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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:

  1. $\det(A^2+B^2)+\det(A^2-B^2)=2(\det A^2+\det B^2)$. (Is this well known?)
  2. (1) $\implies \det(A^2-B^2)=0$.
  3. If $A,B\in M_2(\mathbb{C})$ satisfy $A^2+B^2+2AB=0$, then $AB=BA$.
  4. $(A+B)^2=0 \implies \det(A^2-B^2)=0$.

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), $\det(A^2-B^2)= 2(\det(A^2)-2\det(AB)+\det(B^2)) = 2((\det(A)^2-\det(B)^2) = 0$.

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Of course, the OP did not understand much about the proposed exercise. In particular, in $M_2(\mathbb{C})$, $A^2 +B^2+2AB=0$ implies $\det(A)=\det(B)$.

In fact, the interesting result is

Proposition. Let $A,B\in M_2(\mathbb{C})$ be such that $(1)$ $A^2+B^2+2AB=0$; then $AB=BA$. Note that the result works only in dimension $2$.

Proof. If $H=A+B$, then $H^2=HA-AH$ and $trace(H^2)=0$. Moreover $H^3=H^2A-HAH=HAH-AH^2$ implies that $trace(H^3)=0$. We deduce that the eigenvalues of $H$ are $0$, $H$ is nilpotent, $H^2=0$, $AH=HA$, $AB=BA$. $\square$

Remark. According to Proposition, $(1)$ implies that $(A+B)^2=0$, $\det(A+B)=0$, $\det(A^2-B^2)=0$; moreover $AB=BA$ implies that $A,B$ are simultaneously triangularizable; since $A+B$ is nilpotent, we may assume that $A=\begin{pmatrix}a&c\\0&b\end{pmatrix},B=\begin{pmatrix}-a&d\\0&-b\end{pmatrix}$ and, consequently, $\det(A)=\det(B)$.

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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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