Let $A$ be a square real matrix. We know that if $A$ is orthogonal matrix then $A^TA=I$. Consequently also in this case $(A^TA)^n=I$.
I would like to know whether it is possible to have expression of type $B=A^TA$ (I would call it transquare of $A$ - btw it is a little strange that such important expression, it seems, has no own name..) when $A$ is not orthogonal matrix, but for some natural $n$ equality $B^n=I$ is satisfied.
- Is it possible?
Of course $B$ is full rank matrix, but on the other hand - in general - the equation $B^n=I$ can have even infinite number of solutions. Could one of them have decomposition $A^TA$ without $A$ being orthogonal?
- Can the possible answer (probably negative) be also extended for the case of $ m \times n$ matrices ?