Question

how to prove matrix AB=BA if A^2=B^2=identity

Answer 1

Can we assume:
\[
A^{-1}A=AA^{-1}=e
\]Or shall I prove that?

Answer 2

It's commutativity of the inverse and associativity of the operation. Not a terrible proof.

Answer 3

There's an even simpler proof:

We wish to show
\[
AB=BA
\]Note that:
\[
A=A^{-1}
\]And:
\[B=B^{-1}\]

We have:
\[
AB(AB)^{-1}=AB(A^{-1}B^{-1})=A(BA^{-1})B^{-1}=e
\]Then:
\[
A^{-1}A(BA^{-1})B^{-1}=A^{-1}=(BA^{-1})B^{-1}
\]And:
\[
(BA^{-1})B^{-1}B=A^{-1}B=BA^{-1}e=BA^{-1}
\]So, we have:
\[
A^{-1}B=AB=BA^{-1}=BA
\]Which is what we wished to prove.