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

Let $U$ be the generalized eigenspace of $B$ with respect to a particular eigenvalue $\lambda$. Since $A B = B A$, we have $A U \subseteq U$. 
(If $v \in U$, so that $(B - \lambda I)^{s} v = 0$ for some $s$ and vector $v$, then $(B - \lambda I)^{s} A v = A (B - \lambda I)^{s} v = 0$, that is, $A v \in U$.)
Now on $U$ we will have $B = \lambda I + N$, where $N$ is nilpotent, and $I$ is the identity on $U$. From $A B = BA$ we get $A N = N A$, so $N+A$ is nilpotent. (If $A^{t} = N^{t} = 0$, then $(A+N)^{2t-1} = 0$.)
Now on $U$ we have $A + B = \lambda I + (A + N)$. So we see that $U$ is also the generalized eigenspace of $A + B$ with respect to the eigenvalue $\lambda$. (This is because $(A + B - \lambda I)^{2t-1} = (A + N)^{2t-1} = 0$ for the same $t$ as above. Also, we are actually showing that the $\lambda$-generalized eigenspace of $B$ is contained in the $\lambda$-generalized eigenspace of $A+B$, but the situation is symmetric, as $B = (A+B) - A$, with $A (A+B) = (A+B) A$.)
So $B$ and $A + B$ have the same eigenvalues with the same multiplicities, hence they have the same determinant.
This actually shows that they have the same characteristic polynomial. But they need not have the same minimal polynomial, as the example $A = \begin{bmatrix}0&1\\0&0\end{bmatrix}$, $B = \begin{bmatrix}1&0\\0&1\end{bmatrix}$ shows.
