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.

  • 2
    $\begingroup$ If $A^2 B^2 =I$ then $B^2 A^2 = I$. $\endgroup$ – copper.hat Jan 8 '15 at 16:44

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.

  • $\begingroup$ 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

$A^2 \cdot B^2 = I \Rightarrow A, B$ 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 $B.$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.