Show that no non-trivial $T$ -invariant subspace has a complementary $T$ -invariant subspace

Question

Let $T$ be a linear operator on a finite-dimensional vector space $V$. Suppose that
(a) the minimal polynomial for $T$ is a power of an irreducible polynomial;
(b) the minimal polynomial is equal to the characteristic polynomial.
Show that no non-trivial $T$ -invariant subspace has a complementary $T$ -invariant subspace.

Solution:
$T$ has a cyclic vector.

We first prove that every $T$ -invariant
subspace is $T$ -cyclic. Suppose, to get a contradiction, that the restriction of $T$ to a $T$ -invariant subspace of dimension $d$ was annihilated by a polynomial $p$ with degree less than $d$. Then the image $W$ of $p$ would be a $T$ -invariant subspace of dimension at most $\dim (V) − d$ (by Rank-Nullity Theorem) and annihilated by some polynomial $q$ with degree at most $\dim (W)$, such as the characteristic polynomial for $T$ restricted to $W$. Then $qp$ would annihilate $T$ because $p$ maps $V$ into $W$ which is annihilated by $q$.

But $\deg (qp) = \deg (q) + \deg (p) < \dim (W) + d \leq \dim (V) – d + d = \dim (V) $, which
contradicts that the minimal polynomial for $T$ must equal its characteristic polynomial, which has degree $\dim (V)$.

Hence, the minimal polynomial of the $T$-invariant subspace has degree $d$. But that minimal polynomial also divides
the characteristic polynomial for the $T$-invariant subspace and so must equal the characteristic polynomial. Hence, the $T$-invariant subspace is $T$-cyclic.

Now suppose, to get a contradiction, that $V = W_1 \oplus W_2$ where $W_1$, $W_2$ are non-trivial $T$-invariant subspaces. By the previous paragraph, $W_1$, $W_2$ are also $T$-cyclic, so the minimal polynomials for $T$ restricted to $W_1$, $W_2$ have degrees
equal to $\dim (W_1)$, $\dim (W_2)$, which are each less than $\dim (V)$ because $W_1$, $W_2$ are non-trivial. Let $s^m$ be the minimal
polynomial for $T$ where $s$ is irreducible. Then because the minimal polynomials for $T$ restricted to $W_1$, $W_2$ divides
$s^m$, they are $s^j$ and $s^k$ where j and k are each less than m. The minimal polynomial for $T$ is the least common multiple of $s^j$ and $s^k$, which is $s^{\max(j,k)}$, but that contradicts that the minimal polynomial is $s^m$.
Thus, no non-trivial $T$-invariant subspace has a complementary $T$-invariant subspace.

This is the proof which I have written and it would be great to have some feedback on it.

Answer 1

If $U = V\oplus W$ with $V,W$ invariant (and non-zero) then $charpoly(T)=charpoly(T|_V)charpoly(T|_W)$ and $minpoly(T)=lcm(minpoly(T|_V),minpoly(T|_W))$

So $charpoly(T)=minpoly(T)$ implies that $minpoly(T|_V),minpoly(T|_W)$ are coprime contradicting that $minpoly(T)$ is the power of an irreducible polynomial.