Let T be a linear operator on a finite-dimensional vector space V. Prove that if the characteristic polynomial of T splits, then so does the characteristic polynomial of the restriction of T to any T-invariant subspace of V.
Theorem: Let T be a linear operator on a finite-dimensional vector space V, and let W be a T-invariant subspace of V. Then the characteristic polynomial of $T_W$ divides the characteristic polynomial of T.
Can I use this theorem to argue since $T_W$ is a factor of the polynomial of T, so it splits?
Let $T$ be a linear operator on a finite-dimensional vector space $V$.
Deduce that if the characteristic polynomial of $T$ splits, then any non-trivial $T$-invariant subspace of $V$ contains an eigenvector of $T.$
Let $W$ be a $T$-Invariant subspace. $W\neq\{0\}$($\because$ Given that $W$ is non-trivial). The characteristic polynomial of $T$ restricted to $W$ divides the characteristic polynomial of $T$. Then since nontrivial, there exists an eigenvalue for $det(W_1-tI)=0$ for every $W_1 \in T_{|W}$, hence it has at least one eigenvector.
Is this reasoning correct?