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Q19E

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Found in: Page 224

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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# Using paper and pencil, find the QR factorization of the matrices in Exercises 15 through 28. Compare with Exercises 1 through 14.19.$\left[\begin{array}{cc}2& 1\\ 2& 1\\ 1& 5\end{array}\right]$

The QR factorization of the matrix is $\left[\begin{array}{cc}2& 1\\ 2& 1\\ 1& 5\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}2& -1/\sqrt{2}\\ 2& -1/\sqrt{2}\\ 1& 4/\sqrt{2}\end{array}\right]\left[\begin{array}{cc}3& 0\\ 3& \sqrt[3]{2}\end{array}\right]$.

See the step by step solution

## Step 1: Determine column u→1 and entries r11 of R

Consider the matrix $M=\left[\begin{array}{cc}2& 1\\ 2& 1\\ 1& 5\end{array}\right]$ where ${\stackrel{\to }{v}}_{1}=\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$ and ${\stackrel{\to }{v}}_{2}=\left[\begin{array}{c}1\\ 1\\ 5\end{array}\right]$

By the theorem of QR method, the value of ${\stackrel{\to }{u}}_{1}$ and ${r}_{11}$ is defined as follows:

${r}_{11}=\left|\left|\stackrel{\to }{{v}_{1}}\right|\right|\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{u}}_{1}=\frac{1}{{r}_{11}}{\stackrel{\to }{v}}_{1}$.

Simplify the equation ${r}_{11}=||{\stackrel{\to }{v}}_{1}||$ as follows:

${r}_{11}=||{\stackrel{\to }{v}}_{1}||\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{r}_{11}=\left|\left|\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]\right|\right|\phantom{\rule{0ex}{0ex}}{r}_{11}=\sqrt{{2}^{2}+{2}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}{r}_{11}=3$

Substitute the values 7 for ${r}_{11}$ and $\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$ for ${\stackrel{\to }{v}}_{1}$ in the equation ${\stackrel{\to }{u}}_{1}=\frac{1}{{r}_{11}}{\stackrel{\to }{v}}_{1}$ as follows:

${\stackrel{\to }{u}}_{1}=\frac{1}{{r}_{11}}{\stackrel{\to }{v}}_{1}\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{u}}_{1}=\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{u}}_{1}=\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right]$

Therefore, the values localid="1659437134701" ${\stackrel{\to }{u}}_{1}=\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right]\mathrm{and}{\mathrm{r}}_{11}=3$ .

## Step 2: Determine column  v→2⊥ and entries r12 of R

As ${r}_{12}=\stackrel{\to }{{u}_{1}}·\stackrel{\to }{{v}_{2}}$, substitute the values $\left[\begin{array}{c}1\\ 1\\ 5\end{array}\right]$ for $\stackrel{\to }{{V}_{2}}$ and ${\left[\begin{array}{ccc}\frac{2}{3}& \frac{2}{3}& \frac{1}{3}\end{array}\right]}^{T}$ for $\stackrel{\to }{{u}_{1}}$ in the equation

localid="1659437594586" ${r}_{11}=\stackrel{\to }{{u}_{1}}.\stackrel{\to }{{V}_{2}}$ as follows:

${r}_{11}=\stackrel{\to }{{u}_{1}}.\stackrel{\to }{{V}_{2}}$

${r}_{12}={\left[\begin{array}{ccc}\frac{2}{3}& \frac{2}{3}& \frac{1}{3}\end{array}\right]}^{T}.\left[\begin{array}{c}1\\ 1\\ 5\end{array}\right]$

${r}_{12}=\frac{2}{3}+\frac{2}{3}+\frac{5}{3}\phantom{\rule{0ex}{0ex}}{r}_{12}=3$

Substitute the values $\left[\begin{array}{c}1\\ 1\\ 5\end{array}\right]$ for$\stackrel{\to }{{V}_{2}}$ ,3 for ${r}_{12}$ and $\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right]$ for $\stackrel{\to }{{u}_{1}}$ in the equation ${\stackrel{\to }{v}}_{2}^{\perp }=\stackrel{\to }{{V}_{2}}-{r}_{12}{\stackrel{\to }{u}}_{1}$as follows:

${\stackrel{\to }{v}}_{2}^{\perp }=\stackrel{\to }{{V}_{2}}-{r}_{12}{\stackrel{\to }{u}}_{1}$

${\stackrel{\to }{v}}_{2}^{\perp }=\left[\begin{array}{c}1\\ 1\\ 5\end{array}\right]-3\left[\begin{array}{c}2/2\\ 2/3\\ 1/3\end{array}\right]\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{v}}_{2}^{\perp }=\left[\begin{array}{c}1\\ 1\\ 5\end{array}\right]-\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{v}}_{2}^{\perp }=\left[\begin{array}{c}-1\\ -1\\ 4\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},{\stackrel{\to }{v}}_{2}^{\perp }=\left[\begin{array}{c}-1\\ -1\\ 4\end{array}\right]\mathrm{and}{r}_{12}=3$

## Step 3: Determine column u→2  and entries r22 of R

The value of $\stackrel{\to }{{u}_{2}}$ and ${r}_{22}$ is defined as follows.

$\begin{array}{ll}{r}_{22}=\left|\left|\stackrel{\to }{{V}_{2}^{\perp }}\right|\right|& \phantom{\rule{0ex}{0ex}}\end{array}\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{u}}_{2}=\frac{1}{{r}_{22}}\stackrel{\to }{{V}_{2}^{\perp }}$

Simplify the equation localid="1659438712674" $\begin{array}{ll}{r}_{22}=\left|\left|\stackrel{\to }{{V}_{2}^{\perp }}\right|\right|& \phantom{\rule{0ex}{0ex}}\end{array}$as follows.

${r}_{22}=\left|\left|\stackrel{\to }{{V}_{2}^{\perp }}\right|\right|$

${r}_{22}=\left|\left|\left[\begin{array}{c}-1\\ -1\\ 4\end{array}\right]\right|\right|$

${r}_{22}=\sqrt{{\left(-1\right)}^{2}+{\left(-1\right)}^{2}+{\left(4\right)}^{2}}\phantom{\rule{0ex}{0ex}}{r}_{22}=3\sqrt{2}$

Substitute the values $3\sqrt{2}$ for ${r}_{22}$ and $\left[\begin{array}{c}-1\\ -1\\ 4\end{array}\right]$ for $\stackrel{\to }{{V}_{2}^{\perp }}$ in the equation $\stackrel{\to }{{u}_{2}}=\frac{1}{{r}_{22}}\stackrel{\to }{{V}_{2}^{\perp }}$ as follows.

${\stackrel{\to }{u}}_{2}=\frac{1}{{r}_{22}}{\stackrel{\to }{v}}_{2}^{\perp }\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{u}}_{2}=\frac{1}{3\sqrt{2}}\left[\begin{array}{c}-1\\ -1\\ 4\end{array}\right]\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{u}}_{2}=\left[\begin{array}{c}-1/3\sqrt{2}\\ -1/3\sqrt{2}\\ 4/3\sqrt{2}\end{array}\right]$

The values localid="1659439047262" ${\stackrel{\to }{u}}_{2}=\left[\begin{array}{c}-1/3\sqrt{2}\\ -1/3\sqrt{2}\\ 4/3\sqrt{2}\end{array}\right]$ and ${r}_{22}=3\sqrt{2}$

Therefore, the matrices $Q=\left[\begin{array}{cc}2/3& -1/3\sqrt{2}\\ 2/3& -1/3\sqrt{2}\\ 1/3& 4/3\sqrt{2}\end{array}\right]$ and $R=\left[\begin{array}{cc}3& 3\\ 0& 3\sqrt{2}\end{array}\right]$.

Hence, the QR factorization of the matrix is localid="1659439350112" $\left[\begin{array}{cc}2& 1\\ 2& 1\\ 1& 5\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}2& -1/\sqrt{2}\\ 2& -1/\sqrt{2}\\ 1& 4/\sqrt{2}\end{array}\right]\left[\begin{array}{cc}3& 3\\ 0& 3\sqrt{2}\end{array}\right]$.

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