Diagonalization

93. Diagonalization

A square matrix $𝐴$ is diagonalizable if it can be written as

𝐴 = 𝑃 𝐷 𝑃^{- 1}

where $𝐷$ is a diagonal matrix and $𝑃$ is an invertible matrix.

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

93.1. When is $𝐴$ diagonalizable?

An $𝑛 \times 𝑛$ matrix is diagonalizable iff it has $𝑛$ linearly independent eigenvectors.

Equivalent conditions:

$𝐴$ has $𝑛$ distinct eigenvalues → automatically diagonalizable
Every eigenvalue’s geometric multiplicity equals its algebraic multiplicity (see Characteristic Polynomial)
$ℝ^{𝑛}$ has a basis consisting of eigenvectors of $𝐴$ (an eigenbasis)

Some matrices are not diagonalizable — e.g., $[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$ has only one eigenvalue ( $1$ ) of algebraic multiplicity $2$ but geometric multiplicity $1$ . For non-diagonalizable matrices, see Jordan Canonical Form.

93.2. How to diagonalize

Find the eigenvalues $𝜆_{1}, \dots, 𝜆_{𝑛}$ from $det (𝐴 - 𝜆 𝐼) = 0$ .
For each $𝜆_{𝑖}$ , find a basis for $ker (𝐴 - 𝜆_{𝑖} 𝐼)$ — the eigenvectors.
Assemble eigenvectors into columns of $𝑃$ , eigenvalues into the diagonal of $𝐷$ (in matching order).
Verify by computing $𝑃 𝐷 𝑃^{- 1}$ .

Example

$𝐴 = [\begin{matrix} 4 & - 2 \\ 1 & 1 \end{matrix}]$ .

Characteristic polynomial: $det (𝐴 - 𝜆 𝐼) = (4 - 𝜆) (1 - 𝜆) - (- 2) (1) = 𝜆^{2} - 5 𝜆 + 6 = (𝜆 - 2) (𝜆 - 3)$ .

Eigenvalues: $𝜆_{1} = 2, 𝜆_{2} = 3$ .

Eigenvectors: solve $(𝐴 - 2 𝐼) \vec{𝑣} = 𝟎$ → ${\vec{𝑣}}_{1} = [\begin{matrix} 1 \\ 1 \end{matrix}]$ . And $(𝐴 - 3 𝐼) \vec{𝑣} = 𝟎$ → ${\vec{𝑣}}_{2} = [\begin{matrix} 2 \\ 1 \end{matrix}]$ .

𝑃 = [\begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix}] 𝐷 = [\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}]

93.3. Why it matters: power and exponential

In diagonal form, matrix arithmetic is trivial:

𝐴^{𝑘} = {(𝑃 𝐷 𝑃^{- 1})}^{𝑘} = 𝑃 𝐷^{𝑘} 𝑃^{- 1}

with $𝐷^{𝑘}$ just powering each diagonal entry. This makes:

$𝐴^{𝑘}$ for large $𝑘$ tractable (Markov chains, dynamics)
Matrix exponential $𝑒^{𝐴} = 𝑃 𝑒^{𝐷} 𝑃^{- 1}$ (linear ODE solutions)
Solving linear recurrences in closed form

93.4. Spectral decomposition (real symmetric case)

If $𝐴$ is real and symmetric, then it’s always diagonalizable and can be done with an orthogonal $𝑃$ :

𝐴 = 𝑄 𝐷 𝑄^{𝑇}, where 𝑄^{𝑇} 𝑄 = 𝐼

This is the Spectral Theorem — the gold standard of diagonalization.

93.5. Failure: non-diagonalizable matrices

𝐴 = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]

has only the eigenvalue $1$ (with algebraic multiplicity $2$ ) and only one independent eigenvector $[\begin{matrix} 1 \\ 0 \end{matrix}]$ . Not diagonalizable — the best you can do is the Jordan form.

93.6. See also

Eigenvectors & Eigenvalues
Characteristic Polynomial
Spectral Theorem — symmetric case
Jordan Canonical Form — non-diagonalizable case
Change of Basis — what $𝑃$ does geometrically