Diagonalizabilty of a restriction implies diagonalizability of original operator I'm stuck trying to solve this problem. 
V is finite dimensional vector space over the complex numbers. Suppose $L: V \longrightarrow V$ is linear, and furthermore suppose$U \subset V$ subspace such that $U \subset \ker(L)$ and $im(L)\cap U=\{0\}$. Show that if the induced linear map $L': V/U \longrightarrow V/U $ is diagonalizable, then L is diagonalizable. 
First I have tried to consider $L'$ and its basis consisting of eigenvectors such that it is diagonal, also it's minimal polynomial $f(t)=(t-\lambda_1)\cdots(t-\lambda_k)$. Then I can write $L$ in this basis as a block matrix where
L in the given basis is
\begin{bmatrix}
    L' & B &  \\
   0 & C &  \\
\end{bmatrix}
Still, I am not sure how to proceed. Do I need to show that $U$ is invariant under $L$? Thank you
This is not the same as the problem where you are given a diagonal operator and trying to show that its restriction is diagonal.

If $L'$ is diagonalizable then its minimal polynomial $P(t)$ is a product of distinct linear factors, and for all $v\in V$ we have $(P(L))(v)\in U$. Distinguish two cases:


*

*If $P(0)\neq0$ then $tP(t)$ is a product of distinct linear factors, and $L(P(L))(v)=0$ for all $v\in V$ because $(P(L))(v)\in U\subset\ker L$, so the minimal polynomial of $L$ divides $tP(t)$.

*If $P(0)=0$ then $(P(L))(v)\in\operatorname{im}(L)$ for all $v\in V$. From $\operatorname{im}(L)\cap U=0$ it follows that $P(L)=0$, so the minimal polynomial of $L$ divides $P(t)$.
