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?
Thank you in advance.
 A: 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.
A: The answer by Makoto Kato really captures the essence of the answer best, but you can also do this without using a field extension. The condition that the minimal polynomial $p_T$ be irreducible is a rather strong condition: it means that for every nonzero vector$~v$ the minimal degree monic polynomial$~P$ such that $P[T](v)=0$ is equal to$~p_T$ (in general monic divisors of$~p_T$ are candidates, but here the only such proper divisor is$~1$, but it only annihilates the zero vector). In particular $v,T(v),\ldots,T^{d-1}(v)$ are always linearly independent, where $d=\deg(p_T)$. Consequently, the only $T$-invariant subspaces of the span$~S$ of these $d~$vectors are the zero space and$~S$ itself: this is because for every nonzero vector$~v'$, its repeated images $v',T(v'),\ldots,T^{d-1}(v')$ are linearly independent and therefore necessarily span$~S$. 
Now a fundamental property is that in $V$, and by the same argument in any $T$-stable subspace of it (because the restriction of$~T$ to it has the same minimal polynomial, unless it has dimension$~0$), one can find a set of vectors $v_1,\ldots,v_k$ such that the $dk$ vectors $v_1,T(v_1),\ldots,T^{d-1}(v_1)$, $v_2,\ldots,T^{d-1}(v_2)$, $\ldots,v_k,T(v_k),\ldots,T^{d-1}(v_k)$ form a basis. In fact they can be chosen fairly freely, with the choice of $v_i$ being restricted only by being outside the span $S_{<i}$ the vectors before it in the list, as in the incomplete basis theorem. (In fact this is the incomplete basis theorem with respect to the $K$-vector space structure in the other answer I referred to.) To see this, it suffices to observe that, with $S_i$ being the span of the repeated images $v_i,T(v_i),\ldots,T^{d-1}(v_i)$, one has $S_{<i}\cap S_i=\{0\}$ (as this is a $T$-stable subspace of $S_i$ that does not contain $v_i$), whence one has a direct sum $S_{\leq i}=S_{<i}\oplus S_i$. The process ends when $S_{\leq i}=V$ in which case one sets $k=i$. (Note that as a consequence of this we obtain the information that $\dim(V)=kd$ is necessarily a multiple of$~d=\deg(p_T)$.)
Now for the result of the question: choose such a set of vectors $v_1,\ldots,v_k$ for the $T$-stable subspace$~W$, and complete to such a set of vectors $v_1,\ldots,v_l$ for the whole space; then it is easy to see that the vectors $v_{k+1},T(v),\ldots,T^{d-1}(v_{k+1})$, $\ldots,v_l,T(v_l),\ldots,T^{d-1}(v_l)$ are the basis of a $T$-stable complement of$~W$.
