FP2 3.3 - The Cayley-Hamilton theorem | Edexcel A-Level Further Mathematics

Use a matrix's characteristic equation as an identity involving the matrix itself. Reduce high powers to lower ones and find inverses when they exist, for two by two and three by three matrices, including matrices with repeated or complex eigenvalues.

A matrix satisfies its own characteristic equation

Can the hundredth power of $\mathbf{J}=\begin{pmatrix}2&1\\0&2\end{pmatrix}$ be found without ninety-nine full matrix multiplications? Its only eigenvalue is $2$ , with only one independent eigenvector direction. A method that works directly with the characteristic equation will turn the whole calculation into two terms.

You already know

The characteristic equation is $\det(\mathbf{A}-\lambda\mathbf{I})=0$ , equivalently $\det(\lambda\mathbf{I}-\mathbf{A})=0$ . Repeated and complex roots are allowed. Matrix powers mean repeated matrix multiplication; $\mathbf{A}^0=\mathbf{I}$ , and an inverse exists only when $\det\mathbf{A}\ne0$ .

For $\mathbf{A}=\begin{pmatrix}3&1\\2&2\end{pmatrix}$ , the characteristic equation is $\lambda^2-5\lambda+4=0$ . Direct multiplication gives $\mathbf{A}^2=\begin{pmatrix}11&5\\10&6\end{pmatrix}$ , which is exactly $5\mathbf{A}-4\mathbf{I}$ . The polynomial relation survives when the scalar variable is replaced by the matrix.

Cayley-Hamilton theorem

Every two by two or three by three matrix satisfies its own characteristic equation. Write its monic characteristic polynomial as $p(\lambda)=\det(\lambda\mathbf{I}-\mathbf{A})$ . Then

p(\mathbf{A})=\mathbf{0}

For example, if $p(\lambda)=\lambda^3+b\lambda^2+c\lambda+d$ , then

\mathbf{A}^3+b\mathbf{A}^2+c\mathbf{A}+d\mathbf{I}=\mathbf{0}

Constants multiply the identity matrix of the same size; the right-hand side is the zero matrix.

Not in the formula booklet: learn it.

For two by two matrices, you can see the theorem directly. If $\mathbf{A}=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , then

\mathbf{A}^2 =\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix} =(a+d)\mathbf{A}-(ad-bc)\mathbf{I}

This matches the characteristic polynomial $\lambda^2-(a+d)\lambda+ad-bc$ . The three by three theorem is used in the same way; no diagonalisation or invertibility is required. It works for repeated and complex eigenvalues as well as distinct real ones.

Key idea

Replace each power of the scalar variable by the corresponding matrix power, and replace the constant term by that constant times $\mathbf{I}$ .

If a question supplies the characteristic equation, use it directly and state that you are applying Cayley-Hamilton. If it asks you to find that equation, show the determinant calculation first.

Build the identity for a two by two matrix

Find the characteristic polynomial, then make the matrix substitution. Keep the determinant stage and the matrix-identity stage visibly separate: they produce different types of object.

Worked example 1: a quadratic matrix identity

Question. Find the characteristic equation of $\mathbf{A}=\begin{pmatrix}3&1\\2&2\end{pmatrix}$ , and use Cayley-Hamilton to express $\mathbf{A}^2$ in terms of $\mathbf{A}$ and $\mathbf{I}$ .

Plan. Expand $\det(\lambda\mathbf{I}-\mathbf{A})$ , then substitute the matrix.

\begin{aligned} p(\lambda)&=(\lambda-3)(\lambda-2)-2 && \text{determinant}\\ &=\lambda^2-5\lambda+4 && \text{expand}\\ \mathbf{A}^2-5\mathbf{A}+4\mathbf{I}&=\mathbf{0} && \text{Cayley-Hamilton}\\ \mathbf{A}^2&=5\mathbf{A}-4\mathbf{I} && \text{rearrange} \end{aligned}

Why this step? The equation for a scalar root becomes an identity for the whole matrix; its constant term acts equally on every coordinate.

Answer. The characteristic equation is $\lambda^2-5\lambda+4=0$ , and $\mathbf{A}^2=5\mathbf{A}-4\mathbf{I}$ .

Check. $5\mathbf{A}-4\mathbf{I}=\begin{pmatrix}11&5\\10&6\end{pmatrix}$ , agreeing with direct squaring.

Change only the lower-left entry from $2$ to $6$ . The matrix becomes singular, but the theorem still applies.

Worked example 2: a zero constant term is allowed

Question. Find the characteristic equation of $\mathbf{B}=\begin{pmatrix}3&1\\6&2\end{pmatrix}$ , and use it to express $\mathbf{B}^2$ .

Plan. Repeat the determinant calculation and retain its zero constant term correctly.

\begin{aligned} p(\lambda)&=(\lambda-3)(\lambda-2)-6=\lambda^2-5\lambda && \text{changed entry}\\ \mathbf{B}^2-5\mathbf{B}&=\mathbf{0} && \text{Cayley-Hamilton}\\ \mathbf{B}^2&=5\mathbf{B} && \text{rearrange} \end{aligned}

Why this step? A zero eigenvalue means the matrix is singular. It does not invalidate a polynomial identity involving positive powers.

Answer. The characteristic equation is $\lambda^2-5\lambda=0$ , and $\mathbf{B}^2=5\mathbf{B}$ .

Check. Both $\mathbf{B}^2$ and $5\mathbf{B}$ equal $\begin{pmatrix}15&5\\30&10\end{pmatrix}$ .

Multiplying $\mathbf{B}(\mathbf{B}-5\mathbf{I})=\mathbf{0}$ does not imply either factor is the zero matrix. Matrices can multiply to zero without either factor being zero, as this example shows.

Watch out

Wrong: cancel $\mathbf{B}$ from $\mathbf{B}^2=5\mathbf{B}$ to conclude $\mathbf{B}=5\mathbf{I}$ . Right: cancellation would require an inverse, but $\mathbf{B}$ is singular. Keep the valid identity without cancelling.

Change the same entry again, giving a non-zero negative determinant.

Reduce a cubic identity and the next power

For three by three matrices, use a monic polynomial with leading term $\lambda^3$ . Expanding $\det(\mathbf{A}-\lambda\mathbf{I})$ instead gives its negative, which has the same zero equation. Multiply the whole equation by $-1$ if you change convention.

Worked example 3: reduce a fourth power to a quadratic

Question. For $\mathbf{T}=\begin{pmatrix}2&1&0\\1&3&1\\0&1&2\end{pmatrix}$ , find its characteristic equation and express $\mathbf{T}^4$ using only $\mathbf{T}^2,\mathbf{T},\mathbf{I}$ .

Plan. Use the first-row determinant, then replace every cubic term using Cayley-Hamilton.

\begin{aligned} p(\lambda)&=(\lambda-2)\big((\lambda-3)(\lambda-2)-2\big) && \text{expand the determinant}\\ &=(\lambda-2)(\lambda-1)(\lambda-4) && \text{factor}\\ &=\lambda^3-7\lambda^2+14\lambda-8 && \text{monic polynomial}\\ \mathbf{T}^3&=7\mathbf{T}^2-14\mathbf{T}+8\mathbf{I} && \text{Cayley-Hamilton}\\ \mathbf{T}^4&=7\mathbf{T}^3-14\mathbf{T}^2+8\mathbf{T} && \text{multiply by }\mathbf{T}\\ &=35\mathbf{T}^2-90\mathbf{T}+56\mathbf{I} && \text{replace the cubic again} \end{aligned}

Why this step? Multiplication creates another cubic term; substituting the identity again brings the degree below three.

Answer. The characteristic equation is $p(\lambda)=0$ , and $\mathbf{T}^4=35\mathbf{T}^2-90\mathbf{T}+56\mathbf{I}$ .

Check. At eigenvalues $1,2,4$ , the scalar expression $35\lambda^2-90\lambda+56$ gives $1,16,256$ , respectively, agreeing with $\lambda^4$ .

Change only the central entry from $3$ to $1$ . The same reduction now has no constant term in its characteristic polynomial.

Worked example 4: cubic reduction for a singular matrix

Question. Express $\mathbf{U}^4$ using powers below three, where $\mathbf{U}=\begin{pmatrix}2&1&0\\1&1&1\\0&1&2\end{pmatrix}$ .

Plan. Find the changed polynomial and apply the same substitution twice.

\begin{aligned} p(\lambda)&=(\lambda-2)\big((\lambda-1)(\lambda-2)-2\big) && \text{changed determinant}\\ &=\lambda(\lambda-2)(\lambda-3)=\lambda^3-5\lambda^2+6\lambda && \text{factor and expand}\\ \mathbf{U}^3&=5\mathbf{U}^2-6\mathbf{U} && \text{Cayley-Hamilton}\\ \mathbf{U}^4&=5\mathbf{U}^3-6\mathbf{U}^2 && \text{multiply by }\mathbf{U}\\ &=19\mathbf{U}^2-30\mathbf{U} && \text{substitute and collect} \end{aligned}

Why this step? Singularity stops us taking an inverse, but positive-power reduction still works exactly as before.

Answer. $\mathbf{U}^4=19\mathbf{U}^2-30\mathbf{U}$ .

Check. At eigenvalues $0,2,3$ , the reduced polynomial gives $0,16,81$ , the required fourth powers.

For any two by two matrix, repeated substitution reduces powers to a linear combination of $\mathbf{A}$ and $\mathbf{I}$ . For three by three matrices it reduces them to a combination of $\mathbf{A}^2,\mathbf{A},\mathbf{I}$ . These are consequences of the theorem, not additional booklet formulas.

Watch out

Wrong: keep the original constant $8\mathbf{I}$ when multiplying the cubic identity by $\mathbf{T}$ . Right: it becomes $8\mathbf{T}$ , because $\mathbf{I}\mathbf{T}=\mathbf{T}$ . Multiply every term before reducing again.

Change the central entry to $2$ , and follow the same two stages.

Reduce powers without a basis of eigenvectors

Return to the opening matrix $\mathbf{J}=\begin{pmatrix}2&1\\0&2\end{pmatrix}$ . Its repeated eigenvalue gives only one independent eigenvector, but its characteristic polynomial $(\lambda-2)^2$ gives

\mathbf{J}^2-4\mathbf{J}+4\mathbf{I}=\mathbf{0}

That identity is all we need for successive powers.

Worked example 5: a cubic power from a quadratic identity

Question. Use Cayley-Hamilton to find $\mathbf{J}^3$ .

Plan. Multiply $\mathbf{J}^2=4\mathbf{J}-4\mathbf{I}$ by $\mathbf{J}$ , then reduce again.

\begin{aligned} \mathbf{J}^3&=4\mathbf{J}^2-4\mathbf{J} && \text{multiply the identity}\\ &=4(4\mathbf{J}-4\mathbf{I})-4\mathbf{J} && \text{replace the square}\\ &=12\mathbf{J}-16\mathbf{I} && \text{collect coefficients}\\ &=\begin{pmatrix}8&12\\0&8\end{pmatrix} && \text{insert the matrix} \end{aligned}

Why this step? Reducing after each multiplication keeps the calculation linear in $\mathbf{J}$ , regardless of the exponent.

Answer. $\mathbf{J}^3=\begin{pmatrix}8&12\\0&8\end{pmatrix}$ .

Check. Directly, $\mathbf{J}^2=\begin{pmatrix}4&4\\0&4\end{pmatrix}$ , and multiplication by $\mathbf{J}$ gives the same cube.

Change only the exponent, from $3$ to $4$ .

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FP2 3.3 - The Cayley-Hamilton theorem

A matrix satisfies its own characteristic equation

Build the identity for a two by two matrix

Reduce a cubic identity and the next power

Reduce powers without a basis of eigenvectors

Continue this lesson.

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