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?
Best Answer
Yes, you did it right. Formally:
Named $p(t)$ the characteristic polynomial of $T$ and $p_W(t)$ che characteristic polynomial of $T_{W}$. By hypotesis we have $p(t)=(t-a_1)...(t-a_n)$ (eventually with $a_i=a_j$ for some $i,j$). Thanks to the theorem you mentioned above we have $$ p_W(t) \ | \ p(t) $$ And this implies that $p_W(t)$ is just a product of some factors of $p(t)$; hence $p_W(t)=(t-a_{i_1})...(t-a_{i_k})$ for some indices $i_1,...,i_k$ where $k=\dim W$.
The second part follows directly: since $k>0$, we have an eingenvalue $a_{i_1}$ of $T_W$ and so an eingenvector relative to it.