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Q21E

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Found in: Page 331

### 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

# 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 thatA = {Q_1}\left[ {\begin{aligned}{{}{}}R\\0\end{aligned}} \right]The MATLAB qr command supplies this “full” QR factorizationwhen 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 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].

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.