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

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.

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 $$S_{ the vectors before it in the list, as in the incomplete basis theorem. (In fact this is the incomplete basis theorem with respect to the $$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_{ (as this is a $$T$$-stable subspace of $$S_i$$ that does not contain $$v_i$$), whence one has a direct sum $$S_{\leq i}=S_{. The process ends when $$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$$~W$$.