Diagonalization

A diagonal matrix is the easiest kind of matrix: it just stretches each coordinate axis. If a matrix has enough eigenvectors, then in the basis made of those eigenvectors it is diagonal. Writing that down is diagonalization.

Key idea

$A$ ($n \times n$) is diagonalizable if $A = PDP^{-1}$ with $D$ diagonal. This happens exactly when $A$ has $n$ linearly independent eigenvectors; then the columns of $P$ are those eigenvectors and the diagonal of $D$ holds their eigenvalues, in the same order.

Why it works

Write $P = [\,\mathbf{v}_1\ \cdots\ \mathbf{v}_n\,]$ and $D = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$. Column by column, $AP = [\,A\mathbf{v}_1\ \cdots\ A\mathbf{v}_n\,]$ and $PD$ $= [\,\lambda_1\mathbf{v}_1\ \cdots\ \lambda_n\mathbf{v}_n\,]$. So $AP = PD$ says exactly that each $\mathbf{v}_i$ is an eigenvector with eigenvalue $\lambda_i$. If the $\mathbf{v}_i$ are independent, $P$ is invertible and $A = PDP^{-1}$. In words: change to eigenvector coordinates ($P^{-1}$), stretch each axis ($D$), change back ($P$).

Worked example: Diagonalizing a 2 × 2 matrix

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

Eigenvalues (from the characteristic equation): $5$ and $2$.
Eigenvectors: $(1, 1)$ for $5$ and $(1, -2)$ for $2$.
$P = \begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix}$ and $D = \begin{bmatrix} 5 & 0 \\ 0 & 2 \end{bmatrix}$, in matching order.
Check $AP = PD$: both equal $\begin{bmatrix} 5 & 2 \\ 5 & -4 \end{bmatrix}$. ✓

Answer: $A = PDP^{-1}$ with $P = \begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix}$, $D = \begin{bmatrix} 5 & 0 \\ 0 & 2 \end{bmatrix}$

In the basis of the two eigenvector lines, $A$ is just "stretch by 5 along one line, by 2 along the other".

When it works, and when it does not

$n$ distinct eigenvalues: always diagonalizable (eigenvectors for different eigenvalues are independent).
A repeated eigenvalue: diagonalizable exactly when each eigenspace's dimension equals that eigenvalue's algebraic multiplicity, so the eigenspace dimensions add up to $n$.

Worked example: A repeated eigenvalue that is fine

Is $C = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 1 & 2 & 1 \end{bmatrix}$ diagonalizable?

Eigenvalues $3, 3, 1$ (lower triangular).
From the previous topic, $E_3$ has basis $(-2, 1, 0), (2, 0, 1)$ (dimension 2), and $E_1$ is spanned by $(0, 0, 1)$.
$2 + 1 = 3$ independent eigenvectors, so yes: $P = \begin{bmatrix} -2 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}$, $D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

Answer: Yes.

Worked example: One that fails

Is $\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$ diagonalizable?

Eigenvalue $2$, twice.
$A - 2I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ has rank 1, so $E_2$ is only a line (spanned by $(1, 0)$).
Only one independent eigenvector for a $2 \times 2$ matrix: not diagonalizable.

Answer: No.

The payoff comes in the next topics: powers $A^{k} = PD^{k}P^{-1}$, long-run behavior, and systems of differential equations, which split into separate one-variable equations in eigenvector coordinates. Whenever a problem involves applying the same matrix again and again, diagonalizing first is the natural move.

Common mistake

Thinking a repeated eigenvalue always prevents diagonalizing, or never does. Both happen: $3I$ and the matrix $C$ above are diagonalizable, the shear is not. Decide by computing the dimension of each repeated eigenvalue's eigenspace. Distinct eigenvalues are the safe case.

Diagonalization

When it works, and when it does not

Practice

Builds on

Leads to

Sources and further reading