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 ≤ 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 $W1$, $W2$ 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 sm 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.
