Left inverse of a matrix with an extra condition

Question

Let $A, B\in\mathbb{R}^{m\times n}$ be matrices with $m>n$ and suppose that $A$ and $B$ have orthonormal columns. Is there a matrix $C\in\mathbb{R}^{n\times m}$ with orthonormal columns such that

$C^T A = I_n = identity$

and

$C^T B\in\mathbb{R}^{n\times n}$

is orthogonal?

Is there a way of numerically constructing the matrix $C$?

Answer 1

I'm presuming $C$ in the question should be $m \times n$, for else the products $C^TA$ and $C^TB$ would not be defined... So then, if this is the case, any $C$ with orthonormal columns will satisfy the second condition, as $C^TB(C^TB)^T=(C^TB)^TC^TB=I$. And the first condition is satisfied by $C=A$. So the answer is yes, there is such a $C$, the matrix $A$...presuming $A$ was given, there is no need to numerically determine $C$.