A matrix, $Q \in \mathbb{M}_{mn}$, s.t. $QQ^T = I_m$ and $Q^TQ = I_n$?

Question

Of course, where $m \neq n$. Is this possible?

Answer 1

Consider $m < n$ without loss of generality. What is the maximum dimension of the range of $Q$? Can applying $Q^T$ to the range of $Q$ give you $n$ linearly independent vectors?

Answer 2

That would mean $A:=Q^T$ is both a left and right inverse of $Q.$

Note that rank $(B \cdot Q) \leq$ rank $Q$ for every $k$ by $m$ matrix $B.$ Then prove that since $Q$ has a left inverse $A=Q^T,$ then rank $Q = n.$ Similarly, since $Q$ has a right inverse $A,$ then rank $Q = m$ and thus $m = n.$