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An endomorphism is diagonalisable if and only if its minimal polynomial is a product of distinct linear factors

Statement

An endomorphism $T$ of a finite-dimensional vector space over $F$ is diagonalisable if and only if

$μ_{T} = \prod_{i < r} (x - λ_{i})$

for a finite list of distinct scalars $λ_{i} \in F$ . For the zero space this is the empty product $μ_{T} = 1$ .

Facts & Assumptions

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

[L1]

A polynomial annihilates $T$ exactly when it is divisible by $μ_{T}$ (The annihilator ideal is nonzero and has a unique monic generator; $p (T) = 0$ if and only if $μ_{T} ∣ p$ ).

[L2]

If $p (x) = \sum_{k \geq 0} a_{k} x^{k}$ , then $p (T) = \sum_{k \geq 0} a_{k} T^{k}$ (Polynomial evaluation at an endomorphism: $p (T) = \sum_{k} a_{k} T^{k}$ ).

[L3]

For a polynomial over a field, $p (λ) = 0$ exactly when $x - λ$ divides $p$ (Factor theorem over a commutative ring).

[L4]

Irreducible-power factors of $μ_{T}$ give the primary direct-sum decomposition (Primary decomposition: the irreducible-power factors of $μ_{T}$ split $V$ into their invariant kernels).

[L5]

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

Proof

technique · direct

1.1L5L1algebra

Suppose $T$ is diagonalisable, and let $λ_{0}, \dots, λ_{r - 1}$ be its distinct eigenvalues. On an eigenbasis the polynomial $p = \prod_{i} (x - λ_{i})$ vanishes entrywise, so $p (T) = 0$ and [L1] gives $μ_{T} ∣ p$ .

1.2L4L5

Conversely, suppose $μ_{T}$ is the displayed product of distinct linear factors. In [L4] every exponent is one, so its primary summands are $\ker (T - λ_{i} I) = E_{λ_{i}} (T)$ . Thus $V$ is their direct sum, and [L5] makes $T$ diagonalisable.

2.1step 1.1L1L2L3choosealgebra

For each $i$ , choose a nonzero eigenvector $v_{i}$ for $λ_{i}$ . Formula [L2] and induction give $μ_{T} (T) v_{i} = μ_{T} (λ_{i}) v_{i}$ ; the left side is zero by [L1], so $μ_{T} (λ_{i}) = 0$ . By [L3], every distinct factor $x - λ_{i}$ of $p$ divides $μ_{T}$ . Together with step 1.1 and monicity, this gives $μ_{T} = p$ .

3.1step 2.1step 1.2L1∎

Steps 2.1 and 1.2 prove both directions. When $V = 0$ , the empty basis gives $μ_{T} = 1$ by [L1].

Depends on

Used by

Dependency tree · two levels

23 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 4.11 (standard reference, not scraped)
Anthony W. Knapp, Basic Algebra, 2nd ed., Ch. V, §3, Theorem 5.14 (standard reference, not scraped)