If a linear operator $T$ has minimal, characteristic polynomial $q^k$ for $q$ irreducible, a $T$-invariant subspace has no $T$-invariant complement. Let $T$ be a linear transformation on a finite dimensional vector space $V$ over the field $F$. Let $p_T$ be the minimal polynomial of $T$ and $f_T$ be the characteristic polynomial of $T$. If $p_T = f_T = q^k$ for some irreducible $q$ with $k &gt; 1$, show that no nonzero proper $T$-invariant subspace can have a $T$-invariant complement.


My approach so far is the following: Let $W$ be a nonzero $T$-invariant proper subspace of $V$. Suppose that $W$ has a $T$-invariant complement $W'$.

We know that since $p_T = f_T$, $T$ has a cyclic vector $\alpha$ and that $V$ has a basis $\{\alpha, T\alpha, \dotsc, T^{n-1}\alpha\}$. If $\alpha$ is in $W$ then $W' = \{0\}$ and we are done (and vice versa), so suppose $\alpha = w + w'$ where $w \in W$ and $w' \in W'$.

From here, though, I am not sure how to proceed. Is this the best approach? Is there a better one?
As per Christian's observation in a comment on the origin question:

Let $p_W$ and $p_{W'}$ be the minimal polynomials of $W$ and $T$, respectively. Since $p_T = q^k$ with $q$ irreducible ($k = \dim V$), $p_W = q^m$ and $p_{W'} = q^{m'}$.

Now, since both $W$ and $T'$ are $T$-invariant, i.e., $TW \subseteq W$ and $TW' \subseteq W'$, we must have $m &lt; k$ and $m' &lt; k'$ for otherwise $W = 0$ or $W' = 0$, contrary to our assumptions. 

Let $n = \max{m,m'}$; $q^n$ annihilates both $W$ and $W'$, so $q^n$ annihilates $W \oplus W' = V$. However, $n &lt; k$, and $q^k$ is the minimal polynomial of $W$, which is a contradiction. 

Therefore, if $W$ is a $T$-invariant subspace, then it cannot have a nontrivial $T$-invariant complementary subspace.
As per Christian's observation in a comment on the origin question:

Let $p_W$ and $p_{W'}$ be the minimal polynomials of $W$ and $T$, respectively. Since $p_T = q^k$ with $q$ irreducible ($k = \dim V$), $p_W = q^m$ and $p_{W'} = q^{m'}$.

Now, since both $W$ and $T'$ are $T$-invariant, i.e., $TW \subseteq W$ and $TW' \subseteq W'$, we must have $m &lt; k$ and $m' &lt; k'$ for otherwise $W = 0$ or $W' = 0$, contrary to our assumptions. 

Let $n = \max{m,m'}$; $q^n$ annihilates both $W$ and $W'$, so $q^n$ annihilates $W \oplus W' = V$. However, $n &lt; k$, and $q^k$ is the minimal polynomial of $W$, which is a contradiction. 

Therefore, if $W$ is a $T$-invariant subspace, then it cannot have a nontrivial $T$-invariant complementary subspace.