<s>[INST] What can be said if $A^2+B^2+2AB=0$ for some real $2\times2$ matrices $A$ and $B$? Let $A,B\in M_2(\mathbb{R})$ be such that $A^2+B^2+2AB=0$ and $\det A= \det B$. Our goal is to compute $\det(A^2 - B^2)$.  According to the chain of comments on Art of Problem Solving, the following statements are true:


*

*$\det(A^2+B^2)+\det(A^2-B^2)=2(\det A^2+\det B^2)$. (Is this well known?)

*(1) $\implies \det(A^2-B^2)=0$.

*If $A,B\in M_2(\mathbb{C})$ satisfy $A^2+B^2+2AB=0$, then $AB=BA$.

*$(A+B)^2=0 \implies \det(A^2-B^2)=0$.


Can someone help me with justifying these statements?
Edit:
Doug M provided an explanation for (1) in the answers. Here is an explanation for (2):
$A^2+B^2+2AB=O_2 \implies A^2+B^2=-2AB$. So $\det(A^2+B^2)=4\det(AB)$. Now using (1), $\det(A^2-B^2)= 2(\det(A^2)-2\det(AB)+\det(B^2)) = 2((\det(A)^2-\det(B)^2) = 0$.
 [/INST] If $A = \pmatrix {a&b\\c&d}, B = \pmatrix {e&f\\g&h}$
$\det (A+B) + \det (A-B)\\(a+e)(c+h) - (b+f)(c+g) + (a-e)(c-h) - (b-f)(c-g)\\ 2ac+2eh - 2bc-2fg\\2(\det A + \det B)$
$\det (A+B) + \det (A-B) = 2(\det A + \det B)$
In which case the initial premise indeed holds.
I am pretty sure this only holds in the $2\times2$ case.
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