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I can't find the solution of this problem:

Given two $n\times n$ square matrices $A,B$ such that $A^2\cdot B^2=I_n$, show that $B^{-1}\cdot A^{-1}=B\cdot A$.

Thanks in advance.

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    $\begingroup$ If $A^2 B^2 =I$ then $B^2 A^2 = I$. $\endgroup$ – copper.hat Jan 8 '15 at 16:44
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Since $A^{2}B^{2}=I_{n}$ note that taking $A^{-1}=ABB$ we have that $A\cdot A^{-1}=I_n$. For square matrices this is enough to show $A$ is invertible. The same follows for $B$ by taking $B^{-1}=AAB$. Using the facts that $A^{-1}\cdot A=I_{n}$ and $B\cdot B^{-1}=I_{n}$ we get that \begin{eqnarray*} A^{2}B^{2}=I_{n} & \iff & AB^{2}=A^{-1}I_{n},\mbox{ by left multiplication}\\ & \iff & AB=A^{-1}I_{n}B^{-1},\,\mbox{ by right multiplication}\\ & \iff & AB=A^{-1}B^{-1}. \end{eqnarray*} Taking inverses of both sides and using the property $(AB)^{-1}=B^{-1}A^{-1}$ gives the desired result.

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  • $\begingroup$ Can you clarify? $\endgroup$ – Phanu9000 Jan 8 '15 at 16:54
  • $\begingroup$ Well I should have really said, take inverses of both sides, and then use socks-shoes property? $\endgroup$ – Phanu9000 Jan 8 '15 at 16:57
  • $\begingroup$ I can't judge that. I don't know what is called socks-shoes property. $\endgroup$ – Pp.. Jan 8 '15 at 16:58
  • $\begingroup$ The usual property of inverses: $(AB)^{-1}=B^{-1}A^{-1}$ $\endgroup$ – Phanu9000 Jan 8 '15 at 17:00
  • $\begingroup$ Cool name! Yes, that does make it. And actually it makes it a nicer answer than the other. Just add also the observation that $A^2B^2=I$ implies $A,B$ are invertible. $\endgroup$ – Pp.. Jan 8 '15 at 17:05
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$A^2 \cdot B^2 = I \Rightarrow A, B$ are invertible. The set of all invertible matrices forms a group. So we also have $B^2 \cdot A^2 = I.$ Hence $B \cdot A = B^{-1}\cdot A^{-1}.$

I'm assuming that the entries of the matrices are from a field.

EDIT: $A^2 \cdot B^2 = I \Rightarrow \text{det}(A^2) \cdot \text{det}(B^2) = \text{det}(A^2 \cdot B^2) = 1 \Rightarrow \text{det}(A^2) \neq 0.$ Now $(\text{det}(A))^2 = \text{det}(A^2) \Rightarrow \text{det}(A) \neq 0.$ Similarly for $B.$

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  • $\begingroup$ Thank you for your answer. Can you please explain why $A^{2}⋅B^{2}=I\Rightarrow A,B$ are invertible? $\endgroup$ – igil Jan 8 '15 at 16:57
  • $\begingroup$ ok! thanks for the remark! $\endgroup$ – igil Jan 8 '15 at 17:06
  • $\begingroup$ @ioll: you are welcome. :) $\endgroup$ – Krish Jan 8 '15 at 17:07

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