# 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 $$A^\top = \tilde{Q} \tilde{R}$$?

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).

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.

• Thank you for your comment. I needed a new QR decomposition of A', not a solution for A'x = b necessarily. Q inv(R') is not a new QR decomposition. – Mageek Nov 16 '19 at 22:09