Exercise 5.A.6

Proof. The statement is true. We will give a proof of the contrapositive.

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$ such that $U$ is neither $\{0\}$ or $V$. Let $u\in U$ and $v\in V$ such that $u$ is non-zero and $v\notin U$. Choose any operator $T\in L(V)$ that maps $u$ to $v$. Clearly $U$ is not invariant under $T$. □