Determinant invariance by multiplication Is it true that if $ m\ge n $, $ U\in M_{mn} $ such that $ U^TU=\mathrm{id}_n $, then
$$
\det\left(U^TAU\right)=\det\left(A\right)
$$
for any $ A\in M_{mm} $?
(With $ M_{mn} $ I am denoting the space of real valued matrices with $ m $ rows and $ n $ columns.)
Notice that in the case $ m=n $ the statement is trivially true.

I don't think so, $U=(1, 0)$, $A=e_{11}$, then $U^{T}U=U^TAU=1$ but $\det{A}=0$, right?
