Suppose you have the QR decomposition of a square matrix $A$ (of full rank) such that $A = QR$ where $Q$ is an orthogonal matrix and $R$ is upper triangular. Is there an efficient way to get a QR decomposition of the transpose of $A$?


IE, given $A = QR$ find some orthogonal matrix $\tilde{Q}$ and some upper triangular matrix $\tilde{R}$ such that $A^\top = \tilde{Q} \tilde{R}$?



Translate between QR decomposition of A and A transpose
No, the two are not obviously related. Transposing everything gives an LQ decomposition, which clearly is not the same, and as far as I know there is no simple trick to convert one into the other.


If you want a decomposition that is "robust by transposition" and can be used to solve least squares problems and identify ranges / nullspaces / etc., consider the singular value decomposition (SVD). It is more expensive than QR, but it has the same order of complexity (and it is generally more accurate in rank determination).


