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 $$\det(A+B)=\det B.$$

We shall consider the following two cases:
Case I: $\det B=0$. Then, there exists a vector $u\ne 0$, such that $Bu=0$, and as $AB=BA$, then
$$
(A+B)^ku=\sum_{i=0}^k\binom{k}{i}A^{k-i}B^ju=A^ku+\sum_{i=1}^k\binom{k}{i}A^{k-i}B^iu=0,
$$
and hence $A+B$ is not invertible and thus $\det(A+B)=\det B=0$.
Case II: $\det B\ne 0$. Then it suffices to show that $\det (I+B^{-1}A)=\det I$. Note that $A$ commutes with $B^{-1}$ as well and $(B^{-1}A)^k=(B^{-1})^kA^k=0$. So it suffices to show that if $C^k=0$, then $\det (I+C)=\det I$. Let $F=I+C$. We know that $$(F-I)^k=C^k=0.$$ This means that ALL the eigenvalues of $F$ are equal to $1$, as $F$ is annihilated by the polynomial $p(x)=(x-1)^k$, and hence its determinant (product of its eigenvalues) is equal to $1$, i.e., $$\det (I+C)=\det F=1.$$
