$T$ is a linear operator on a finite dimensional vector space $V$, and $W$ be a $T$-invariant subspace of $V$. Define $\bar T: V/W\to V/W$ by $\bar T(v+W)=T(v)+W$. It can be proved that $\bar T$ is linear. Prove that if both $T_W$ and $\bar T$ are diagonalizable and have no common eigenvalues, then $T$ is diagonalizable.
I don't know how to use the assumption "no common eigenvalues". I don't know whether eigenvalues of $\bar T$ is also eigenvalues of $T$. My attempt: suppose $T(v)+W=\bar T(v+W)=\lambda(v+W)=\lambda v+W$, then $T(v)-\lambda v \in W$. Can we show that this is the zero vector?
By diagonalizability of $\overline{T}: V/W\to V/W$, we can find a basis of eigenvectors $\{v_1+W, \cdots, v_k+W\}$ for $\overline{T}$. Let $\{v_{k+1}, \cdots, v_n\}$ be the basis of eigenvectors of $T_W: W\to W$.
I claim that there exist $v_1', \cdots, v_k'$ such that they form the basis of eigenvectors of $T$ together with $v_{k+1}, \cdots, v_n$.
Note that $\overline{T}(v_i+W)=\lambda_iv_i+W$, and hence $T(v_i)-\lambda_iv_i\in W$. Consider $T_W-\lambda_iI_W:W\to W$. By assumption $\lambda_i$ is not an eigenvalue of $T_W$ and thus it is invertible. So there exists $w_i\in W$ such that $T_W(w_i)-\lambda_iw_i=\lambda_iv_i-T(v_i)\in W$. Let $v_i'=v_i+w_i$. Consequently $v_i'$ is an eigenvector of $T$ with eigenvalue $\lambda_i$. It is easy to show that $\{v_1', \cdots, v_k', v_{k+1}, \cdots, v_n\}$ form a basis of $V$.