This is not for homework, and I would just a like a hint please. The question asks
Prove or give a counterexample: If $U$ is a subspace of $V$ that is invariant under every operator on $V$, then $U = \{ 0 \}$ or $U = V$.
In the question, $V$ is finite dimensional. My gut feeling is that this claim is true, so I tried to begin a proof. I began by assuming $U \neq \{ 0 \}$. I want to try to exploit the fact that $U$ is invariant under every operator to try and force every basis vector of $V$ into $U$. I would like just a hint please!
Best Answer
Hint: assume $U\ne\{0\}$ and $U\ne V$; then if $\{u_1,\dots,u_k\}$ is a basis for $U$, you can extend it to a basis $\{u_1,\dots,u_k,u_{k+1},\dots,u_n\}$ of $V$. You can define a linear map $V\to V$ by telling the action on a basis. Can you make $f$ that shows $U$ is not $f$-invariant?
Note: you don't need finite dimensionality, but of course the proof in the non finite dimensional case requires the axiom of choice.
Details (added after accept)
We know that $0<\dim U=k<\dim V=n$, so in the argument we do have $u_1$ and $u_{k+1}$. Thus the linear map $f\colon V\to V$ defined by $$ f(u_i)=\begin{cases} u_{k+1} & \text{if $i=1$}\\ 0 & \text{if $1<i<n$} \end{cases} $$ shows $U$ is not $f$-invariant.
The same idea can be used in the general case: take a non-zero vector $u\in U$ and extend it to a basis $B$ of $U$; now take $v\notin U$ and extend $B\cup\{v\}$ to a basis $C$ of $V$ (this is possible using Zorn's lemma). Then define the linear map $f\colon V\to V$ by $f(u)=v$ and $f(w)=0$ for $w\in C$, $w\ne u$.
Another possibility is to define $f(u_i)=u_{i+1}$ for $1\le i<n$ and $f(u_n)=u_1$. Since $f(u_k)=u_{k+1}\notin U$, $U$ is not $f$-invariant.