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Let $A$ and $B$ be square matrices of the same order so that $ABA = B$ and $BAB = A$. If $A$ is invertible, prove that $A^4 = I$.

I already proved that $A^2=B^2$. How can I prove $A^4=I$?

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You have that $A$ and $B$ are square matrices of the same order with

$$ABA = B \tag{1}\label{eq1}$$

$$BAB = A \tag{2}\label{eq2}$$

You're asking that, if $A$ is invertible, to then prove that $A^4 = I$.

You've already proven that $A^2 = B^2$, but I am doing it here as well, just in case it's a different method than yours or if somebody reading this doesn't know how to do it. With \eqref{eq1}, multiply on the right by $B$, then use associativity of matrix multiplication, and finally \eqref{eq2}, to get

$$\begin{equation}\begin{aligned} ABA(B) & = B(B) \\ A(BAB) & = B^2 \\ A(A) & = B^2 \\ A^2 & = B^2 \end{aligned}\end{equation}\tag{3}\label{eq3}$$

With \eqref{eq2}, left & right multiply by $B$, then use \eqref{eq3}, the associativity of matrix multiplication, and finally the invertibility of $A$ to multiply on the right by $A^{-1}$, to get

$$\begin{equation}\begin{aligned} B(BAB)B & = B(A)B \\ (BB)A(BB) & = A \\ (AA)A(AA) & = A \\ (A^4)AA^{-1} & = AA^{-1} \\ A^4 & = I \end{aligned}\end{equation}\tag{4}\label{eq4}$$

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