I$αA+βI for some scalars. To climb from one power to the next, multiply the current expression by $A$A and use the relation again to eliminate the $A^2$A2 that appears.

Worked Example 3 — reduce $A^3$A3 to $\alpha A + \beta I$αA+βI (with mark scheme)

Express $A^3$A3 for $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$A=(13​24​) in the form $\alpha A + \beta I$αA+βI, and hence evaluate it.

From Worked Example 1, $A^2 = 5A + 2I$A2=5A+2I. Then

$$A^3 = A\cdot A^2 = A(5A + 2I) = 5A^2 + 2A = 5(5A + 2I) + 2A = 27A + 10I.$$A3=A⋅A2=A(5A+2I)=5A2+2A=5(5A+2I)+2A=27A+10I.

So $\alpha = 27,\ \beta = 10$α=27, β=10, and

$$A^3 = 27\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} + 10\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 27+10 & 54 \\ 81 & 108+10 \end{pmatrix} = \begin{pmatrix} 37 & 54 \\ 81 & 118 \end{pmatrix}.$$A3=27(13​24​)+10(10​01​)=(27+1081​54108+10​)=(3781​54118​).

Check by direct multiplication: $A^3 = A^2 A = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 37 & 54 \\ 81 & 118 \end{pmatrix}$A3=A2A=(715​1022​)(13​24​)=(3781​54118​) ✓.

Mark scheme: M1 use $A^2 = 5A + 2I$A2=5A+2I; M1 substitute to eliminate $A^2$A2 from $A^3$A3; A1 $A^3 = 27A + 10I$A3=27A+10I; A1 evaluate to $\binom{37\ \ 54}{81\ 118}$(81 11837  54​). The reduction $A^3 = \alpha A + \beta I$A3=αA+βI is the examinable skill; the numerical check protects the final mark.

A faster route for a specific high power uses the eigenvalues directly. Since $A^n = \alpha_n A + \beta_n I$An=αn​A+βn​I, the same scalars satisfy the scalar equation $\lambda^n = \alpha_n\lambda + \beta_n$λn=αn​λ+βn​ at each eigenvalue $\lambda$λ (because the eigenvalues satisfy the same characteristic relation). With two distinct eigenvalues you get two equations for $\alpha_n,\beta_n$αn​,βn​; solve and you have $A^n$An without iterating. (For a repeated eigenvalue, differentiate $\lambda^n = \alpha_n\lambda + \beta_n$λn=αn​λ+βn​ with respect to $\lambda$λ to get the second equation.)

---

6. Application 3 — reducing polynomial expressions

The same idea evaluates any polynomial in $A$A. Asked for $f(A)$f(A) where $f$f has degree $\ge 2$≥2 (for a $2\times2$2×2), use the characteristic relation to keep replacing $A^2$A2 until only $A$A and $I$I remain: every polynomial in a $2\times2$2×2 matrix collapses to $\alpha A + \beta I$αA+βI. Concretely, treat the polynomial as an ordinary polynomial in $\lambda$λ, divide it by the characteristic polynomial, and the remainder (degree $< 2$<2) is exactly $\alpha\lambda + \beta$αλ+β; then $f(A) = \alpha A + \beta I$f(A)=αA+βI. This polynomial-division viewpoint is the cleanest way to handle a high-degree request such as "find $A^7 - 3A^5 + 2I$A7−3A5+2I".

As a quick illustration, suppose $A$A satisfies $A^2 = 4A - 3I$A2=4A−3I (eigenvalues $1$1 and $3$3) and you must evaluate $f(A) = A^3 - 2A^2 + A$f(A)=A3−2A2+A. The eigenvalue shortcut is fastest: $f(A) = \gamma A + \delta I$f(A)=γA+δI where $f(\lambda) = \gamma\lambda + \delta$f(λ)=γλ+δ holds at $\lambda = 1$λ=1 and $\lambda = 3$λ=3. At $\lambda=1$λ=1: $1 - 2 + 1 = 0 = \gamma + \delta$1−2+1=0=γ+δ; at $\lambda=3$λ=3: $27 - 18 + 3 = 12 = 3\gamma + \delta$27−18+3=12=3γ+δ. Subtracting, $2\gamma = 12\Rightarrow\gamma = 6$2γ=12⇒γ=6, $\delta = -6$δ=−6, so $f(A) = 6A - 6I$f(A)=6A−6I. No matrix multiplication is needed at all — the eigenvalues do the work. (You could equally reduce step by step: $A^3 = 13A - 12I$A3=13A−12I from before, so $f(A) = (13A - 12I) - 2(4A - 3I) + A = 6A - 6I$f(A)=(13A−12I)−2(4A−3I)+A=6A−6I ✓, agreeing exactly.)

---

7. The 3×3 case

For a $3\times3$3×3 matrix the characteristic polynomial is a cubic, $\lambda^3 + a\lambda^2 + b\lambda + c$λ3+aλ2+bλ+c (after fixing signs), and Cayley–Hamilton gives a relation between $A^3,A^2,A$A3,A2,A and $I$I:

$$A^3 + aA^2 + bA + cI = O.$$A3+aA2+bA+cI=O.

Worked Example 4 — verify Cayley–Hamilton for a 3×3 (with mark scheme)

Verify the theorem for $A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{pmatrix}$A=​201​010​102​​.

Expand $\det(A-\lambda I)$det(A−λI) along the middle row (only one non-zero entry, $1-\lambda$1−λ):

$$\det(A-\lambda I) = (1-\lambda)\begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} = (1-\lambda)\bigl[(2-\lambda)^2 - 1\bigr].$$det(A−λI)=(1−λ)​2−λ1​12−λ​​=(1−λ)[(2−λ)2−1].

Now $(2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda-1)(\lambda-3)$(2−λ)2−1=λ2−4λ+3=(λ−1)(λ−3), so

$$\det(A-\lambda I) = (1-\lambda)(\lambda-1)(\lambda-3) = -(\lambda-1)^2(\lambda-3) = -\lambda^3 + 5\lambda^2 - 7\lambda + 3.$$det(A−λI)=(1−λ)(λ−1)(λ−3)=−(λ−1)2(λ−3)=−λ3+5λ2−7λ+3.

(Check: eigenvalues $1,1,3$1,1,3 give sum $5 = \operatorname{tr}A$5=trA ✓ and product $3 = \det A$3=detA ✓.) Cayley–Hamilton then asserts $-A^3 + 5A^2 - 7A + 3I = O$−A3+5A2−7A+3I=O, i.e. $A^3 = 5A^2 - 7A + 3I$A3=5A2−7A+3I. Computing the powers, $A^2 = \begin{pmatrix} 5 & 0 & 4 \\ 0 & 1 & 0 \\ 4 & 0 & 5 \end{pmatrix}$A2=​504​010​405​​ and $A^3 = \begin{pmatrix} 14 & 0 & 13 \\ 0 & 1 & 0 \\ 13 & 0 & 14 \end{pmatrix}$A3=​14013​010​13014​​; then $5A^2 - 7A + 3I = \begin{pmatrix} 25-14+3 & 0 & 20-7 \\ 0 & 5-7+3 & 0 \\ 20-7 & 0 & 25-14+3 \end{pmatrix} = \begin{pmatrix} 14 & 0 & 13 \\ 0 & 1 & 0 \\ 13 & 0 & 14 \end{pmatrix} = A^3$5A2−7A+3I=​25−14+3020−7​05−7+30​20−7025−14+3​​=​14013​010​13014​​=A3 ✓.