$U$ is invariant under all $T\in\mathcal{L}(V)$ I have written the following Proof but i am unsure about whether it is valid could you please verify that the argument is right
NOTE: $\mathcal{L}(V)$ denotes the set of all linear transfomations from $V$ to $V$ and $I_n = \{j\in\mathbf{Z^+}|j\leq n\}$
Theorem. If V is finite-dimensional and $U$ is a
subspace of $V$ that is invariant under every operator on $V$, then $U = \{0\}$ or $U=V$.
Proof. Assume on the contrary that $U$ is invariant under all $T\in\mathcal{L}(V)$ and that $U\neq \{0\}$ and $U\neq V$ which implies that $1\leq\dim U=n< \dim V = m$, furthermore since $U$ is a subspace of $V$ we may invoke a basis namely $u_1,u_2,...,u_n$ for it which could be extended to a basis for $V$ namely
$$u_1,u_2...,u_n,u_{n+1},u_{n+2},...,u_m$$
Now consider the Linear operator $T:V\to V$ defined such that $Tu_j = u_{m-j+1}$ for $1\leq j\leq m$. Let $u\in U$  such  that $u = \alpha_1u_1+\alpha_2u_2+...+\alpha_nu_n$ where $\forall j\in I_n(\alpha_j\neq 0)$, consequently by applying $T$ to $u$ we have
$$Tu = \alpha_1Tu_1+\alpha_2T_2+...+\alpha_nTu_n = \alpha_1u_m+\alpha_2u_{m-2}+...+\alpha_{m-n+1} = w$$ but $w\not\in U$ because with the above choice of $u$, $\alpha_mu_m\not\in U$ for every possible dimension of $U$  contradicting our original assumption of $U$ being invariant under all $T\in\mathcal{L}(V)$.
 A: I'd say your proof is good, but you're doing too much work.
Suppose $U\ne\{0\}$ and $U\ne V$. Then there are $u\in U$, $u\ne0$ and $v\in V$, $v\notin U$.
Extend $u$ to a basis $\{u=u_1,\dots,u_n\}$ of $V$. Then there exists $T\in\mathcal{L}(V)$ such that $T(u_1)=v$ and $T(u_i)=0$ for $i=2,\dots,n$. Since $v\notin U$, but $u=u_1\in U$, we see that $U$ is not $T$-invariant.
(The idea is similar to yours, actually, but the proof is much simpler.)
A: Here is another way of proving it, but without using a contradiction. Let $T$ be an operator on $V$. Suppose $U \neq \{0\}$ and let $u \in U$ be nonzero. If $(v_1, v_2, ...., v_n)$ is a basis for V, then we can find $a_1, a_2, ..., a_n \in \mathbb{C} \text{ (or } \mathbb{R} \text{)}$, such that $u = a_1v_1 + .... + a_nv_n$. Since $u$ is nonzero, there must be a nonzero $a_j$ as well. 
Let $v_0$ be any vector in $V$. Then we can define $S_{j,v_0}$ on $V$ by
$$S_{j,v_0}(b_1v_1 + .... + b_nv_n) = \frac{b_j}{a_j}v_0,$$
where $\frac{1}{a_j}$ exists because $a_j \neq 0$. Notice that this is a linear map and it also is clearly an operator on $V$.
Now, by our hypothesis, $U$ is invariant under $S_{j,v_0}$. Thus, since $u \in U$, it follows that $S_{j,v_0}u \in U$. But $S_{j,v_0} = v_0$ and so $v_0 \in U$. Since this process can be repeated for any $v_0 \in V$, it follows that $V \subset U$ and so $U = V$ as desired.
