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Let $A,B$ be two $n\times n$ matrices with real entries, where $n$ is odd, such that $A\cdot A^{t}=I_n$ and $B\cdot B^{t}=I_n$. Prove that $$\det(A+B)\det(A-B)=0$$ It is obvious that $A^{-1}=A^{t}$ and $B^{-1}=B^{t}$, so $\det A, \det B = \pm 1$. Then I tried to write $\det(A+B)\det(A-B)=\det(A^2-AB+BA-B^2)$, but I didn't get anything useful.

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$$\begin{aligned}\det(A+B)\det(A-B)&=\det(A+B)\det(A^T-B^T) \\ &= \det(AA^T+BA^T-AB^T-BB^T)\\&=\det(BA^T-AB^T)\end{aligned}$$

Now use the assumption that $n$ is odd and the fact that $C=BA^T-AB^T$ is skew-symmetric: $C=-C^T$.

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    $\begingroup$ Very nice argument! Note that it uses only $AA^T = BB^T $, rather than the orthogonality of the two matrices. $\endgroup$ – darij grinberg Oct 30 '17 at 19:03
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We know that... $(A+B)(A-B)=A^2-AB+BA-B^2$ So $det[(A+B)(A-B)]=d(A^2-AB+BA-B^2)≠0 $ in general. Which is possible when $A,B$ be the involuntary matrix. i.e $A=A^{-1},B=B^{-1}$,then $A$ and $B$ are commutative i.e $AB=BA$ and $A^2=I_n,B^2=I_n$. Hence in this case... $det[(A+B)(A-B)]=det(A+B)det(A-B)=det(I_n-AB+AB-I_n)=0$. If I mistake please tell me.

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  • $\begingroup$ We don't have $A=A^{-1}$, but $A^{t}=A^{-1}$ $\endgroup$ – Shroud Oct 30 '17 at 20:14

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