# Irreducible minimal polynomial implies every invariant subspace has an invariant complement

Full version of the problem is following:

Let T be a linear transformation on a finite dimensional vector space $V$ over a field $\mathbb{F}$. If the minimal polynomial $p_t$ of T is irreducible, then every T invariant subspace $W$ has a T-invariant complement $W'$

I used Cyclic decomposition Theorem which states that

"The finite dimensional vector space V can be expressed a s a decomposition of T-cyclic subspaces $Z(\alpha_1;T)\oplus Z(\alpha_2;T)\oplus...\oplus Z(\alpha_k;T)$ and their annihilators $p_1,...p_k$ have properties; (1) $p_k|...|p_1$, (2)$p_T=p_1, f_T=p_1\cdot p_2\cdot ...\cdot p_k$ where $f_T$ is the characteristic polynomial of T."

Can I say this? Since $p_T$ is irreducible, there is a cyclic vector $\alpha$ such that $V=Z(\alpha;T)$ and V=W and $W'=\{0\}$. Therefore $W'$ for each W is T-invariant.

Is my way correct?

Certainly the $W=V$ step doesn't make any sense. If, say, $W$ had 5 dimensions, then it would have many $T$ invariant subspaces of dimension 1,2,3,4... are those all $V$ as well? Please also consider explaining some of your notation. –  rschwieb Aug 12 '12 at 0:47
Let $\mathbb{F}[X]$ be the polynomial ring with one variable. $V$ can be regarded as an $\mathbb{F}[X]$-module by defining $Xv = T(v)$ for every $v \in V$. $\mathbb{F}[X]$-submodules of $V$ are none other than $T$-invariant subspaces of $V$. Let $K = \mathbb{F}[X]/(p_t)$. Since $p_t$ is irreducible, $K$ is a field. Since $p_t V = 0$, $V$ can be regarded as a $K$-module. Let $W$ be a $T$-invariant subspace of $V$. $W$ can be regarded as a $\mathbb{F}[X]$-submodule of $V$. Since $p_t W = 0$, $W$ can be regarded as a $K$-submodule of $V$. Hence there exists a $K$-submodule $W'$ such that $V = W \oplus W'$. Since $W'$ is a $\mathbb{F}[X]$-submodule, it is $T$-invariant. This completes the proof.