Question

Using paper and pencil, find the QR factorization of the matrices in Exercises 15 through 28. Compare with Exercises 1 through 14.

21.$\left[\begin{array}{ccc}2& -2& 18\\ 2& 1& 0\\ 1& 2& 0\end{array}\right]$

Short answer

The QR factorization of the matrix is$\left[\begin{array}{ccc}2& -2& 18\\ 2& 1& 0\\ 1& 2& 0\end{array}\right]=\left[\begin{array}{ccc}\frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{1}{3}& \frac{2}{3}& \frac{2}{3}\end{array}\right]\left[\begin{array}{ccc}3& 0& 12\\ 0& 3& -12\\ 0& 0& 6\end{array}\right]$.

Step-by-step solution

Determine column u→1 and entries r11 of R

Consider the matrix M$M=\left[\begin{array}{ccc}2& -2& 18\\ 2& 1& 0\\ 1& 2& 0\end{array}\right]$and role="math" localid="1659936948232" ${\overrightarrow{v}}_1=\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$, and ${\overrightarrow{v}}_3=\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$.

By the theorem of QR method, the value of${\overrightarrow{u}}_1$and$r_{11}$is defined as follows:

$$\begin{gathered} r_{11}=\left|\right|{\overrightarrow{v}}_1\left|\right| \\ {r}_{11}=\left|\left|{\overrightarrow{v}}_{11}\right|\right| \\ {\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1 \end{gathered}$$

Simplify the equation $r_{11}=\left|\right|{\overrightarrow{v}}_1\left|\right|$as follows:

$$\begin{gathered} r_{11}=\left|\right|{\overrightarrow{v}}_1\left|\right| \\ {r}_{11}=\left|\left|\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]\right|\right| \\ {r}_{11}=\sqrt{2^2+2^2+1^2} \\ {r}_{11}=3 \end{gathered}$$

Substitute the values 3 for $r_{11}$and $\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$for ${\overrightarrow{v}}_1$in the equation ${\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1$as follows:

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1 \\ {\overrightarrow{u}}_1=\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \\ {\overrightarrow{u}}_1=\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right] \end{gathered}$$

Therefore,${\overrightarrow{u}}_1=\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$ and$r_{11}=3$ .

Determine column v→2⊥ and entries r12 of R

As $r_{12}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2$, substitute the values $\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$for ${\overrightarrow{v}}_2$and${\left[\begin{array}{ccc}\frac{2}{3}& \frac{2}{3}& \frac{2}{3}\end{array}\right]}^T$for in the equation

$r_{12}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2$as follows:

$r_{12}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2$

$$\begin{gathered} r_{12}={\left[\begin{array}{ccc}\frac{2}{3}& \frac{2}{3}& \frac{2}{3}\end{array}\right]}^T.\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right] \\ {r}_{12}=-\begin{array}{ccc}\frac{4}{3}& +\frac{2}{3}& \frac{2}{3}\end{array} \\ {r}_{12}=0 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$for ${\overrightarrow{v}}_2$,0 for $r_{12}$and$\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$ for in the equation

${\overrightarrow{v}}_2^{\perp }=\overrightarrow{v_2}-r_{12}{\overrightarrow{u}}_1$as follows :

${\overrightarrow{v}}_2^{\perp }=\overrightarrow{v_2}-r_{12}{\overrightarrow{u}}_1$

${\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]-\left(0\right)\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$

${\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$

Therefore ,${\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]\mathrm{and}{r}_{12}=0$

Determine column u→2 and entries r22 of R

The value of${\overrightarrow{u}}_2$and$r_{22}$is defined as follows:

$$\begin{gathered} r_{22}=||{\overrightarrow{v}}_2^{\perp }|| \\ {\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp } \end{gathered}$$

Simplify the equation $r_{22}=||{\overrightarrow{v}}_2^{\perp }||$as follows:

$$\begin{gathered} r_{22}=||{\overrightarrow{v}}_2^{\perp }|| \\ {r}_{22}||\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]|| \\ {r}_{22}=\sqrt{{\left(-2\right)}^2+1^2+2^2} \\ {r}_{22}=3 \end{gathered}$$

Substitute the values 3 for $r_{22}$and $\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$for ${\overrightarrow{v}}_2^{\perp }$in the equation role="math" localid="1659938646006" ${\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp }$as follows:

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp } \\ {\overrightarrow{u}}_2=\frac{1}{3}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right] \\ {\overrightarrow{u}}_2=\left[\begin{array}{c}\frac{-2}{3}\\ \frac{1}{3}\\ \frac{2}{3}\end{array}\right] \end{gathered}$$

Therefore ,${\overrightarrow{u}}_2=\left[\begin{array}{c}\frac{-2}{3}\\ \frac{1}{3}\\ \frac{2}{3}\end{array}\right]\mathrm{and}{\mathrm{r}}_{22}=3$

Determine column v→3⊥ , entries r13 and r23 of R

As $r_{13}={\overrightarrow{u}}_1.{\overrightarrow{v}}_3$, substitute the values $\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$for ${\overrightarrow{v}}_3$and ${\left[\frac{2}{3}\frac{2}{3}\frac{1}{3}\right]}^T$for ${\overrightarrow{u}}_1$in the equation

$r_{13}={\overrightarrow{u}}_1.{\overrightarrow{v}}_3$as follows:

$r_{13}={\overrightarrow{u}}_1.{\overrightarrow{v}}_3$

$r_{12}={\left[\frac{2}{3}\frac{2}{3}\frac{1}{3}\right]}^T.\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$

$r_{12}=12$

Substitute the values $\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$for ${\overrightarrow{v}}_3$and ${\left[-\frac{2}{3}\frac{1}{3}\frac{2}{3}\right]}^T$for ${\overrightarrow{u}}_1$in the equation $r_{23}={\overrightarrow{u}}_2.{\overrightarrow{v}}_3$as follows:

$$\begin{gathered} r_{23}={\overrightarrow{u}}_2.{\overrightarrow{v}}_3 \\ {r}_{23}={\left[-\frac{2}{3}\frac{1}{3}\frac{2}{3}\right]}^T.\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right] \\ {r}_{23}=-12 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$for $\overrightarrow{v_3}$, 12 for $r_{13}$, -12 for $r_{23}$, $\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$for ${\overrightarrow{u}}_1$and $\left[\begin{array}{c}-\frac{2}{3}\\ \frac{1}{3}\\ \frac{2}{3}\end{array}\right]$for ${\overrightarrow{u}}_2$in the equation ${\overrightarrow{v}}_3^{\perp }={\overrightarrow{v}}_3-r_{13}\overrightarrow{u}-r_{23}{\overrightarrow{u}}_2$as follows:

role="math" localid="1659939547497" ${\overrightarrow{v}}_3^{\perp }={\overrightarrow{v}}_3-r_{13}\overrightarrow{u}-r_{23}{\overrightarrow{u}}_2\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$

$$\begin{gathered} {\overrightarrow{v}}_3^{\perp }=\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]-12\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]+12\left[\begin{array}{c}-\frac{2}{3}\\ \frac{1}{3}\\ \frac{2}{3}\end{array}\right] \\ {\overrightarrow{v}}_3^{\perp }=\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]-\left[\begin{array}{c}8\\ 8\\ 4\end{array}\right]+\left[\begin{array}{c}-8\\ 4\\ 8\end{array}\right] \\ {\overrightarrow{v}}_3^{\perp }=\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right] \end{gathered}$$

Therefore ,${\overrightarrow{v}}_3^{\perp }=\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right]$,$r_{12}=12\mathrm{and}{\mathrm{r}}_{23}=-12$

Determine column u→3 and entries r33 of R

The value of${\overrightarrow{u}}_3$and$r_{33}$is defined as follows:

$$\begin{gathered} r_{33}=||{\overrightarrow{v}}_3^{\perp }|| \\ {\overrightarrow{u}}_3=\frac{1}{r_{33}}{\overrightarrow{v}}_3^{\perp } \end{gathered}$$

Simplify the equation $r_{33}=||{\overrightarrow{v}}_3^{\perp }||$as follows:

$$\begin{gathered} r_{33}=||{\overrightarrow{v}}_3^{\perp }|| \\ {r}_{33}=||\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right]|| \\ {r}_{33}=\sqrt{{\left(-2\right)}^2+{\left(-4\right)}^2+{\left(4\right)}^2} \\ {r}_{33}=6 \end{gathered}$$

Substitute the values 6 for $r_{33}$and $\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right]$for ${\overrightarrow{v}}_3^{\perp }$in the equation ${\overrightarrow{u}}_3=\frac{1}{r_{33}}{\overrightarrow{v}}_3^{\perp }$as follows:

$$\begin{gathered} {\overrightarrow{u}}_3=\frac{1}{r_{33}}{\overrightarrow{v}}_3^{\perp } \\ {\overrightarrow{u}}_3=\frac{1}{6}\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right] \\ {\overrightarrow{u}}_3=\left[\begin{array}{c}\frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3}\end{array}\right] \end{gathered}$$

Therefore, the matrices $\mathrm{Q}=\left[\begin{array}{ccc}\frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{1}{3}& \frac{2}{3}& \frac{2}{3}\end{array}\right]\mathrm{and}R=\left[\begin{array}{ccc}3& 0& 12\\ 0& 3& -12\\ 0& 0& 6\end{array}\right]$

Hence, the QR factorization of the matrix is$\left[\begin{array}{ccc}2& -2& 18\\ 2& 1& 0\\ 1& 2& 0\end{array}\right]=\left[\begin{array}{ccc}\frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{1}{3}& \frac{2}{3}& \frac{2}{3}\end{array}\right]\left[\begin{array}{ccc}3& 0& 12\\ 0& 3& -12\\ 0& 0& 6\end{array}\right]$ .