If $\,A^k=0$ and $AB=BA$, then $\,\det(A+B)=\det B$ Assume that the matrices $A,\: B\in \mathbb{R}^{n\times n}$ satisfy 
$$
A^k=0,\,\, \text{for some $\,k\in \mathbb{Z^+}$}\quad\text{and}\quad
AB=BA.
$$
Prove that $$\det(A+B)=\det B.$$

Here's an alternative to the (admittedly excellent) answer by Andreas:
Since $A$ and $B$ commute, they are simulaneously triangularizable. However, since $A$ is nilpotent, for any basis in which $A$ is triangular, the diagonal entries of $A$ will in fact be $0$. Thus in the chosen basis, the diagonal entries of $A + B$ and $B$ agree, so the determinants (being the product of the diagonal entries in this basis) are equal.
