Translate between QR decomposition of A and A transpose 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 $A^\top = \tilde{Q} \tilde{R}$?
 A: 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).
A: There is a way you can use a QR decomposition for a Matrix A to solxe a linear system using A transpose (A').
Ax = b
Recognizing that:
A'=(QR)'=R'Q'
We can write R'Q'x=b
inv(R') is something you can calculate by hand so you can multiply both sides by inv(R') on the left to get:
inv(R')R'Q'x=inv(R')b -> Q'x = inv(R')b
Then, recognizing that, since Q is orthogonal, Q'=inv(Q)
So QQ'=Qinv(Q)=I
Now, multiply both sides by Q on the left and you get:
QQ'x=Qinv(R')b
x=Qinv(R')b
The right hand side can all be calculated with matrix multiplication and you have your answer for x.
Cheers.
