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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

1.1

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 $μ_{T} ∣ p$ .

L5L1algebra

1.2

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.

L4L5

2.1

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 $μ_{T} = p$ .

step 1.1L1L2L3choosealgebra

3.1

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

step 2.1step 1.2L1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 76 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)