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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

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    The map is surjective if and only if it is injective, by finite dimensionality of $M_n(\mathbb{C})$. Maybe this helps. – 2017-01-21

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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.

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    did you use the simultaneous diagonalisability? – 2017-01-21
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    or you used a translation to eigenvalues of the matrix $A$ and $B$ ? – 2017-01-21
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    anyway ,by using my method I found that the necessar and sufficient condition : $A$ and $B$ must have nonzero coefficients out of the diagonal and the elements in the diagonal of $A$ and $B$ have to be different . – 2017-01-21
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    @Seginus that's incorrect. For example, take $A=2I$ and $B=I$. – 2017-01-21
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    can you give your solution please – 2017-01-21
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    Are you aware that commuting operators are simultaneously triangularizable? Also, did you try to come up with a proof? Any promising ideas? – 2017-01-21
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    here's my idea : I took tha canonical basis of $M_n(C)$ and I tried to find a sufficient and necessary condition such that the image of the basis is a basis (which means an independent set ) – 2017-01-21
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    I understand your idea you found that the intersection of the spectra of $A$ and $B$ is emty ? – 2017-01-22
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    I think mine works too :) – 2017-01-22
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    But yours is not equivalent to mine. The example I gave shows that your conditions aren't necessary (but maybe they're sufficient) – 2017-01-22
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    See my latest edit. – 2017-01-22