Suggested languages for you:

Americas

Europe

Q21E

Expert-verifiedFound in: Page 331

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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 \({\mathbb{R}^m}\).

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 \({Q^T}A = R\).

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 1^{st} 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 \({Q_1} = \left[ {\begin{aligned}{{}{}}Q&{{Q_0}}\end{aligned}} \right]\).

Then

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

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

94% of StudySmarter users get better grades.

Sign up for free