@TedShifrin, suppose $A: X \rightarrow X$ is an $n \times k$ (suppose $n > k$), orthogonal matrix, where $X$ is an (arbitrary) $n-$dimensional Euclidean space. Then $A^T A = I_k$. Now the (half-assed) claim is (by a very dubious source) that $A A^T = I$ where $I$ is an identity element on X, *and may not even be a matrix*. This is an interesting claim, especially the latter part, since the product of two matrices somehow gave me an identity element in an entirely brand new ring. I claim
$(AA^T)^2 = A(A^TA)A^T = A \cdot I_k \cdot A^T = AA^T$. That is, $AA^T$ is it's own right/left identity. …