A matrix satisfying $AB-BA=B$

Solution 1:

The most concise answer I can think of does actually involve traces, though for contradiction. Note $$\mathrm{tr}(B^{-1}AB - A) = \mathrm{tr}I \implies 0=n$$

Solution 2:

Suppose that $B$ is invertible, then $A-1=BAB^{-1}$. Hence $A-1$ and $A$ should have the same set of eigenvalues, which is impossible.

Solution 3:

Here is a proof, not by contradiction. Let $C=A+kI$. The given assumption implies that $$CB = B(C+I).\tag{1}$$ When $k$ is sufficiently large, $\det(C)\neq\det(C+I)$ and both determinants are nonzero. Hence $(1)$ implies that $\det(B)=0$ must be zero. Alternatively, pick a (perhaps complex) number $k$ such that $C$ is singular but $C+I$ is not. Again, $(1)$ implies that $\det(B)=0$.