Recipe

Diagonalize a matrix

Use when: you're given a square matrix $A$ and asked to find $P$ and $D$ with $A=PDP^{-1}$ , $D$ diagonal

Goal: the eigenvalues plus enough independent eigenvectors to fill $D$ and $P$ , or a confirmed conclusion that $A$ can't be diagonalized

Taught in Diagonalization · a later section extends this recipe with more cases

Quick recipe

Find the eigenvalues
Find an eigenvector for each eigenvalue
Check for enough independent eigenvectors
Build $P$ and $D$ in matching order
Find $P^{-1}$ and verify

1
Solve the characteristic equation $\det(A-\lambda I)=0$ for $\lambda$ .

Example

Take $A = \begin{bmatrix} 7 & 2 \\ -3 & 2 \end{bmatrix}$ , the matrix from the widget above. Its characteristic equation is $\lambda^2 - \operatorname{tr}(A)\lambda + \det(A) = \lambda^2 - 9\lambda + 20 = 0$ , using $\operatorname{tr}(A) = 7+2=9$ and $\det(A)=(7)(2)-(2)(-3)=20$ . That factors as $(\lambda-5)(\lambda-4)=0$ , so $\lambda = 5 \quad\text{or}\quad \lambda = 4$

Why: $P$ 's columns and $D$ 's diagonal entries are built entirely out of $A$ 's eigenpairs, so those have to be found first.
2
For each $\lambda$ , solve $(A-\lambda I)\mathbf v=\mathbf 0$ ; the nonzero solutions are the eigenvectors for that $\lambda$ .

Example

For $\lambda=5$ : $A - 5I = \begin{bmatrix} 2 & 2 \\ -3 & -3 \end{bmatrix}$ , giving $2x_1+2x_2=0$ , i.e. $x_1=-x_2$ , so $v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$ . For $\lambda=4$ : $A-4I = \begin{bmatrix} 3 & 2 \\ -3 & -2 \end{bmatrix}$ , giving $3x_1+2x_2=0$ , so $v_2 = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$ (pick $x_1=2$ to clear the fraction). Both match the eigenpairs the widget revealed.

Why: these eigenvectors are exactly what fills $P$ 's columns in the next step.
3
Count the independent eigenvectors gathered so far, across every eigenvalue.

Example

$A$ is $2\times2$ , and $v_1=\begin{bmatrix}1\\-1\end{bmatrix}$ , $v_2=\begin{bmatrix}2\\-3\end{bmatrix}$ aren't scalar multiples of each other (their $2\times2$ determinant $(1)(-3)-(2)(-1) = -1 \neq 0$ ), so that's $2$ independent eigenvectors out of the $2$ needed — continue to Step 4.
- an $n\times n$ matrix has $n$ independent eigenvectors total — continue to Step 4.
- fewer than $n$ independent eigenvectors — stop; $A$ is not diagonalizable.
Why: $P$ needs a full set of $n$ independent columns to be invertible, and $A=PDP^{-1}$ is only possible when such a full eigenbasis exists.
4
Put the eigenvectors into the columns of $P$ , and put each eigenvector's matching eigenvalue into the same-position diagonal entry of $D$ .

Example

$P = \begin{bmatrix} 1 & 2 \\ -1 & -3 \end{bmatrix}, \qquad D = \begin{bmatrix} 5 & 0 \\ 0 & 4 \end{bmatrix}$
Column 1 of $P$ (the eigenvector for $\lambda=5$ ) lines up with $D$ 's first diagonal entry, $5$ ; column 2 (for $\lambda=4$ ) lines up with $D$ 's second entry, $4$ .

Why: $Av_i=\lambda_iv_i$ has to hold column by column, so the order eigenvectors go into $P$ has to match the order their eigenvalues go into $D$ exactly.
5
Compute $P^{-1}$ , then confirm that multiplying $PDP^{-1}$ back out reproduces $A$ .

Example

$\det P = (1)(-3)-(2)(-1) = -1$ , so $P^{-1} = \dfrac{1}{-1}\begin{bmatrix} -3 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ -1 & -1 \end{bmatrix}$ . Then $PD = \begin{bmatrix} 5 & 8 \\ -5 & -12 \end{bmatrix}, \qquad PDP^{-1} = \begin{bmatrix} 5 & 8 \\ -5 & -12 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 7 & 2 \\ -3 & 2 \end{bmatrix} = A \ \checkmark$

Why: this checks the whole construction against the actual definition of diagonalizable, rather than trusting the eigenvalue and eigenvector computations alone.

Common mistakes

Writing $A=P^{-1}DP$ instead of $A=PDP^{-1}$ .
Putting an eigenvector into a column of $P$ that doesn't line up with its own eigenvalue's position in $D$ .
Finding an eigenvalue but never actually solving for its eigenvector.
Assuming every square matrix is diagonalizable, and skipping the independent-eigenvector count entirely.