# Invariant subspace and characteristic polynomial spliting

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?

• Hi! With $p(x)$ "splits" do you mean that splits in linear factor (i. e. $p(x) = (x-a_1)...(x-a_k)$)? Commented Apr 14, 2020 at 21:59
• @Menezio yes exactly Commented Apr 14, 2020 at 22:53

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.