Math Core

Diagonal matrices are the easiest matrices there are: you multiply, invert and raise them to powers entry by entry. Diagonalization is the process of rewriting a matrix as a diagonal matrix in disguise, $A = PDP^{-1}$ , so that questions about $A$ (especially about $A^k$ for large $k$ ) become questions about a diagonal matrix.

Why diagonal is easy

If $D = \begin{bmatrix} 5 & 0 \\ 0 & 2 \end{bmatrix}$ , then $D^k = \begin{bmatrix} 5^k & 0 \\ 0 & 2^k \end{bmatrix}$ . Now suppose $A = PDP^{-1}$ for some invertible $P$ . Then

A^2 = (PDP^{-1})(PDP^{-1}) = PD(P^{-1}P)DP^{-1} = PD^2P^{-1},

and in the same way $A^k = PD^kP^{-1}$ for every positive integer $k$ . The inner $P^{-1}P$ pairs cancel, and all the hard work is on the diagonal.

Definition

Diagonalizable

A square matrix $A$ is diagonalizable if it is similar to a diagonal matrix, that is, if $A = PDP^{-1}$ for some invertible matrix $P$ and diagonal matrix $D$ .

Which matrices can be diagonalized?

Write $P = [\mathbf{v}_1\ \cdots\ \mathbf{v}_n]$ and let $D$ have diagonal entries $\lambda_1, \dots, \lambda_n$ . Then

AP = [A\mathbf{v}_1\ \cdots\ A\mathbf{v}_n] \qquad\text{and}\qquad PD = [\lambda_1\mathbf{v}_1\ \cdots\ \lambda_n\mathbf{v}_n].

So $AP = PD$ says precisely that $A\mathbf{v}_i = \lambda_i\mathbf{v}_i$ for each column. If $P$ is also invertible, its columns are linearly independent (in particular nonzero), and $AP = PD$ is the same as $A = PDP^{-1}$ . That proves the central theorem.

The Diagonalization Theorem

An $n \times n$ matrix $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors.

In that case $A = PDP^{-1}$ , where the columns of $P$ are $n$ linearly independent eigenvectors and the diagonal entries of $D$ are the corresponding eigenvalues, in the same order.

Geometrically, the columns of $P$ form an eigenvector basis of $\mathbb{R}^n$ . In that basis, $A$ just scales each coordinate by its eigenvalue. The picture shows this for $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$ , whose eigen-directions are $(1, 1)$ with $\lambda = 5$ and $(1, -2)$ with $\lambda = 2$ . The vector $\mathbf{x} = (2, -1)$ equals $(1, 1) + (1, -2)$ , so $A\mathbf{x} = 5(1, 1) + 2(1, -2) = (7, 1)$ .

In the eigenvector basis, A multiplies the v1-coordinate by 5 and the v2-coordinate by 2. So x = v1 + v2 maps to Ax = 5v1 + 2v2 = (7, 1).Open in grapher →

The recipe

Find the eigenvalues from the characteristic equation.
For each eigenvalue, find a basis of its eigenspace.
If the bases together contain $n$ vectors, they are automatically linearly independent; use them as the columns of $P$ . If there are fewer than $n$ , $A$ is not diagonalizable.
Build $D$ from the eigenvalues, matching the order of the columns of $P$ .
Check by verifying $AP = PD$ (easier than computing $P^{-1}$ ).

Worked example: Diagonalizing a 2 × 2 matrix

Diagonalize $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$ .

The characteristic polynomial is $\lambda^2 - 7\lambda + 10 = (\lambda - 5)(\lambda - 2)$ .

For $\lambda = 5$ : $A - 5I = \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix}$ gives $x_1 = x_2$ , so $\mathbf{v}_1 = (1, 1)$ .

For $\lambda = 2$ : $A - 2I = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}$ gives $2x_1 + x_2 = 0$ , so $\mathbf{v}_2 = (1, -2)$ .

P = \begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix}, \qquad D = \begin{bmatrix} 5 & 0 \\ 0 & 2 \end{bmatrix}.

Check: $AP = \begin{bmatrix} 5 & 2 \\ 5 & -4 \end{bmatrix} = PD$ .

Worked example: A formula for the powers

Use the diagonalization above to find a formula for $A^k$ .

$P^{-1} = \dfrac{1}{-3}\begin{bmatrix} -2 & -1 \\ -1 & 1 \end{bmatrix} = \dfrac{1}{3}\begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}$ , so

A^k = PD^kP^{-1} = \frac{1}{3}\begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix}\begin{bmatrix} 5^k & 0 \\ 0 & 2^k \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 2 \cdot 5^k + 2^k & 5^k - 2^k \\ 2 \cdot 5^k - 2^{k+1} & 5^k + 2^{k+1} \end{bmatrix}.

At $k = 1$ this gives $\frac{1}{3}\begin{bmatrix} 12 & 3 \\ 6 & 9 \end{bmatrix} = A$ , as it should.

When it fails, and when it is guaranteed

Worked example: Not enough eigenvectors

Is $A = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}$ diagonalizable?

The only eigenvalue is $3$ (algebraic multiplicity $2$ ). But $A - 3I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ has one free variable, so the eigenspace is the line spanned by $(1, 0)$ . There are not two independent eigenvectors, so $A$ is not diagonalizable. This matrix is a shear; no basis makes it act by pure scaling.

The dimension of the eigenspace for $\lambda$ is called its geometric multiplicity. It is always at least $1$ and at most the algebraic multiplicity. The theorem can be restated in those terms:

If $A$ has $n$ distinct eigenvalues, it is diagonalizable. (Eigenvectors for distinct eigenvalues are independent, and there are $n$ of them.)
In general, $A$ is diagonalizable if and only if the geometric multiplicities add up to $n$ , which happens exactly when each eigenvalue's geometric multiplicity equals its algebraic multiplicity (assuming all eigenvalues are real).

Worked example: Repeated eigenvalue, still diagonalizable

In the first lesson of this unit, $A = \begin{bmatrix} 4 & -1 & 2 \\ 1 & 2 & 2 \\ 1 & -1 & 5 \end{bmatrix}$ had a two-dimensional eigenspace for $\lambda = 3$ , with basis $(1, 1, 0)$ and $(-2, 0, 1)$ . Since $\operatorname{tr} A = 11$ , the third eigenvalue is $11 - 3 - 3 = 5$ . Solving $(A - 5I)\mathbf{x} = \mathbf{0}$ gives $(1, 1, 1)$ , which you can check: $A(1, 1, 1) = (5, 5, 5)$ . Three independent eigenvectors, so

P = \begin{bmatrix} 1 & -2 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}, \qquad D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}.

A repeated eigenvalue does not prevent diagonalization; a deficient eigenspace does.

Common mistake

Diagonalizable and invertible are unrelated. $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is diagonalizable but not invertible, while $\begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}$ is invertible but not diagonalizable. Invertibility is about whether $0$ is an eigenvalue; diagonalizability is about whether there are enough eigenvectors.

Practice

Practice 1

Let $A = PDP^{-1}$ with $P = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$ and $D = \begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix}$ . Find the $(1, 1)$ entry of $A^5$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

$A = \begin{bmatrix} 1 & 2 \\ 3 & 2 \end{bmatrix}$ . An eigenvector for the larger eigenvalue of $A$ has the form $(2, k)$ . Find $k$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

For $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$ , use the formula $A^k = \dfrac{1}{3}\begin{bmatrix} 2 \cdot 5^k + 2^k & 5^k - 2^k \\ 2 \cdot 5^k - 2^{k+1} & 5^k + 2^{k+1} \end{bmatrix}$ to find the $(1, 2)$ entry of $A^3$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Is $A = \begin{bmatrix} 1 & 5 & -3 \\ 0 & 2 & 7 \\ 0 & 0 & 4 \end{bmatrix}$ diagonalizable?

Practice 5

Is $A = \begin{bmatrix} 5 & 0 & 0 \\ 1 & 5 & 0 \\ 0 & 0 & -2 \end{bmatrix}$ diagonalizable?

Practice 6

Find the geometric multiplicity of $\lambda = 2$ for $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

$A = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 3 & -1 \\ 4 & 2 & 0 \end{bmatrix}$ has eigenvalues $2, 2, 1$ . An eigenvector for $\lambda = 1$ has the form $(0, 1, k)$ . Find $k$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.