Xm5, Xm6 - Cayley-Hamilton and transformation eigenvectors | OCR A-Level Further Mathematics B (MEI)

Use the Cayley-Hamilton theorem to reduce powers and calculate inverses, then interpret eigenvectors as geometric directions in two and three dimensions. You will distinguish an invariant line from fixed points, locate a reflection plane, and find a rotation axis from a given matrix.

A characteristic equation becomes a matrix identity

Could the fourth power of $\mathbf{A}=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ be found using only a multiple of $\mathbf{A}$ and a multiple of the identity?

Its characteristic equation is $\lambda^2-4\lambda+3=0$ . The Cayley-Hamilton theorem turns that scalar equation into an identity involving the matrix itself. You can use it even without finding an eigenvector.

You already know

A characteristic equation is $\det(\mathbf{A}-\lambda\mathbf{I})=0$ . Matrix powers mean repeated multiplication; a scalar constant in a matrix expression must multiply the identity matrix. For a real eigenvector, $\mathbf{A}\mathbf{v}=\lambda\mathbf{v}$ .

A direct square gives $\mathbf{A}^2=\begin{pmatrix}5&4\\4&5\end{pmatrix}=4\mathbf{A}-3\mathbf{I}$ . The characteristic equation has predicted a relationship between powers.

Cayley-Hamilton theorem

Every square matrix satisfies its own characteristic equation. If that equation is

a_n\lambda^n+\cdots+a_1\lambda+a_0=0,

then

a_n\mathbf{A}^n+\cdots+a_1\mathbf{A}+a_0\mathbf{I}=\mathbf{0}

The constant term becomes a multiple of $\mathbf{I}$ , and the right side is the zero matrix. You need to understand and use the theorem; its proof is not required.

Not in the formula booklet: learn it.

Rearranging this identity reduces a high power to lower powers. Multiply matrix equations by $\mathbf{A}$ when advancing to the next power.

Worked example 1: reduce a cubic power

Question. The characteristic equation of $\mathbf{A}=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ is $\lambda^2-4\lambda+3=0$ . Use Cayley-Hamilton to find $\mathbf{A}^3$ .

Plan. Replace the squared matrix using the theorem, then multiply once and reduce again.

\begin{aligned} \mathbf{A}^2-4\mathbf{A}+3\mathbf{I}&=\mathbf{0} &&\text{Cayley-Hamilton}\\ \mathbf{A}^3&=4\mathbf{A}^2-3\mathbf{A} &&\text{multiply by }\mathbf{A}\\ &=4(4\mathbf{A}-3\mathbf{I})-3\mathbf{A} &&\text{reduce the square}\\ &=13\mathbf{A}-12\mathbf{I} =\begin{pmatrix}14&13\\13&14\end{pmatrix} &&\text{collect and evaluate} \end{aligned}

Why this step? Reducing each occurrence of $\mathbf{A}^2$ keeps the result in terms of $\mathbf{A}$ and $\mathbf{I}$ .

Answer. $\mathbf{A}^3=\begin{pmatrix}14&13\\13&14\end{pmatrix}$ .

Check. Direct multiplication of $\begin{pmatrix}5&4\\4&5\end{pmatrix}$ by $\mathbf{A}$ gives the same matrix.

Change only the requested exponent, from $3$ to $4$ . The previous reduced expression is a useful starting point.

Worked example 2: advance the reduction

Question. Use the same characteristic equation to find $\mathbf{A}^4$ .

Plan. Multiply the reduced cubic expression by $\mathbf{A}$ , then substitute for $\mathbf{A}^2$ .

\begin{aligned} \mathbf{A}^4 &=13\mathbf{A}^2-12\mathbf{A} &&\text{advance one power}\\ &=13(4\mathbf{A}-3\mathbf{I})-12\mathbf{A} &&\text{Cayley-Hamilton again}\\ &=40\mathbf{A}-39\mathbf{I} &&\text{collect coefficients}\\ &=\begin{pmatrix}41&40\\40&41\end{pmatrix} &&\text{evaluate} \end{aligned}

Why this step? The number $39$ subtracts from diagonal entries only, because it multiplies $\mathbf{I}$ .

Answer. $\mathbf{A}^4=40\mathbf{A}-39\mathbf{I}$ , answering the opening question.

Check. Squaring $\mathbf{A}^2=\begin{pmatrix}5&4\\4&5\end{pmatrix}$ gives diagonal entries $25+16=41$ and off-diagonal entries $20+20=40$ .

This is more than an isolated trick: further powers can be reduced in the same way, however large the exponent.

Watch out

Wrong: $\mathbf{A}^2-4\mathbf{A}+3=\mathbf{0}$ . Right: $\mathbf{A}^2-4\mathbf{A}+3\mathbf{I}=\mathbf{0}$ . A scalar constant alone cannot be added to a matrix.

Advance the same reduction by one more step.

Rearrange the identity to find an inverse

When a characteristic equation has a non-zero constant term, Cayley-Hamilton can express an inverse using non-negative powers. There is no operation of dividing by a matrix: multiply by its inverse, or construct a matrix whose product is $\mathbf{I}$ .

An inverse from a cubic characteristic equation

If the characteristic equation is

\lambda^3+a\lambda^2+b\lambda+c=0,\qquad c\ne0,

Cayley-Hamilton gives

\mathbf{A}(\mathbf{A}^2+a\mathbf{A}+b\mathbf{I})=-c\mathbf{I}, \qquad \mathbf{A}^{-1}=-\frac1c(\mathbf{A}^2+a\mathbf{A}+b\mathbf{I})

The non-zero constant excludes a zero eigenvalue. If $c=0$ , the matrix is singular and this inverse formula cannot be used, although Cayley-Hamilton still holds.

Not in the formula booklet: learn it.

The identity both supplies the inverse and explains why it works. A supplied characteristic polynomial need not have leading coefficient $1$ ; multiplying the entire equation by $-1$ does not alter its roots or the resulting identity.

Worked example 3: a cubic identity supplies an inverse

Question. The matrix

\mathbf{B}=\begin{pmatrix}1&1&0\\0&2&1\\0&0&3\end{pmatrix}

has characteristic equation $\lambda^3-6\lambda^2+11\lambda-6=0$ . Use Cayley-Hamilton to express $\mathbf{B}^{-1}$ in terms of $\mathbf{B}^2,\mathbf{B},\mathbf{I}$ , and hence calculate it.

Plan. Isolate the constant matrix, factor out $\mathbf{B}$ , and evaluate the resulting polynomial.

\begin{aligned} \mathbf{B}^3-6\mathbf{B}^2+11\mathbf{B}-6\mathbf{I} &=\mathbf{0}&&\text{Cayley-Hamilton}\\ \mathbf{B}(\mathbf{B}^2-6\mathbf{B}+11\mathbf{I}) &=6\mathbf{I}&&\text{factor}\\ \mathbf{B}^{-1} &=\frac16(\mathbf{B}^2-6\mathbf{B}+11\mathbf{I}) &&\text{product is identity}\\ \mathbf{B}^2 &=\begin{pmatrix}1&3&1\\0&4&5\\0&0&9\end{pmatrix} &&\text{square}\\ \mathbf{B}^{-1} &=\frac16\begin{pmatrix}6&-3&1\\0&3&-1\\0&0&2\end{pmatrix} =\begin{pmatrix}1&-1/2&1/6\\0&1/2&-1/6\\0&0&1/3\end{pmatrix} &&\text{substitute} \end{aligned}

Why this step? “Hence calculate” calls for substitution into the derived expression; an unsupported calculator inverse would not show that link.

Answer. The inverse is the displayed matrix.

Check. In $\mathbf{B}\mathbf{B}^{-1}$ , the first-row last entry is $-1/6+1/6=0$ , the second-row last entry is $-2/6+1/3=0$ , and the diagonal entries are all $1$ .

Change only the top-right entry of the matrix from $0$ to $1$ . Its triangular form leaves the characteristic equation unchanged.

Worked example 4: the same identity, a changed matrix

Question. Use Cayley-Hamilton to find the inverse of

\mathbf{C}=\begin{pmatrix}1&1&1\\0&2&1\\0&0&3\end{pmatrix},

whose characteristic equation is again $\lambda^3-6\lambda^2+11\lambda-6=0$ .

Plan. Retain the polynomial identity but recalculate the square of the new matrix.

\begin{aligned} \mathbf{C}^3-6\mathbf{C}^2+11\mathbf{C}-6\mathbf{I} &=\mathbf{0}&&\text{Cayley-Hamilton}\\ \mathbf{C}^{-1} &=\frac16(\mathbf{C}^2-6\mathbf{C}+11\mathbf{I}) &&\text{factor as before}\\ \mathbf{C}^2 &=\begin{pmatrix}1&3&5\\0&4&5\\0&0&9\end{pmatrix} &&\text{new square}\\ \mathbf{C}^{-1} &=\frac16\begin{pmatrix}6&-3&-1\\0&3&-1\\0&0&2\end{pmatrix} =\begin{pmatrix}1&-1/2&-1/6\\0&1/2&-1/6\\0&0&1/3\end{pmatrix} &&\text{evaluate} \end{aligned}

Why this step? Equal characteristic equations do not imply equal matrices or equal inverses; the polynomial is evaluated at the actual matrix.

Answer. The inverse is the displayed matrix.

Check. The first-row last entry of $\mathbf{C}\mathbf{C}^{-1}$ is $-1/6-1/6+1/3=0$ ; the full product is $\mathbf{I}$ .

For comparison, $\operatorname{diag}(0,1,2)$ has a characteristic equation with zero constant term. It satisfies that equation but cannot undo the collapse of its first coordinate.

Watch out

Wrong: “divide the equation by $\mathbf{B}$ ”. Right: factor a matrix product equal to a multiple of $\mathbf{I}$ , as above, or explicitly multiply by $\mathbf{B}^{-1}$ . Keep $\mathbf{I}$ in every constant term.

Change the same entry again and show the requested theorem-based route.

Read a transformation along its eigendirections

An eigenvector tells you about every point on its line through the origin. If $\mathbf{M}\mathbf{v}=\lambda\mathbf{v}$ , then $\mathbf{M}(t\mathbf{v})=t\lambda\mathbf{v}$ . The multiplier therefore applies along the whole direction.

Geometric meaning of a real eigenvalue

For a real non-zero eigenvector $\mathbf{v}$ :

$\lambda=1$ : every point on its origin line is fixed.
$\lambda=-1$ : vectors on that line reverse with unchanged length.
$\lambda=0$ : the entire line is sent to the origin.
General real $\lambda$ : length is multiplied by $|\lambda|$ , with reversal when $\lambda<0$ .

For $\lambda\ne0$ , the line is mapped onto itself. This need not mean its points are fixed.

Not in the formula booklet: learn it.

A reflection fixes its mirror line and reverses the perpendicular direction. A projection can fix the same line but collapse the perpendicular direction instead.

Worked example 5: reflection in a line

Question. The matrix $\mathbf{S}=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ represents a reflection. Use eigenvectors to identify its mirror line.

Plan. Find the fixed direction using $\lambda=1$ , then check the reversed direction using $\lambda=-1$ .

\begin{aligned} \mathbf{S}\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}x\\y\end{pmatrix} &\Longrightarrow y=x&&\text{fixed points}\\ \mathbf{S}\begin{pmatrix}1\\1\end{pmatrix} &=\begin{pmatrix}1\\1\end{pmatrix}&&\text{eigenvalue }1\\ \mathbf{S}\begin{pmatrix}1\\-1\end{pmatrix} &=-\begin{pmatrix}1\\-1\end{pmatrix}&&\text{eigenvalue }-1 \end{aligned}

Why this step? The mirror consists of fixed points, so it is the direction with eigenvalue $1$ , not the reversed direction.

Answer. Reflection in $y=x$ .

Check. The two displayed directions are perpendicular, and the normal direction reverses.

Keep the fixed direction but change the perpendicular multiplier from $-1$ to $0$ .

Worked example 6: projection onto the same line

Question. Describe the transformation with matrix

\mathbf{T}=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}

using its action on the two directions from Worked example 5.

Plan. Calculate the multipliers along the line and its perpendicular.

\begin{aligned} \mathbf{T}\begin{pmatrix}1\\1\end{pmatrix} &=\begin{pmatrix}1\\1\end{pmatrix}&&\text{line is fixed}\\ \mathbf{T}\begin{pmatrix}1\\-1\end{pmatrix} &=\begin{pmatrix}0\\0\end{pmatrix}&&\text{normal is collapsed}\\ \mathbf{T}\begin{pmatrix}x\\y\end{pmatrix} &=\begin{pmatrix}(x+y)/2\\(x+y)/2\end{pmatrix}&&\text{general image} \end{aligned}

Why this step? A zero multiplier removes the perpendicular component instead of reversing it, giving a perpendicular projection.

Answer. Perpendicular projection onto $y=x$ .

Check. The image lies on $y=x$ , and its difference from $\begin{pmatrix}x\\y\end{pmatrix}$ is a multiple of $\begin{pmatrix}1\\-1\end{pmatrix}$ , perpendicular to that line.

An eigenvalue of $3$ , by contrast, would preserve the origin line but move all its non-origin points. “Invariant line” and “line of invariant points” make different claims.

Watch out

Wrong: every eigenvector is unchanged by its matrix. Right: only eigenvalue $1$ fixes the vector; other eigenvalues describe stretching, reversal or collapse.

Change the projection's fixed line and identify both its fixed and collapsed directions.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Core Pure

Mechanics (Major and Minor)

Statistics (Major and Minor)

Modelling with Algorithms

Numerical Methods

Extra Pure

Further Pure with Technology

Xm5, Xm6 - Cayley-Hamilton and transformation eigenvectors

A characteristic equation becomes a matrix identity

Rearrange the identity to find an inverse

Read a transformation along its eigendirections

Continue this lesson.

Learn faster with Nova