Diagonalization: A = PDP⁻¹

What is diagonalization?

A square matrix $A$ is diagonalizable if there exists an invertible $P$ and diagonal $D$ with $A = P D P^{-1}$

The columns of $P$ are eigenvectors of $A$; the diagonal entries of $D$ are the corresponding eigenvalues.

Equivalently, $A$ is diagonalizable iff it has $n$ linearly independent eigenvectors. A matrix is defective (not diagonalizable) if it lacks enough independent eigenvectors for some eigenvalue.

Why diagonalize?

Powers become cheap. Using $A = P D P^{-1}$: $A^n = P D^n P^{-1}$

and $D^n$ is just each diagonal entry raised to the $n$. One matrix multiplication becomes one diagonal multiplication — for big $n$, the savings are huge.

Other uses: solving linear ODEs, matrix exponentials, Markov-chain steady states, principal-axis theorems.

Step-by-step

$A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ (same matrix as problem 355).

Step 1 — Find eigenvalues.

From problem 355: $\lambda_1 = 5$, $\lambda_2 = 2$.

Step 2 — Find eigenvectors.

From problem 355: $\mathbf{v}_1 = (1, 1)^T$ for $\lambda_1 = 5$; $\mathbf{v}_2 = (1, -2)^T$ for $\lambda_2 = 2$.

Step 3 — Build $P$ and $D$.

Columns of $P$ = eigenvectors (in any order, as long as $D$ matches). $P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}, \quad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$

Step 4 — Compute $P^{-1}$.

$\det P = (1)(-2) - (1)(1) = -3$. By the 2×2 inverse formula: $P^{-1} = \dfrac{1}{-3} \begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix} = \dfrac{1}{3} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}$

Step 5 — Verify $P D P^{-1} = A$.

$D P^{-1} = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix} \cdot \dfrac{1}{3} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} = \dfrac{1}{3} \begin{pmatrix} 10 & 5 \\ 2 & -2 \end{pmatrix}$

$P D P^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} \cdot \dfrac{1}{3} \begin{pmatrix} 10 & 5 \\ 2 & -2 \end{pmatrix} = \dfrac{1}{3} \begin{pmatrix} 12 & 3 \\ 6 & 9 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} = A ✓$

Using diagonalization to compute $A^n$

Compute $A^{10}$ without diagonalizing: multiply $A$ by itself 9 times. Slow.

With diagonalization: $A^{10} = P D^{10} P^{-1}$

$D^{10} = \begin{pmatrix} 5^{10} & 0 \\ 0 & 2^{10} \end{pmatrix} = \begin{pmatrix} 9{,}765{,}625 & 0 \\ 0 & 1{,}024 \end{pmatrix}$. One matrix multiplication on each side and you're done.

When diagonalization works

$A$ is diagonalizable if any of these holds:

• $A$ has $n$ distinct eigenvalues (no repeats).
• $A$ is symmetric (then it's orthogonally diagonalizable — see spectral theorem).
• $A$ is normal ($A A^* = A^* A$) — true of symmetric, Hermitian, unitary, orthogonal matrices.

Repeated eigenvalues don't automatically make $A$ defective — you have to check if the eigenspace has the right dimension.

The defective case

Example: $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. Characteristic polynomial: $(1 - \lambda)^2$. Only eigenvalue is $\lambda = 1$ with algebraic multiplicity 2. But the eigenspace is spanned by $(1, 0)^T$ only — geometric multiplicity 1 < 2. Not diagonalizable. For this case, use the Jordan canonical form.

Common mistakes

• Mismatching columns of $P$ with diagonal of $D$. Column $i$ of $P$ must be the eigenvector of the eigenvalue $d_{ii}$.
• Using the wrong $P^{-1}$. Always double-check by computing $P P^{-1} = I$.
• Assuming every square matrix diagonalizes. Many important ones do not (shear matrices, nilpotent matrices).

Try it in the visualization

Slide the matrix entries; eigenvalues/eigenvectors update live. $P, D, P^{-1}$ are displayed and their product is compared against $A$ — red when not diagonalizable, green when verified.