Question

24. (QR Factorization) Suppose \[A = QR\], where Qand R are \[n \times n\], Ris invertible and upper triangular, and Q has the property that \[{Q^T}{\bf{Q}} = I\]. Show that for each b in \[{\mathbb{R}^n}\], the equation \[Ax = b\] has a unique solution. What computations with Q and R will produce the solution?

Short answer

The matrices Q and R are both invertible, so A is invertible. Hence, \[Ax = b\] has a unique solution for each b in \[{\mathbb{R}^n}\]. Moreover, solutionxis obtained by the row reduction of \[\left[ {\begin{array}{*{20}{c}}R&{{Q^T}b}\end{array}} \right]\].

Step-by-step solution

Describe the given data

Given, \[A = QR\] with R is invertible and upper triangular, and \[{Q^T}Q = I\].

Use the fact of an invertible matrix

Here, \[{Q^T}Q = I\]. This implies that Q is invertible and \[{Q^{ - 1}} = {Q^T}\].

Hence, Q and R are both invertible. Therefore, QR is invertible, and A is also invertible. This implies that \[Ax = b\] has aunique solution for each b in \[{\mathbb{R}^n}\].

Use row reduction

Substitute \[A = QR\] in \[Ax = b\] to get:

\[\begin{array}{c}QRx = b\\{Q^T}QRx = {Q^T}b\\IRx = {Q^T}b\\Rx = {Q^T}b\end{array}\]

This implies thatxis obtained by the row reduction of \[\left[ {\begin{array}{*{20}{c}}R&{{Q^T}b}\end{array}} \right]\]. This reduction is fast since R is upper triangular.