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Linear Algebra and its Applications
Found in: Page 331
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Given \(A = QR\) as in Theorem 12, describe how to find an orthogonal\(m \times m\)(square) matrix \({Q_1}\) and an invertible \(n \times n\) upper triangular matrix \(R\) such that

\(A = {Q_1}\left[ {\begin{aligned}{{}{}}R\\0\end{aligned}} \right]\)

The MATLAB qr command supplies this “full” QR factorization

when rank \(A = n\).

We find the square matrix \({Q_1}\) by extending the column vectors of \(Q\) to the orthonormal basis of \({\mathbb{R}^m}\).

See the step by step solution

Step by Step Solution

Step 1: \(QR\) factorization of a Matrix

A matrix with order \(m \times n\) can be written as the multiplication of an upper triangular matrix \(R\) and a matrix \(Q\) which is formed by applying the Gram–Schmidt orthogonalization process to the \({\rm{col}}\left( A \right)\).

The matrix \(R\) can be found by the formula \({Q^T}A = R\).

Step 2: Finding the matrix \(R\) and \({Q_1}\)

Let \({q_1},{q_2}, \ldots ,{q_n}\) be the columns of the matrix \(Q\). Here \(n \le m\), since \(A\) is a \(m \times n\) matrix and it has linearly independent columns.

The columns of \(Q\) can be extended to an orthogonal basis \({\mathbb{R}^m}\).

Let \({v_1}\) be the 1st vector of the standard basis of \({\mathbb{R}^m}\) which is not in the set\({W_n} = {\rm{span}}\left\{ {{q_1},......,{q_n}} \right\}\).

Let \({u_1} = {v_1} - {\rm{pro}}{{\rm{j}}_{{W_n}}}{v_1}\) and let \({q_{n + 1}} = \frac{{{u_1}}}{{\left\| {{u_1}} \right\|}}\). Then \(\left\{ {{q_1},......,{q_n},{q_{n + 1}}} \right\}\) is an orthogonal basis for \(\)\({W_{n + 1}} = {\rm{span}}\left\{ {{q_1},......,{q_n},{q_{n + 1}}} \right\}\)

Now, again \({v_2}\)be the 1st vector of the standard basis of \({\mathbb{R}^m}\) which is not in the set \({W_{n + 1}} = {\rm{span}}\left\{ {{q_1},......,{q_n},{q_{n + 1}}} \right\}\).

Let \({u_2} = {v_2} - {\rm{pro}}{{\rm{j}}_{{W_{n + 1}}}}{v_2}\) and let \({q_{n + 2}} = \frac{{{u_2}}}{{\left\| {{u_2}} \right\|}}\). Then \(\left\{ {{q_1},......,{q_n},{q_{n + 1}},{q_{n + 2}}} \right\}\)is an orthogonal basis for\({W_{n + 2}} = {\rm{span}}\left\{ {{q_1},......,{q_n},{q_{n + 1}},{q_{n + 2}}} \right\}\).

Applying this process to get \(m - n\) such vectors, we get an orthogonal basis

\(\left\{ {{q_1},......,{q_n},{q_{n + 1}},...,{q_m}} \right\}\) for \({\mathbb{R}^m}\).

Let \({Q_0} = \left[ {\begin{aligned}{{}{r}}{{q_{n + 1}}}&.&.&.&.&{{q_m}}\end{aligned}} \right]\) and then let \({Q_1} = \left[ {\begin{aligned}{{}{}}Q&{{Q_0}}\end{aligned}} \right]\).


\(\begin{aligned}{}{Q_1}\left[ {\begin{aligned}{{}{}}R\\0\end{aligned}} \right] &= \left[ {\begin{aligned}{{}{}}Q&{{Q_0}}\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}R\\0\end{aligned}} \right]\\ &= QR\\ &= A\end{aligned}\)

In this way, we can find \({Q_1}\) and the upper triangular matrix.

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