sufficient and necessar condition and surjectivity Let $A$ and $B$ be two matrix from $\mathbb M_n(\mathbb C)$ Give a necessar and sufficient condition on $(A,B)$ such that the linear map : $X\to AX-XB$ from $\mathbb M_n(\mathbb C)$ to $\mathbb M_n(\mathbb C)$is surjective

**Hint:** The maps $X\mapsto AX$ and $X\mapsto XB$ commute. Consider their eigenvalues.

**Answer:** Your map is surjective if and only if $A$ and $B$ have no common eigenvalues.

* * *

**My solution:** Because $X\mapsto AX$ and $X\mapsto XB$ commute, they are simultaneously diagonalizable. Note that the eigenvalues of $X\mapsto AX$ coincide with the eigenvalues of $A$, with possibly a different multiplicity. Similarly, the eigenvalues of $X\mapsto XB$ are the same as those of $B$. Conclude that if $A$ and $B$ have disjoint spectra, their difference can't have $0$ as an eigenvalue

On the other hand: if $u$ and $v$ are eigenvectors associated with the same $\lambda$ for $A$ and $B^T$ respectively, then $uv^T$ is a non-zero element of the kernel.