Transformation of Matrices

Lessons

Senior 6 · Mathematics

Transformation of Matrices

5 lessons

Lesson 5 of 5

Diagonalisation and powers of a matrix

40 min

Why a diagonal matrix is the easiest matrix

Raising a diagonal matrix to a power needs no work at all: $\begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}^{\,n} = \begin{pmatrix} 5^n & 0 \\ 0 & 2^n \end{pmatrix}$ . Raising $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ to the tenth power by hand needs nine matrix products. The eigenvectors of lesson 4 remove the difference. Measured along its eigenvector directions, $A$ is just "stretch by 5 along $(1, 1)$ , stretch by 2 along $(1, - 2)$ ": a diagonal matrix in disguise. This lesson takes the disguise off.

Definition

A square matrix $A$ is diagonalisable if there is an invertible matrix $P$ and a diagonal matrix $D$ such that

P^{-1}AP = D , \quad \text{equivalently} \quad A = PDP^{-1} .

Finding such $P$ and $D$ is called diagonalising $A$ . Matrices $A$ and $B$ with $B = P^{-1}AP$ for some invertible $P$ are called similar; similar matrices have the same characteristic polynomial, hence the same eigenvalues, trace and determinant.

Important

Theorem. A square matrix $A$ of order $n$ is diagonalisable if and only if it has $n$ linearly independent eigenvectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ . Then

P = \big(\mathbf{v}_1 \ \mathbf{v}_2 \ \cdots \ \mathbf{v}_n\big) \ \text{(eigenvectors as columns)}, \qquad D = \operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_n) ,

with $\lambda_i$ the eigenvalue of $\mathbf{v}_i$ , in the same order.

Reason. Column $i$ of $A P$ is $A\mathbf{v}_i = \lambda_i\mathbf{v}_i$ , which is column $i$ of $P D$ . So $A P = P D$ , and, since independent columns make $P$ invertible, $P^{-1}AP = D$ .

Sufficient condition. Eigenvectors for different eigenvalues are linearly independent, so a matrix of order $n$ with $n$ distinct real eigenvalues is always diagonalisable. With a repeated eigenvalue, it is diagonalisable exactly when each eigenspace has dimension equal to the number of times its eigenvalue is repeated.

Important

Powers. If $A = PDP^{-1}$ , then for every positive integer $n$

A^n = PD^nP^{-1} , \qquad D^n = \operatorname{diag}(\lambda_1^n, \dots, \lambda_n^n) .

Reason. $A^2 = PDP^{-1}PDP^{-1} = PD(P^{-1}P)DP^{-1} = PD^2P^{-1}$ : each inner $P^{-1}P$ cancels. Induction gives every $n$ .

Worked example 1: diagonalise, then a formula for $A^{n}$

Diagonalise $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ and find $A^{n}$ .

Solution. From lesson 4: $\lambda = 5$ with $(1, 1)$ and $\lambda = 2$ with $(1, - 2)$ . Two distinct eigenvalues, so $A$ is diagonalisable:

P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} , \quad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix} , \quad P^{-1} = \frac{1}{-3}\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} .

Then

A^n = PD^nP^{-1} = \frac{1}{3}\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} 5^n & 0 \\ 0 & 2^n \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 2\cdot 5^n + 2^n & 5^n - 2^n \\ 2\cdot 5^n - 2^{n+1} & 5^n + 2^{n+1} \end{pmatrix} .

Check with $n = 2$ : $\frac{1}{3}\begin{pmatrix} 54 & 21 \\ 42 & 33 \end{pmatrix} = \begin{pmatrix} 18 & 7 \\ 14 & 11 \end{pmatrix}$ , and directly $A^2 = \begin{pmatrix} 18 & 7 \\ 14 & 11 \end{pmatrix} = \begin{pmatrix} 18 & 7 \\ 14 & 11 \end{pmatrix}$ .

Worked example 2: a repeated eigenvalue that still works

Diagonalise $S = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$ and find $S^{n}$ .

Solution. From lesson 4: $E_4 = \operatorname{span}\{(1, 1, 1)\}$ and $E_1 = \operatorname{span}\{(-1, 1, 0), (-1, 0, 1)\}$ . The double eigenvalue 1 has a 2-dimensional eigenspace, so there are three independent eigenvectors:

P = \begin{pmatrix} 1 & -1 & -1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} , \quad D = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} , \quad P^{-1} = \frac{1}{3}\begin{pmatrix} 1 & 1 & 1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix} .

Multiplying out $PD^nP^{-1}$ gives

S^n = \frac{1}{3}\begin{pmatrix} 4^n + 2 & 4^n - 1 & 4^n - 1 \\ 4^n - 1 & 4^n + 2 & 4^n - 1 \\ 4^n - 1 & 4^n - 1 & 4^n + 2 \end{pmatrix} .

Check with $n = 3$ : diagonal $\frac{66}{3} = 22$ , off-diagonal $\frac{63}{3} = 21$ , which matches $S^{3}$ computed directly.

Worked example 3: a matrix that cannot be diagonalised

Show that the shear $N = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ is not diagonalisable, and find $N^{n}$ another way.

Solution. $N$ is triangular, so its only eigenvalue is 1, repeated twice. But $N - I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has rank 1, so $E_1 = \operatorname{span}\{(1, 0)\}$ has dimension 1. There are not two independent eigenvectors, so $N$ is not diagonalisable. (If it were, $D$ would be $I$ and $N = PIP^{-1} = I$ , which is false.)

Compute: $N^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ , $N^3 = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$ . Suppose $N^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ ; then $N^{k+1} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & k+1 \\ 0 & 1 \end{pmatrix}$ . By induction, $N^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ .

Application: where a system settles

Some quantities change each year by a fixed matrix rule. Suppose (an illustrative model, not census data) that in a district, each year 10% of the rural residents move to town and 20% of the town residents move back, and nobody enters or leaves the district. If $r_{n}$ and $t_{n}$ are the rural and town shares after $n$ years, then

\begin{pmatrix} r_{n+1} \\ t_{n+1} \end{pmatrix} = \begin{pmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{pmatrix}\begin{pmatrix} r_n \\ t_n \end{pmatrix} , \quad \text{so} \quad \begin{pmatrix} r_n \\ t_n \end{pmatrix} = M^n \begin{pmatrix} r_0 \\ t_0 \end{pmatrix} \text{ with } M = \begin{pmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{pmatrix} .

A two-state model. The arrows out of each state carry fractions that add up to 1.

Worked example 4: the long run

In the model above, all residents start rural: $(r_{0}, t_{0}) = (1, 0)$ . Find $(r_{n}, t_{n})$ and its limit as $n \to \infty$ .

Solution. $\operatorname{tr} M = 1.7$ and $\det M = 0.72 - 0.02 = 0.7$ , so $\lambda^2 - 1.7\lambda + 0.7 = (\lambda - 1)(\lambda - 0.7) = 0$ .

$\lambda = 1$ : $M - I = \begin{pmatrix} -0.1 & 0.2 \\ 0.1 & -0.2 \end{pmatrix}$ , so $x = 2 y$ : eigenvector $(2, 1)$ . $\lambda = 0.7$ : $M - 0.7I = \begin{pmatrix} 0.2 & 0.2 \\ 0.1 & 0.1 \end{pmatrix}$ : eigenvector $(1, - 1)$ .

M^n = \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0.7^n \end{pmatrix}\frac{1}{3}\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 2 + 0.7^n & 2 - 2(0.7)^n \\ 1 - 0.7^n & 1 + 2(0.7)^n \end{pmatrix} .

Starting from $(1, 0)$ : $r_n = \frac{2 + 0.7^n}{3}$ and $t_n = \frac{1 - 0.7^n}{3}$ . Since $0 \lt 0.7 \lt 1$ , $0.7^n \to 0$ , so the shares tend to $\frac{2}{3}$ rural and $\frac{1}{3}$ town, whatever the start. The limit is the eigenvector for $\lambda = 1$ , $(2, 1)$ , scaled so that its entries add up to 1. After 3 years, $r_3 = \frac{2 + 0.343}{3} \approx 0.781$ .

Warning

Keep the orders matching. Column $i$ of $P$ must be the eigenvector of the $i$ -th diagonal entry of $D$ . Swapping two columns of $P$ without swapping the eigenvalues gives a wrong $A^{n}$ .

$A^n = PD^nP^{-1}$ , not $P^nD^nP^{-n}$ . Only the middle matrix is raised to the power.

A repeated eigenvalue is not the end. $S$ in worked example 2 has one and is diagonalisable; $N$ in worked example 3 has one and is not. Check the dimension of the eigenspace.

Tip

Test any formula for $A^{n}$ at $n = 1$ (it must give $A$ ) and at $n = 0$ (it must give $I$ ). In worked example 1, $n = 0$ gives $\frac{1}{3}\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = I$ .

MIT OpenCourseWare: Lecture 22 of 18.06 Linear Algebra, Diagonalization and Powers of A, with Gilbert Strang (about 52 min). Watch how he writes AS = SΛ with the eigenvectors as the columns of S, the same as AP = PD here, and uses it for the powers of A and for Fibonacci numbers.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a reason.

Diagonalise $A = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ (its eigenvectors were found in lesson 4, practice 1).
Use question 1 to find a formula for $A^{n}$ , and check it with $n = 3$ .
Show that $\begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$ is not diagonalisable.
Use the formula of worked example 1 to compute $\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}^5$ .
Diagonalise $B = \begin{pmatrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ (see lesson 4, practice 5).
Each week, 80% of a school's pupils who bought lunch at the canteen buy there again and 20% bring food from home; 30% of those who brought food switch to the canteen and 70% bring food again (illustrative). Find the long-run share that buys at the canteen.
Explain why a $3 \times 3$ matrix with eigenvalues $1, 2, 3$ must be diagonalisable, but one with eigenvalues $2, 2, 5$ need not be.

Show answersHide answers

$P = \begin{pmatrix} 2 & 1 \\ 3 & -1 \end{pmatrix}$ , $D = \begin{pmatrix} 4 & 0 \\ 0 & -1 \end{pmatrix}$ , $P^{-1} = \frac{1}{5}\begin{pmatrix} 1 & 1 \\ 3 & -2 \end{pmatrix}$ .
$A^n = \frac{1}{5}\begin{pmatrix} 2\cdot 4^n + 3(-1)^n & 2\cdot 4^n - 2(-1)^n \\ 3\cdot 4^n - 3(-1)^n & 3\cdot 4^n + 2(-1)^n \end{pmatrix}$ ; $A^3 = \begin{pmatrix} 25 & 26 \\ 39 & 38 \end{pmatrix}$ .
Only eigenvalue 3 (double); $E_3 = \operatorname{span}\{(1, 0)\}$ has dimension 1, so there are not two independent eigenvectors.
$\frac{1}{3}\begin{pmatrix} 6282 & 3093 \\ 6186 & 3189 \end{pmatrix} = \begin{pmatrix} 2094 & 1031 \\ 2062 & 1063 \end{pmatrix}$
$P = \begin{pmatrix} 1 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ , $D = \operatorname{diag}(-1, 3, 3)$ .
$M = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}$ ; $\lambda = 1$ gives $0.2 x = 0.3 y$ , eigenvector $(3, 2)$ : 60% at the canteen in the long run.
Three distinct eigenvalues give three independent eigenvectors. With $2, 2, 5$ , the eigenspace of 2 may have dimension only 1, as for $\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{pmatrix}$ .

1Why a diagonal matrix is the easiest matrix

2Application: where a system settles

Why a diagonal matrix is the easiest matrix

Application: where a system settles