Prove that $AB = 0$ implies $\det\left(A + B\right)^2 = \det\left(A - B\right)^2$. Given matrices $A, B \in \mathcal{M}_{n \times n}\left(\mathbb{R}\right)$, prove that if $AB = 0$, $\det\left(A + B\right)^2 = \det\left(A - B\right)^2$.
This is based on a quiz question from a few weeks ago that I never quite solved, and I forgot my instructors' solutions as well.
Here are possible ideas I had:


*

*Show that $AB = BA$ (except it's not true).

*Show that $BA = -BA$—wait, isn't this equivalent to (1)?

*Show that $\forall i,j\, \left(A_{ij} = 0\right) \mathrm{or} \left(B_{ij}=0\right)$ so that $\det\left(A + B\right) = \left(-1\right)^k\det\left(A - B\right)$ for some $k$. This is definitely a sufficient condition, but I'm not sure if it's necessary.


We have learned diagonalization, but we have not learned the Jordan canonical form yet, so it would be ideal if you avoided that kind of answer.
 A: $\renewcommand\im{\operatorname{im}}$Consider a basis of $\mathbb{R}^n$ formed by taking a basis of the image $\operatorname{im}(B)$ and extending to a basis of $\mathbb{R}^n$. Then $\im(B) \subseteq \ker(A)$, so in this basis, the matrices of $A$ and $B$ are block matrices
$$
A = \begin{bmatrix}
0 & A_1 \\ 0 & A_2
\end{bmatrix},\quad
B = \begin{bmatrix}
B_1 & B_2 \\ 0 & 0
\end{bmatrix},
$$
where $B_1$ is $r \times r$ (where $r= \operatorname{rank}(B)$), $A_1$ and $B_2$ are $r \times (n-r)$, and $A_2$ is $(n-r) \times (n-r)$.
Why? If $\{v_1,\cdots v_n\}$ is a basis of $\mathbb{R}^n$ with $\{v_1,\cdots,v_r\}$ a basis of $\im(B)$, then $Bv_i$ is in $\im(B)$ for all $i$, so the last $n-r$ rows of $B$ are zero. Also, $Av_i = 0$ for $i \le r$ since $v_i \in \im(B)$, so the first $r$ columns of $A$ are zero.
Then
$$
A \pm B = \begin{bmatrix}
\pm B_1 & A_1 \pm B_2 \\ 0 & A_2
\end{bmatrix}
$$
has determinant $\pm \det(B_1) \det(A_2)$ (it's block triangular), so the squares have the same determinant.
