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An endomorphism is diagonalisable if and only if its characteristic polynomial splits and every eigenvalue's geometric multiplicity equals its algebraic multiplicity

Statement

Let $T : V \to V$ be an endomorphism of a finite-dimensional vector space over $F$ . Then $T$ is diagonalisable if and only if $χ_{T}$ splits over $F$ and, for every eigenvalue $λ$ ,

$\dim E_{λ} (T) = {mult}_{χ_{T}} (λ) .$

The assertion includes the zero-dimensional case, where $χ_{T} = 1$ and the multiplicity condition is vacuous.

Facts & Assumptions

Given: A finite-dimensional endomorphism $T : V \to V$ .

[L1]

The algebraic multiplicity is the exponent of $x - λ$ in $χ_{T}$ , and the geometric multiplicity is $\dim E_{λ} (T)$ (Algebraic multiplicity as the exponent of $x - λ$ in $χ_{T}$ , and geometric multiplicity as $\dim E_{λ} (T)$ ).

[L2]

An endomorphism is diagonalisable exactly when its distinct eigenspaces have direct sum $V$ (An endomorphism is diagonalisable exactly when $V = ⨁_{i < r} E_{λ_{i}} (T)$ for some finite list of distinct scalars $λ_{i}$ ).

[L3]

Eigenvectors belonging to distinct eigenvalues are linearly independent (Eigenvectors belonging to pairwise distinct eigenvalues are linearly independent).

[L4]

A split polynomial is a product of linear factors, with repetitions allowed (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

[L5]

In any ordered basis, $χ_{T}$ is the characteristic polynomial of the representing matrix, independently of the chosen basis (The basis-independent characteristic polynomial $χ_{T}$ of an endomorphism of a finite-dimensional space, including $χ_{T} = 1$ in dimension zero).

[L6]

The characteristic polynomial of an endomorphism on an $n$ -dimensional space is monic of degree $n$ , and it is $1$ for $n = 0$ ( $χ_{A} (x)$ is monic of degree $n$ ; for $n \geq 1$ its $x^{n - 1}$ coefficient is $- tr (A)$ and its constant coefficient is $(- 1)^{n} \det (A)$ , while $χ_{0 \times 0} = 1$ ).

[L7]

For $n \geq 1$ , the determinant of a triangular matrix in $M_{n} (R)$ over a commutative ring is the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries).

[L8]

For an endomorphism of a finite-dimensional space, $σ_{F} (T) = {λ \in F : χ_{T} (λ) = 0}$ (For every finite-dimensional space, $σ_{F} (T)$ is exactly the set of roots in $F$ of $χ_{T}$ ).

Proof

technique · direct

1.1L1L2L4L5L7choose

Suppose $T$ is diagonalisable and $\dim V \geq 1$ , and choose an eigenbasis. Its matrix is diagonal, so $x I - [T]$ is triangular of positive size and [L5] and [L7] show that $χ_{T}$ is the product of $x - λ$ over the diagonal entries. For each $λ$ , the basis vectors carrying that diagonal entry form a basis of $E_{λ} (T)$ ; hence their number is both multiplicities in [L1].

1.2L1L4L6L8algebra

Conversely, suppose $χ_{T}$ splits and the multiplicities agree. By [L8] the roots of $χ_{T}$ in $F$ are exactly the eigenvalues of $T$ , so by [L4] the sum of the algebraic multiplicities of the eigenvalues is the degree of $χ_{T}$ , and [L6] identifies that degree with $\dim V$ . Hence the sum of the dimensions of the distinct eigenspaces is $\dim V$ .

2.1step 1.2L2L3choose

Choose a basis of each eigenspace. By [L3] their concatenation is independent, and step 1.2 says it has $\dim V$ vectors, so it is a basis. Therefore [L2] makes $T$ diagonalisable.

3.1step 1.1step 2.1L2L4L6∎

Steps 1.1 and 2.1 prove the equivalence when $\dim V \geq 1$ . If $\dim V = 0$ , the empty basis is an eigenbasis, so $T$ is diagonalisable by [L2]; [L6] gives $χ_{T} = 1$ , which splits as an empty product by [L4]; and $T$ has no eigenvalue, so the multiplicity condition holds vacuously. Both sides therefore hold, which proves the equivalence in every finite dimension.

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Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., §§8A–8B (standard reference, not scraped)