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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

Used by

Dependency tree · two levels

17 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

Anthony W. Knapp, Basic Algebra, 2nd ed., Ch. V, §5, Theorem 5.19 (standard reference, not scraped)

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