Diagonalizability Criterion

Statement

Let $K$ be a field and $A$ a square matrix over $K$ (or equivalently, a linear endomorphism of a finite-dimensional $K$-vector space). Then:

In particular:

• If $A$ has $n$ distinct eigenvalues (over $K$), then $A$ is diagonalizable.
• $A$ is diagonalizable iff $\mu_A$ is squarefree (no repeated irreducible factor), i.e.\ $\gcd(\mu_A, \mu_A') = 1$.
• Over an algebraically closed field, diagonalizable $\Leftrightarrow$ $\mu_A = (x - \lambda_1)\cdots(x - \lambda_k)$ for distinct $\lambda_i$.

Visualization

Three canonical $2 \times 2$ examples:

Diagonalizable (distinct eigenvalues):
A = ( 1  1 )    eigenvalues 1, 2 (distinct)
    ( 0  2 )    mu_A = (x-1)(x-2)  -- squarefree
                P = ( 1  1 )  P^{-1} A P = diag(1, 2)
                    ( 0  1 )

Not diagonalizable (repeated, non-trivial Jordan block):
B = ( 1  1 )    eigenvalue 1 (double)
    ( 0  1 )    mu_B = (x-1)^2  -- NOT squarefree
                (B - I)^1 ≠ 0, so min poly has degree 2

Diagonalizable (repeated but semisimple):
C = ( 1  0 )    eigenvalue 1 (double)
    ( 0  1 )    mu_C = x - 1  -- squarefree!
                C = I, already diagonal

Matrix	Eigenvalues	Min poly	Squarefree?	Diagonalizable?
$A = \begin{pmatrix}1&1\\0&2\end{pmatrix}$	$1, 2$	$(x-1)(x-2)$	Yes	Yes
$B = \begin{pmatrix}1&1\\0&1\end{pmatrix}$	$1, 1$	$(x-1)^2$	No	No
$C = I_2$	$1, 1$	$x - 1$	Yes	Yes

Proof Sketch

1. Squarefree implies diagonalizable. If $\mu_A = p_1 \cdots p_k$ with distinct monic irreducibles, the Chinese Remainder Theorem for $K[x]$-modules gives $V = \ker p_1(A) \oplus \cdots \oplus \ker p_k(A)$. Over an algebraically closed field, each $p_i = x - \lambda_i$ and each kernel is an eigenspace.
2. Diagonalizable implies squarefree. If $A = P \, \text{diag}(\lambda_1, \ldots, \lambda_n) P^{-1}$, then $(A - \lambda_i I)|_{\text{eigenspace of }\lambda_i} = 0$, so $\mu_A$ vanishes at each $\lambda_i$ with multiplicity $1$ — hence $\mu_A = \prod_{\text{distinct}} (x - \lambda_i)$, which is squarefree.
3. Distinct eigenvalues are sufficient. $n$ distinct eigenvalues $\Rightarrow$ $n$ linearly independent eigenvectors (by the linear independence of eigenvectors for distinct eigenvalues) $\Rightarrow$ $A$ is diagonalizable.
4. Semisimple endomorphism perspective. An endomorphism $f$ is semisimple iff $\mu_f$ is squarefree; this is Module.End.IsSemisimple in Mathlib.

Connections

• Minimal PolynomialMinimal Polynomial$$\mu_A \mid p \iff p(A) = 0$$The minimal polynomial of a matrix is the monic polynomial of least degree that annihilates it.Read more → — the squarefree condition on $\mu_A$ is precisely the distinction between the diagonal and non-diagonal Jordan forms
• Jordan Canonical FormJordan Canonical Form$$A = P J P^{-1},\quad J = \bigoplus_i J_{n_i}(\lambda_i)$$Every square matrix over an algebraically closed field is similar to a direct sum of Jordan blocks.Read more → — the Jordan form is fully diagonal iff every Jordan block has size $1$, which happens iff $\mu_A$ is squarefree
• Spectral TheoremSpectral Theorem$$A = Q \Lambda Q^T \;\text{(real symmetric case)}$$Every real symmetric matrix is orthogonally diagonalizableRead more → — for Hermitian matrices over $\mathbb{C}$, all eigenvalues are real and the spectral theorem provides an orthonormal eigenbasis, giving diagonalizability unconditionally

import Mathlib.LinearAlgebra.Semisimple
import Mathlib.LinearAlgebra.Eigenspace.Basic

/-- Eigenvectors corresponding to distinct eigenvalues are linearly independent.
    This is the key sufficiency lemma: n distinct eigenvalues give n independent
    eigenvectors and hence diagonalizability.
    Mathlib's direct alias is `Module.End.eigenvectors_linearIndependent`. -/
theorem distinct_eigenvalues_independent
    {K : Type*} [Field K]
    {V : Type*} [AddCommGroup V] [Module K V]
    (f : Module.End K V)
    (μs : Set K) (xs : μs → V)
    (h_eigenvec : ∀ μ : μs, f.HasEigenvector μ (xs μ)) :
    LinearIndependent K xs :=
  Module.End.eigenvectors_linearIndependent f μs xs h_eigenvec