Cayley-Hamilton theorem

At a Glance

• Frequency: 3 sub-parts across 3 of 14 years (2014, 2019, 2023)
• Priority tier: T3
• Marks (count): 10 (1), 15 (1), 20 (1)
• Average solve time: ~14 min
• Difficulty mix: medium 3
• Section: A | Dominant type: computation

Why This Chapter Matters

Cayley-Hamilton delivers 10–20 marks across three different question styles: compute a high matrix power, find the inverse via the characteristic equation, or evaluate a matrix polynomial by polynomial division. All three reduce to the same workflow — compute the characteristic polynomial, apply Cayley-Hamilton to reduce powers, then evaluate numerically. The $3\times 3$ matrix with repeated eigenvalue (appearing in both 2019 and 2023) adds one step: the double-root derivative condition. A student who has practised this template can solve any variant in under 15 minutes.

Minimum Theory

Cayley-Hamilton theorem. Every $n\times n$ matrix $A$ satisfies its own characteristic equation: if $p(\lambda)=\det(\lambda I-A)$ then $p(A)=O$. This means every matrix power $A^k$ (for $k\ge n$) can be expressed as a polynomial in $A$ of degree $\le n-1$.

Power reduction via Lagrange interpolation. To find $A^N$ for large $N$, write $\lambda^N=q(\lambda)p(\lambda)+r(\lambda)$ where $\deg r<n$. Since $p(A)=O$, $A^N=r(A)$. Determine the $n$ coefficients of $r$ by substituting each eigenvalue (with multiplicity): a simple root $\lambda_0$ gives $\lambda_0^N=r(\lambda_0)$; a double root $\lambda_1$ gives both $\lambda_1^N=r(\lambda_1)$ and $N\lambda_1^{N-1}=r'(\lambda_1)$ (differentiate both sides of $\lambda^N=q(\lambda)p(\lambda)+r(\lambda)$ and use $p(\lambda_1)=p'(\lambda_1)=0$).

Inverse via Cayley-Hamilton. For $2\times 2$: from $A^2-(\text{tr}\,A)A+(\det A)I=O$, rearrange to isolate $I$ on one side and factor out $A$: $A^{-1}=\bigl((\text{tr}\,A)I-A\bigr)/\det A$.

Question Archetypes

cayley-hamilton-power (2 question(s); 2019, 2023)

Recognition Cues

• “State the Cayley-Hamilton theorem and use it to find $A^{100}$” (or $A^{40}$, $A^n$).
• The matrix is $3\times 3$ with characteristic polynomial $(\lambda-1)^2(\lambda+1)$.
• Either a direct remainder computation (2019) or a recurrence + telescope (2023).

Solution Template

1. State the theorem. Every $n\times n$ matrix satisfies its characteristic equation $p(A)=O$.
2. Compute $p(\lambda)=\det(\lambda I-A)$. Expand and factorise.
3. Write $\lambda^N=q(\lambda)p(\lambda)+r(\lambda)$ with $r$ of degree $\le 2$ (i.e.\ $r=a\lambda^2+b\lambda+c$). So $A^N=aA^2+bA+cI$.
4. Find $a,b,c$ from eigenvalue conditions. Simple root: substitute; double root: substitute and differentiate.
5. Compute $A^2$ entry-by-entry. Assemble $A^N=aA^2+bA+cI$.

Worked Example

2019 Paper 1, 2019-P1-Q4a (15 marks)

State the Cayley-Hamilton theorem. Use it to find $A^{100}$ for

$A=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}.$

Step 1 — State theorem. Every square matrix satisfies its own characteristic equation: $p(A)=O$.

Step 2 — Characteristic polynomial.

$p(\lambda)=\det\begin{pmatrix}\lambda-1&0&0\\-1&\lambda&-1\\0&-1&\lambda\end{pmatrix}=(\lambda-1)(\lambda^2-1)=(\lambda-1)^2(\lambda+1).$

So $p(\lambda)=\lambda^3-\lambda^2-\lambda+1$ and by CH, $A^3=A^2+A-I$.

Step 3 — Write remainder. $\lambda^{100}=q(\lambda)(\lambda-1)^2(\lambda+1)+(a\lambda^2+b\lambda+c)$, so $A^{100}=aA^2+bA+cI$.

Step 4 — Eigenvalue conditions.

$\lambda=1$ (value): $1=a+b+c. \qquad(1)$

$\lambda=1$ (derivative, double root): $100=2a+b. \qquad(2)$

$\lambda=-1$ (simple root): $(-1)^{100}=1=a-b+c. \qquad(3)$

From $(1)-(3)$: $2b=0$, so $b=0$. From $(2)$: $a=50$. From $(1)$: $c=1-50=-49$.

Step 5 — Compute $A^2$ and assemble.

$A^2=\begin{pmatrix}1&0&0\\1&1&0\\1&0&1\end{pmatrix}.$

$A^{100}=50A^2-49I=50\begin{pmatrix}1&0&0\\1&1&0\\1&0&1\end{pmatrix}-49\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=\begin{pmatrix}1&0&0\\50&1&0\\50&0&1\end{pmatrix}.$

$\boxed{\;A^{100}=\begin{pmatrix}1&0&0\\50&1&0\\50&0&1\end{pmatrix}.\;}$

2023 Paper 1, 2023-P1-Q3a (20 marks)

(i) Verify CH for $A=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}$. (ii) Show $A^n=A^{n-2}+A^2-I$ for $n\ge 3$ and find $A^{40}$.

Part (i) — Verify CH.

Characteristic polynomial: $p(\lambda)=(\lambda-1)^2(\lambda+1)=\lambda^3-\lambda^2-\lambda+1$.

Compute $A^2$ and $A^3$ entry-by-entry:

$A^2=\begin{pmatrix}1&0&0\\1&1&0\\1&0&1\end{pmatrix},\qquad A^3=\begin{pmatrix}1&0&0\\2&0&1\\1&1&0\end{pmatrix}.$

$A^3-A^2-A+I=\begin{pmatrix}1-1-1+1&0&0\\2-1-1+0&0-1-0+1&1-0-1+0\\1-1-0+0&1-0-1+0&0-1-0+1\end{pmatrix}=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}.\;\checkmark$

Part (ii) — Recurrence and $A^{40}$.

From CH: $A^3=A^2+A-I$, i.e.\ $A^3-A=A^2-I$, i.e.\ $A^3-A^1=A^2-I$.

Claim: $A^n-A^{n-2}=A^2-I$ for all $n\ge 3$.

Proof by induction. Base $n=3$: $A^3-A=A^2-I$ follows directly from CH. Inductive step: assume $A^k-A^{k-2}=A^2-I$. Multiply both sides on the left by $A$: $A^{k+1}-A^{k-1}=A^3-A=A^2-I$ (using the base case again). This proves $A^{k+1}-A^{k-1}=A^2-I$.

Telescope to $A^{40}$: sum the identity $A^n-A^{n-2}=A^2-I$ for $n=4,6,8,\ldots,40$ (19 terms, all even steps):

$A^{40}-A^2=19(A^2-I)\qquad\Rightarrow\qquad A^{40}=20A^2-19I.$

$A^{40}=20\begin{pmatrix}1&0&0\\1&1&0\\1&0&1\end{pmatrix}-19I=\begin{pmatrix}1&0&0\\20&1&0\\20&0&1\end{pmatrix}.$

$\boxed{\;A^{40}=\begin{pmatrix}1&0&0\\20&1&0\\20&0&1\end{pmatrix}.\;}$

Common Traps

• The double root $\lambda=1$ gives two conditions — value and derivative. A student who only substitutes $\lambda=1$ once has one equation too few and cannot determine all three unknowns.
• Evaluate $(-1)^{100}=+1$ (not $-1$). An even-power sign error corrupts $a$ and $c$.
• Always reduce to a remainder of degree $\le n-1$ (here $\le 2$, i.e.\ $aA^2+bA+cI$), not degree $\le n$.
• Triple-check $A^2$ entry-by-entry. A single error in $A^2$ propagates to the final answer.
• For the telescope in 2023: there are exactly 19 even steps from $n=4$ to $n=40$ (i.e.\ $A^{40}-A^2$, with each step contributing $A^2-I$), giving coefficient $19$, not $20$. The final result is $A^{40}=20A^2-19I$, not $21A^2-20I$.

cayley-hamilton-application (1 question(s); 2014)

Recognition Cues

• “Verify CH; find $A^{-1}$; evaluate the polynomial expression $p(A)$.”
• $2\times 2$ matrix; characteristic polynomial is quadratic $\lambda^2-(\text{tr}\,A)\lambda+\det A$.
• Inverse: factor the CH equation; polynomial evaluation: reduce all powers via the recurrence $A^2=(\text{tr}\,A)A-(\det A)I$.

Solution Template

1. Characteristic polynomial. $p(\lambda)=\lambda^2-(\text{tr}\,A)\lambda+\det A$.
2. Verify CH. Compute $A^2$ directly and check $A^2-(\text{tr}\,A)A+(\det A)I=O$.
3. Inverse. Rearrange CH: $A(A-(\text{tr}\,A)I)=-(\det A)I$, so $A^{-1}=((\text{tr}\,A)I-A)/\det A$.
4. Higher powers. Use $A^2=(\text{tr}\,A)A-(\det A)I$ as the recurrence; compute $A^3,A^4,A^5,\ldots$ iteratively as $\alpha_k A+\beta_k I$.
5. Sum the polynomial. Collect all $A$-coefficients and all $I$-coefficients. Write the result as a single matrix.

Worked Example

2014 Paper 1, 2014-P1-Q2b-ii (10 marks)

Verify CH for $A=\begin{pmatrix}1&4\\2&3\end{pmatrix}$; find $A^{-1}$; evaluate $A^5-4A^4-7A^3+11A^2-A-10I$.

Step 1 — Characteristic polynomial. $\text{tr}\,A=4$, $\det A=3-8=-5$. So $p(\lambda)=\lambda^2-4\lambda-5$.

Step 2 — Verify CH. $A^2=\begin{pmatrix}9&16\\8&17\end{pmatrix}$. Check: $A^2-4A-5I=\begin{pmatrix}9-4-5&16-16-0\\8-8-0&17-12-5\end{pmatrix}=O$. $\checkmark$

Step 3 — Inverse. From $A^2-4A-5I=O$: $A(A-4I)=5I$, so

$A^{-1}=\frac{A-4I}{5}=\frac{1}{5}\begin{pmatrix}-3&4\\2&-1\end{pmatrix}.$

Step 4 — Reduce higher powers. Recurrence: $A^2=4A+5I$.

$A^3=4A^2+5A=4(4A+5I)+5A=21A+20I.$

$A^4=A\cdot A^3=21A^2+20A=21(4A+5I)+20A=104A+105I.$

$A^5=A\cdot A^4=104A^2+105A=104(4A+5I)+105A=521A+520I.$

Step 5 — Evaluate polynomial. Collect:

Result: $A+5I=\begin{pmatrix}6&4\\2&8\end{pmatrix}$.

$\boxed{\;A^5-4A^4-7A^3+11A^2-A-10I=\begin{pmatrix}6&4\\2&8\end{pmatrix}.\;}$

(Alternative: polynomial division of $\lambda^5-4\lambda^4-7\lambda^3+11\lambda^2-\lambda-10$ by $\lambda^2-4\lambda-5$ gives remainder $\lambda+5$, so $p(A)=A+5I$.)

Common Traps

• From $A^2-4A-5I=O$: factor as $A(A-4I)=5I$ (not $A^2-4A=5I$ then “solve”); the factored form directly gives $A^{-1}$.
• For the $2\times 2$ case: the characteristic polynomial uses $+\det A$ (not $-\det A$); be careful with the sign, especially when $\det A$ is negative.
• When building the table in Step 5, keep separate $A$-coefficient and $I$-coefficient columns to avoid sign errors.

Marks-Aware Writing

10-mark question (2014): Verification (Steps 1–2), inverse (Step 3), and polynomial evaluation (Steps 4–5) each carry roughly $3+3+4$ marks. An answer that verifies CH but miscomputes the inverse loses 3 marks; missing the polynomial evaluation entirely loses 4.

15-mark question (2019): Stating CH (2 marks), characteristic polynomial (3 marks), setting up and solving the $a,b,c$ system (5 marks including the double-root derivative condition), computing $A^2$ and the final matrix (5 marks).

20-mark question (2023): Verification (10 marks) and recurrence + $A^{40}$ (10 marks). For verification: $A^2$ and $A^3$ computed correctly (5 marks), substitution and confirmation of zero (5 marks). For part (ii): proving the recurrence by induction (5 marks), telescoping correctly to get $A^{40}=20A^2-19I$ (3 marks), final matrix (2 marks).

Practice Set

(No additional practice items beyond the three worked examples.)