Diagonalization and Powers of a Matrix

Quick answer

A square matrix with n linearly independent eigenvectors can be written as A = PDP⁻¹, where the columns of P are the eigenvectors and D is the diagonal matrix of matching eigenvalues. In the coordinates given by the eigenvectors, A does nothing but scale each coordinate. That turns powers into a single step, Aᵏ = PDᵏP⁻¹, and explains long-run behavior: the largest eigenvalues dominate. Distinct eigenvalues guarantee diagonalizability; a repeated eigenvalue with too few eigenvectors prevents it.

What you'll learn

Diagonalize a matrix as A = PDP⁻¹
Compute powers of a matrix from its diagonalization
Decide whether a matrix is diagonalizable
Find the long-run state of a Markov chain

A matrix in its own coordinates

The matrix $B = \begin{bmatrix} 4 & -2\\ 1 & 1 \end{bmatrix}$ from the previous lesson has eigenvectors $(1, 1)$ for $\lambda = 2$ and $(2, 1)$ for $\lambda = 3$ . Put the eigenvectors into the columns of $P$ and the eigenvalues, in the same order, on the diagonal of $D$ :

P = \begin{bmatrix} 1 & 2\\ 1 & 1 \end{bmatrix} \qquad D = \begin{bmatrix} 2 & 0\\ 0 & 3 \end{bmatrix} \qquad B = PDP^{-1}

This is diagonalization. Read right to left, $PDP^{-1}\mathbf{x}$ first finds the coordinates of $\mathbf{x}$ along the eigenvectors, then scales each coordinate by its eigenvalue, then assembles the result.

Why AP = PD

Column $j$ of $AP$ is $A\mathbf{v}_j$ , and column $j$ of $PD$ is $\lambda_j\mathbf{v}_j$ , since $D$ multiplies the $j$ th column of $P$ by $\lambda_j$ . Those are equal exactly because $\mathbf{v}_j$ is an eigenvector. So $AP = PD$ , and when the eigenvectors are independent, $P$ is invertible and $A = PDP^{-1}$ . Diagonalizing is the eigenvector equation written for all $n$ eigenvectors at once. It works exactly when there are $n$ independent eigenvectors to fill the columns of $P$ .

Two facts decide that:

$n$ distinct eigenvalues always give $n$ independent eigenvectors, so the matrix is diagonalizable.
A repeated eigenvalue is fine only if its eigenspace has as many dimensions as the number of times the eigenvalue repeats.

Powers in one step

Multiplying $A = PDP^{-1}$ by itself, the inner $P^{-1}P$ pairs cancel:

A^k = PDP^{-1}\,PDP^{-1}\cdots PDP^{-1} = PD^kP^{-1}

and $D^k$ is diagonal with entries $\lambda_1^k, \ldots, \lambda_n^k$ . A hundredth power costs the same two multiplications as a square.

Worked examples

Example 4 — Where a Markov chain settles

Each year $10\%$ of city residents move to the suburbs and $20\%$ of suburban residents move to the city. Starting with everyone in the city, what fraction lives there in the long run?

The shares evolve by $\mathbf{x}_{k+1} = M\mathbf{x}_k$ with $M = \begin{bmatrix} 0.9 & 0.2\\ 0.1 & 0.8 \end{bmatrix}$ . Its eigenvalues are $1$ , with eigenvector $(2, 1)$ , and $0.7$ , with eigenvector $(1, -1)$ . Writing the start in those coordinates, $(1, 0) = \tfrac{1}{3}(2, 1) + \tfrac{1}{3}(1, -1)$ , so

M^k\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 2 \\ 1 \end{bmatrix} + \frac{1}{3}(0.7)^k\begin{bmatrix} 1 \\ -1 \end{bmatrix} \longrightarrow \begin{bmatrix} 2/3 \\ 1/3 \end{bmatrix}

The factor $0.7^k$ dies away, and two thirds of the population ends up in the city, whatever the starting split.

long run: 2/3

The city's share, year by year, approaches 2/3

Common mistakes

Practice problems

Diagonalize $\begin{bmatrix} 3 & 1\\ 2 & 2 \end{bmatrix}$ .

Answer

$P = \begin{bmatrix} 1 & 1\\ 1 & -2 \end{bmatrix}$ , $D = \begin{bmatrix} 4 & 0\\ 0 & 1 \end{bmatrix}$

Full solution

The eigenvalues are $4$ and $1$ , with eigenvectors $(1, 1)$ and $(1, -2)$ from the previous lesson. Place them as columns in the same order as the eigenvalues.
Is $\begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix}$ diagonalizable?

Answer

No

Full solution

The eigenvalue $2$ repeats, and $A - 2I = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$ gives only one independent eigenvector.
Is $\begin{bmatrix} 1 & 2\\ 0 & 3 \end{bmatrix}$ diagonalizable?

Answer

Yes

Full solution

Two distinct eigenvalues, $1$ and $3$ , guarantee two independent eigenvectors.
For $B = \begin{bmatrix} 4 & -2\\ 1 & 1 \end{bmatrix}$ , use the formula to find $B^2$ , and check by multiplying.

Answer

$\begin{bmatrix} 14 & -10\\ 5 & -1 \end{bmatrix}$

Full solution

With $2^2 = 4$ and $3^2 = 9$ : $-4 + 18 = 14$ , $8 - 18 = -10$ , $-4 + 9 = 5$ and $8 - 9 = -1$ . Multiplying $B$ by itself gives the same matrix.
$A = PDP^{-1}$ with $P = \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$ and $D = \begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix}$ . Find $A^3$ .

Answer

$\begin{bmatrix} 1 & 7\\ 0 & 8 \end{bmatrix}$

Full solution

$A^3 = PD^3P^{-1}$ with $D^3 = \text{diag}(1, 8)$ and $P^{-1} = \begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix}$ . Then $PD^3 = \begin{bmatrix} 1 & 8\\ 0 & 8 \end{bmatrix}$ , and multiplying by $P^{-1}$ gives the answer.
A $3 \times 3$ matrix has eigenvalues $1$ , $2$ and $2$ , and the eigenspace for $2$ is a line. Is it diagonalizable?

Answer

No

Full solution

The eigenvalue $2$ repeats twice but contributes only one independent eigenvector, so there are only two independent eigenvectors in all.
A Markov chain has $M = \begin{bmatrix} 0.8 & 0.3\\ 0.2 & 0.7 \end{bmatrix}$ . Find its long-run state.

Answer

$(0.6, 0.4)$

Full solution

$M - I = \begin{bmatrix} -0.2 & 0.3\\ 0.2 & -0.3 \end{bmatrix}$ gives $2x_1 = 3x_2$ , so $(3, 2)$ is an eigenvector for $1$ . Scaled to add to $1$ , it is $(0.6, 0.4)$ . The other eigenvalue is $0.5$ , whose contribution dies away.
$A = PDP^{-1}$ with $D = \text{diag}(2, 3)$ . Find $\det A$ .

Answer

$6$

Full solution

$\det A = \det P \cdot \det D \cdot \det P^{-1} = \det D = 2 \cdot 3$ .
$A = PDP^{-1}$ with $D = \text{diag}(1, 0.5)$ . What does $A^k$ approach as $k$ grows?

Answer

$P\,\text{diag}(1, 0)\,P^{-1}$

Full solution

$D^k = \text{diag}(1, 0.5^k)$ , and $0.5^k \to 0$ . Only the eigenvector for $1$ survives.
A student says $\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$ is diagonalizable because it is triangular and its eigenvalues are on the diagonal. What went wrong?

Hint

Count independent eigenvectors.

Answer

It has only one independent eigenvector, so it is not diagonalizable.

Full solution

Triangular form gives the eigenvalues, here $1$ twice. Diagonalizing needs two independent eigenvectors, and $A - I = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$ has a one-dimensional null space.

Frequently asked questions

What does it mean to diagonalize a matrix?

To write A = PDP⁻¹ with D diagonal. The columns of P are eigenvectors of A, and the diagonal of D holds the matching eigenvalues in the same order.

When is a matrix diagonalizable?

When it has n linearly independent eigenvectors. An n × n matrix with n distinct eigenvalues always does.

Why does diagonalization help with powers?

Aᵏ = PDᵏP⁻¹, and a power of a diagonal matrix raises each diagonal entry to the kth power.

Is every matrix diagonalizable?

No. [[1, 1], [0, 1]] has the single eigenvalue 1 but only one independent eigenvector, so there are not enough columns for P.

What is the steady state of a Markov chain?

The eigenvector for eigenvalue 1, scaled so its entries add to 1. The chain approaches it because every other eigenvalue has size less than 1.

What to learn next

Orthogonal Projections and Orthogonal BasesWith an orthogonal basis, coordinates come from dot products alone. Project a vector onto a line or subspace, and see why the projection is the closest point.

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