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

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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).

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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.

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  • $\begingroup$ 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. $\endgroup$ – Mageek Nov 16 '19 at 22:09

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