Diagonalisation of matricesAQA A-Level Further Maths: Revision notes

Section 1

Diagonal matrices and their powers

A diagonal matrix has zeros everywhere except on the leading diagonal. It is easy to work with: to raise it to a power, raise each diagonal entry to that power, $\begin{pmatrix} d_1 & 0 \\ 0 & d_2 \end{pmatrix}^n=\begin{pmatrix} d_1^n & 0 \\ 0 & d_2^n \end{pmatrix}.$ The same holds for a $3\times3$ diagonal matrix. Diagonalisation uses this: it rewrites a matrix so that the hard work is done on a diagonal matrix instead.

Key termsdiagonal matrix

Section 2

Diagonalising a matrix

If a matrix $M$ has real eigenvalues $\lambda_1,\lambda_2,\dots$ with eigenvectors $\mathbf{v}_1,\mathbf{v}_2,\dots$ , then $M=UDU^{-1},$ where the columns of $U$ are the eigenvectors and $D$ is the diagonal matrix of the matching eigenvalues in the same order. Example: $M=\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$ has eigenvalues $2$ and $5$ with eigenvectors $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ . Then $U=\begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix}$ , $D=\begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}$ and $U^{-1}=\frac13\begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix}$ . Check: $U^{-1}MU=D$ .

Key termsdiagonalisation

Common mistake

Putting the eigenvalues in $D$ in a different order from the eigenvectors in $U$ . The $k$ th column of $U$ must go with the $k$ th diagonal entry of $D$ .

Common mistake

Writing the eigenvectors as rows of $U$ . They must be columns.

Section 3

Method

Find the eigenvalues from $\det(M-\lambda I)=0$ .
Find an eigenvector for each eigenvalue from $(M-\lambda I)\mathbf{v}=\mathbf{0}$ .
Write the eigenvectors as the columns of $U$ and the eigenvalues, in the same order, on the diagonal of $D$ .
Find $U^{-1}$ . For a $2\times2$ matrix, $\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ . Any non-zero multiples of the eigenvectors work, and any order works, provided $U$ and $D$ agree. So $U$ and $D$ are not unique.

Key termseigenvectoreigenvalue

Exam tip

Check $UDU^{-1}$ by testing one entry of $M$ , or check $MU=UD$ , which avoids finding $U^{-1}$ .

Section 4

Powers of a matrix

Because the inner $U^{-1}U$ pairs cancel, $M^2=UDU^{-1}UDU^{-1}=UD^2U^{-1},\qquad M^n=UD^nU^{-1}.$ So $M^n$ is found by raising the two diagonal entries to the power $n$ , which gives a formula for any $n$ . Example: $M=\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$ gives $M^n=UD^nU^{-1}=\frac13\begin{pmatrix} 2\times2^n+5^n & 5^n-2^n \\ 2\times5^n-2\times2^n & 2^n+2\times5^n \end{pmatrix}$ . Check $n=1$ : the entries are $\frac{9}{3},\frac{3}{3},\frac{6}{3},\frac{12}{3}$ , which are $3,1,2,4$ , the original matrix. The eigenvalues of $M^n$ are $\lambda^n$ , and the eigenvectors stay the same.

Key termsmatrix power

Common mistake

Raising $U$ or $U^{-1}$ to the power $n$ . Only $D$ is raised to the power: $M^n=UD^nU^{-1}$ , not $U^nD^nU^{-n}$ .

Exam tip

Check a general formula for $M^n$ by substituting $n=1$ (you should get $M$ ) and $n=0$ (you should get $I$ ).

Section 5

3×3 matrices and real eigenvalues

The method is identical for a $3\times3$ matrix, with a $3\times3$ matrix $U$ of eigenvectors and a diagonal $D$ with three eigenvalues. Example: $B=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 1 & 3 \end{pmatrix}$ has $D=\mathrm{diag}(1,2,4)$ and $U=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end{pmatrix}$ , so $B^5=U\,\mathrm{diag}(1,32,1024)\,U^{-1}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 528 & 496 \\ 0 & 496 & 528 \end{pmatrix}$ . This method works with real eigenvalues. A matrix with no real eigenvalues, such as the $90^\circ$ rotation $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ , cannot be written as $UDU^{-1}$ with real $U$ and $D$ .

Key termsreal eigenvalues

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Diagonalisation of matrices

The matrix $M=\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$ has eigenvalues $2$ and $5$ , with corresponding eigenvectors $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ .
Taking $U=\begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix}$ , find $U^{-1}$ .2 marks
The matrix $A=\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ can be written as $A=UDU^{-1}$ , where $U=\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ and $D$ is a diagonal matrix.
Explain why $A^4$ has the same eigenvectors as $A$ , and state the eigenvalues of $A^4$ .2 marks
The matrix $B=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 1 & 3 \end{pmatrix}$ has eigenvalues $1$ , $2$ and $4$ , with corresponding eigenvectors $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ , $\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ .
Write down a matrix $U$ and a diagonal matrix $D$ such that $B=UDU^{-1}$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).