TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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A family of diagonalisable endomorphisms of a finite-dimensional space is simultaneously diagonalisable if and only if its members commute pairwise

Statement

Let $T$ be a family of diagonalisable endomorphisms of a finite-dimensional vector space. Then $T$ is simultaneously diagonalisable if and only if its members commute pairwise.

Facts & Assumptions

Given: A family $T$ of diagonalisable endomorphisms of a finite-dimensional space $V$ .

[L1]

Commuting endomorphisms preserve each other's eigenspaces (Commuting endomorphisms preserve each other's eigenspaces).

[L2]

The restriction of a diagonalisable endomorphism to an invariant subspace is diagonalisable (The restriction of a diagonalisable endomorphism to an invariant subspace is diagonalisable).

[L3]

The space $L (V, V)$ is finite-dimensional, with dimension $(\dim V)^{2}$ ( $\dim_{F} M_{m \times n} (F) = m n$ and $\dim_{F} L (V, W) = (\dim_{F} V) (\dim_{F} W)$ for finite-dimensional $V, W$ ).

[L4]

$span (S)$ is the intersection of all subspaces containing $S$ , hence the smallest such subspace (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ); and $span (S) = L (S)$ , the set of finite linear combinations of elements of $S$ ( $span (S)$ is exactly the set of linear combinations of finite lists of elements of $S$ , and $span (\emptyset) = {0_{V}}$ ).

[L5]

A diagonalisable endomorphism's distinct eigenspaces have direct sum equal to the whole space (An endomorphism is diagonalisable exactly when $V = ⨁_{i < r} E_{λ_{i}} (T)$ for some finite list of distinct scalars $λ_{i}$ ).

[L6]

Simultaneous diagonalisability means that one basis diagonalises every member of the family (Simultaneous diagonalisability: one basis that diagonalises every endomorphism in a family).

Proof

technique · direct

1.1L1L2L5choose

First suppose $T$ is finite and pairwise commuting. Induct on its size. For the empty family any basis works. For a nonempty family, choose one member $T$ . Its eigenspaces have direct sum $V$ by [L5]; by [L1] every remaining member preserves each eigenspace, and by [L2] every restriction is diagonalisable. Applying the induction hypothesis within each eigenspace and concatenating the resulting bases gives one common eigenbasis.

1.2L3L4choose

For an arbitrary pairwise commuting family, [L3] makes $U = span (T)$ finite-dimensional. Choose a finite basis of $U$ ; by [L4], each basis vector is a finite linear combination of members of $T$ . The union of the finitely many supports is a finite subfamily $T_{0}$ spanning $U$ .

1.3L6givenalgebra

Conversely, if one basis diagonalises every member of $T$ as in [L6], then every pair is represented by diagonal matrices, which commute. The represented endomorphisms therefore commute.

2.1step 1.1step 1.2L6

Step 1.1 gives a common eigenbasis for $T_{0}$ . Every member of $T$ lies in its span, so it is represented by a linear combination of diagonal matrices in that basis and is diagonal too. Hence [L6] makes $T$ simultaneously diagonalisable.

3.1step 2.1step 1.3∎

Steps 2.1 and 1.3 prove both directions, including empty families and the zero space.

Depends on

Used by

Dominant characters form the representation-ring basis Corollary
If two commuting endomorphisms are diagonalisable, then every finite linear combination of products of their powers is diagonalisable; in particular, their sum and product are diagonalisable Corollary
Commuting alone does not imply simultaneous diagonalisability Counterexample
Roots of a compact connected Lie group Definition
Two commuting non-scalar matrices simultaneously diagonalised in one explicit basis Example
Finite-dimensional modules decompose into weight spaces Proposition
Cayley transforms connect theta-stable Cartans in the classification Theorem
Classification of real forms by Vogan diagrams Theorem
Compact roots form a reduced crystallographic root system Theorem
Conjugacy of Cartan subalgebras Theorem
Conjugacy of maximal tori Theorem
Existence of Cartan subalgebras Theorem
Maximal abelian subspaces of p are conjugate by K Theorem
Restricted root space decomposition Theorem
Root-space decomposition Theorem

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Keith Conrad, The Minimal Polynomial and Some Applications, Theorem 5.2 and Corollary 5.4 (standard reference, not scraped)