TITLE: A product on the square roots of unit matrix
QUESTION [11 upvotes]: There is a strange product that takes two square roots of unit matrix, say $A$ and  $B$, $A^2=I$, $B^2=I$ to a square root again, 
$$ A\star B=(A+B)^{-1}(A-B+2I), \qquad (A\star B)^2=I$$
Could anybody help me with identifying this structure? Where it comes from?
It was obtained from the Caley transform $C(A)=(1-A)^{-1}(1+A)$, $C(C(A))=A$, by expanding $C(C(A),C(B))$ and using $A^2=B^2=I$. Ones we imposed $A^2=B^2=I$ the inverse transform does not exists anymore, so we are at the singular point of the Caley transform. Somehow it looks like adding some points at infinity and extending the action on them.

REPLY [8 votes]: Let me denote $E_\pm(M)$ the eigenspaces of $M$ associated with the eigenvalues $\pm1$. Let me assume that the characteristic of the scalar field $k$ is not $2$. Then your assumption is that 
$$k^n=E_+(A)\oplus E_-(A)=E_+(B)\oplus E_-(B).$$
In addition, the assumption that $A+B$ is non-singular means $E_+(A)\cap E_-(B)=(0)$ and $E_+(B)\cap E_-(A)=(0)$, from which it follows that $\dim E_+(A)=\dim E_+(B)$.
Then one verifies easily
$$E_+(A\star B)=E_+(B),\qquad E_-(A\star B)=E_-(A).$$
Because 
 the dimensions of $E_-(A)$ and  $E_+(B)$ sum up to $n$, one deduces 
$$k^n= E_+(A\star B)\oplus E_-(A\star B),$$
and therefore $(A\star B)^2= I_n$.