# If $ABA = B$ and $BAB = A$ and $A$ is invertible, then $A^4 = I$

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 $$A^4=I$$?

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{aligned} ABA(B) & = B(B) \\ A(BAB) & = B^2 \\ A(A) & = B^2 \\ A^2 & = B^2 \end{aligned}\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{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}\tag{4}\label{eq4}