The following stumps me:
Let $\mathbb K$ be a field. Let $A, B \in \mathbb K^{n \times n}$ where $B$ is nilpotent and commutes with $A$, i.e., $A B = B A$. Show that $$ \det(A+B)=\det(A) $$
I have no idea how to approach this I thought perhaps raise both sides to a power but nothing works. Thanks for all help.
Here's a way to do it without the machinery of JNF or triangulization, although that machinery is absolutely worth learning. You really only need to know that $\det(M) = 0$ if and only if there exists $x\ne 0$ such that $Mx = 0$ (for an $n\times n$ matrix $M$), that $\det$ is multiplicative, and that $\det(M)$ is the product of the roots (in an algebraic closure) of the characteristic polynomial of $M$.
Let $m>0$ be such that $B^m = 0$. Since $A$ and $B$ commute, we may expand as follows \begin{equation}\tag{$*$}(A+B)^m = \sum_{i=0}^m \binom{m}{i} B^{m-i} A^{i}.\end{equation}
Suppose that $\det(A) = 0$, and let $v\ne 0$ be such that $Av = 0$. Observe that $(A+B)^m(v) = 0$, since each term in the expansion $(*)$ kills $v$: for $0< i \le m$, we have $B^{m-i}(A^i(v)) = B^{m-i}A^{i-1}(Av) = 0$, and for $i=0$, we have $B^m(v) = 0$ since $B^m = 0$. Therefore $0\ne v\in \ker (A+B)^m$, so $(\det(A+B))^m = \det((A+B)^m) = 0$, so $\det(A+B)=0=\det(A)$.
Now suppose that $\det(A) \ne 0$. Let $C = A^{-1}B$. It suffices to show that $\det(I+C) = 1$. Let $\lambda \in \bar K$ be a root of the characteristic polynomial of $I+C$, so that $\det((1-\lambda) I + C) = \det(I + C - \lambda I) = 0$. Since $A$ commutes with $B$, so does $A^{-1}$, and thus $-C = -A^{-1}B$ is nilpotent. Now since $-C$ is nilpotent and commutes with $(1-\lambda)I + C$, which has determinant $0$, by the above paragraph we have $$(1-\lambda)^n = \det((1-\lambda)I) = \det(((1-\lambda)I+C)+(-C)) = \det((1-\lambda)I+C) = 0,$$ so $\lambda = 1$. Therefore, the characteristic polynomial of $I+C$ has only $1$ as a root, and thus $\det(I+C) = 1$.