Let $V$ be a finite dimensional vector space over $\mathbb{C}$ and $L: V \rightarrow V$ be a linear operator. Suppose that $U$ is a subspace of $V$ such that $U \subset \ker L$ and $\text{im}(L) \cap U = \{0 \}$. Show that if the induced linear map $\overline{L} : V/U \rightarrow V/U$ is diagonalizable, then $L$ is diagonalizable. Find a counterexample to the statement in case $\text{im}(L) \cap U \neq \{0 \}$.
This is from a qualifying exam I took an hour ago. So I am more interested in the critique of my argument although I will also appreciate a correct alternative solution.
Here is how I argued:
Note that $L$ is diagonalizable if and only if $V$ has a basis of eigenvectors of $L$. Since $\overline{L}$ is diagonalizable, $V/U$ has a basis of eigenvectors of $\overline{L}$, say $\mathfrak{B} = \{v_{1} + U, ..., v_{k} + U \}$. Since $\mathfrak{B}$ is a basis, $\{v_{1},..., v_{k}\}$ is linearly independent. We extend it to a basis of $V$, say $\mathfrak{C} = \{v_{1},..., v_{k}, v_{k+1},..., v_{n}\}$.
Since $\mathfrak{B}$ is a basis, $v_{j} + U = U$ for all $j > k$ so that $v_{j} \in U \implies v_{j} \in \ker L$. So that $Lv_{j} = 0$ for all $j >k$. Also, since $v_{j} \notin U$ for all $j = 1, ..., k$, $v_{j} \notin \ker L$. We have $\lambda_{j}(v_{j} + U) = \overline{L}(v_{j} +U) = L(v_{j}) + U \implies \lambda_{j}v_{j} - L(v_{j}) \in U$ and since $\lambda_{j}v_{j} \in \text{im}(L)$, $\lambda_{j}v_{j}- Lv_{j} \in U \cap \text{im}(L) = \{0 \} \implies Lv_{j} = \lambda_{j}v_{j}$. Hence, $\mathfrak{C}$, is a basis of $V$ consisting of eigenvectors of $V$.
I could not come up with a counterexample after playing with a few two by two matrices and worrying about moving on to other problems I gave up, but my idea was to take $U = \ker L$ and find $L$ such that $\text{rank} L = \text{rank} L^{2}$.
Your argument is on the right track. Note, however, that it is not the case that $v \notin U \implies v \notin \ker L$. You already know that $v_j$ is an eigenvector for $j\leq k$, so it's moot.
As for a counterexample: take $$ L = \pmatrix{0&1\\0&0} $$ and confirm that with $U = \ker L$, we have $\bar L = 0$. Up to similarity, this is the only $2 \times 2$ example.