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Primary decomposition: the irreducible-power factors of $μ_{T}$ split $V$ into their invariant kernels

Statement

Let $T : V \to V$ be an endomorphism of a finite-dimensional vector space over $F$ , and factor its minimal polynomial in the UFD $F [x]$ as

$μ_{T} = \prod_{i < r} q_{i}^{e_{i}},$

where the $q_{i}$ are distinct monic irreducibles and $e_{i} \geq 1$ . Then the subspaces $V_{i} : = \ker q_{i} (T)^{e_{i}}$ are $T$ -invariant and

$V = ⨁_{i < r} V_{i} .$

Moreover, the minimal polynomial of $T ∣_{V_{i}}$ is exactly $q_{i}^{e_{i}}$ . If $V = 0$ , then $r = 0$ and this is the empty direct sum.

Facts & Assumptions

Given: A finite-dimensional endomorphism $T$ and the displayed irreducible factorisation of $μ_{T}$ .

[L1]

Every polynomial ring over a field is a unique factorisation domain (For every field $F$ , $F [x]$ is a unique factorisation domain).

[L2]

If coprime $f, g$ satisfy $(f g) (T) = 0$ , then $V = \ker f (T) \oplus \ker g (T)$ (If $\gcd (f, g) = 1$ and $(f g) (T) = 0$ , then $V = \ker f (T) \oplus \ker g (T)$ ).

[L3]

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

Proof

technique · direct

1.1L1L2L3

By [L1], the distinct powers $f_{i} = q_{i}^{e_{i}}$ are pairwise coprime. Since $(\prod_{i} f_{i}) (T) = μ_{T} (T) = 0$ , repeated application of [L2] gives $V = ⨁_{i < r} \ker f_{i} (T)$ .

2.1step 1.1L3algebra

Each summand is $T$ -invariant because $T$ commutes with $f_{i} (T)$ . The restriction $T_{i} = T ∣_{V_{i}}$ is annihilated by $f_{i}$ , so [L3] gives $μ_{T_{i}} ∣ f_{i}$ .

3.1step 1.1step 2.1algebra

Suppose for some $i$ that $μ_{T_{i}}$ is a proper divisor of $f_{i}$ . Put $h = μ_{T_{i}} \prod_{j \neq i} f_{j}$ . On $V_{i}$ , the first factor annihilates; on $V_{j}$ with $j \neq i$ , the factor $f_{j}$ annihilates. Step 1.1 therefore gives $h (T) = 0$ .

4.1step 3.1L3∎

The polynomial $h$ has smaller degree than $μ_{T}$ and so cannot be divisible by $μ_{T}$ , contradicting [L3]. Hence $μ_{T_{i}} = f_{i}$ by monicity. If $r = 0$ , then $μ_{T} = 1$ , [L3] gives $I_{V} = 0$ , and $V = 0$ , which is precisely the empty direct sum.

Depends on

Dependency tree · 2 levels

Sources

Anthony W. Knapp, Basic Algebra, 2nd ed., Ch. V, §5, Theorem 5.19

Primary decomposition: the irreducible-power factors of μTμT​ split VV into their invariant kernels