If $W$ is $T$-invariant space of $V$ and $\left\{0\right\}\ne W\subseteq V$ then $\exists\ w\in W$ such that $w$ is an eigenvector of $T$

Question

Assume $V$ is a finite dimensional vector space over $\mathbb{C}$. and $T:V\to V$ is a linear transformation.

Show that if $W$ is $T$-invariant space of $V$ and $\left\{0\right\}\ne W\subseteq V$ then $\exists\ w\in W$ such that $w$ is an eigenvector of $T$.

If $W=V$ then the correctness of the theorem is trivial since $V$ is a trivial $T$-invariant subspace of $V$.

I tried to show this by induction on the dimension of $W$ in the case when $W\subset V$.

If $\dim W=1$ and $\left\{0\right\}\ne W\subset V$ and $\operatorname{span}\{w\}= W$ then it's also trivial to show that $w$ is an eigenvector of $T$.

From here I need guidance please for the induction step or for another direction for the proof. Thanks.

Answer 1

Let $p$ be the characteristic polynomial of $T |_W$. Since your vector spaces are assumed to be complex, you can always find a root $c$ of $p$, which is an eigenvalue of $T |_W$. Hence, there is an eigenvector $w \in W$ of $T |_W$ with eigenvalue $c$. But, $T(w) = T|_W(w) = cw$. Hence, proved.

Answer 2

I assume you are considering vector spaces over an algebraically closed field $k$? If not, the statement is not correct.

Let $S:U\rightarrow U$ be any endomorphism of a finite-dimensional vector space $U$ over $k$. Then $\det(S-X\cdot Id)$ is a polynomial over $k$ and hence has a root $\lambda$. It follows that the linear system $SY=\lambda Y$ has a non-zero solution. Such a non-zero solution is an eigenvector of $S$.

Now consider the original map $T:V\rightarrow V$ and consider the restriction $T_{\mid W}:W\rightarrow W$. Since $W$ is $T$-invariant, this restriction is well-defined. Applying the above to $T_{\mid W}$, we conclude that $T_{\mid W}$ has an eigenvector.

Answer 3

You apparently already dealt with the case $W=V$ (though I don't quite understand what you write after "since"; I would say, "since any linear operator on a complex space of finite nonzero dimension has eigenvectors"). Then it suffices to consider the restriction of$~T$ to$~W$, which can be viewed as a linear operator on$~W$ since $W$ is $T$-stable; then for this restriction you are in the situation already dealt with. (And an eigenvector for the restriction $T|_W$ is obviously an eigenvector for$~T$ lying inside$~W$.)