Question

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

18.$\left[\begin{array}{ccc}4& 25& 0\\ 0& 0& -2\\ 3& 25& 0\end{array}\right]$

Short answer

The QR factorization of the matrix is $\left[\begin{array}{ccc}4& 25& 0\\ 0& 0& -2\\ 3& -25& 0\end{array}\right]=\left[\begin{array}{ccc}4/5& 21/35& 0\\ 0& 0& -1\\ 3/5& -28/35& 0\end{array}\right]\left[\begin{array}{ccc}5& 5& 0\\ 0& 35& 0\\ 0& 0& 2\end{array}\right]$.

Step-by-step solution

Determine column u→1 and entries r11 of R

Consider the matrix$M=\left[\begin{array}{ccc}4& 25& 0\\ 0& 0& -2\\ 3& -25& 0\end{array}\right]$and ${\overrightarrow{v}}_1=\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right]$,${\overrightarrow{v}}_2=\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right]$and ${\overrightarrow{v}}_3=\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right]$.

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

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

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

$$\begin{gathered} r_{11}=\left|\left|{\overrightarrow{v}}_1\right|\right| \\ {r}_{11}=\left|\left|\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right]\right|\right| \\ {r}_{11}=\sqrt{4^2+0^2+3^2} \\ {r}_{11}=5 \end{gathered}$$

Substitute the values 5 for$r_{11}$and$\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right]$for${\overrightarrow{v}}_1$in the equationlocalid="1659958842441" ${\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1$as follows:

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1 \\ {\overrightarrow{u}}_1=\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right] \\ {\overrightarrow{u}}_1=\left[\begin{array}{c}4/5\\ 0\\ 3/5\end{array}\right] \end{gathered}$$

Therefore,${\overrightarrow{u}}_1=\left[\begin{array}{c}4/5\\ 0\\ 3/5\end{array}\right]$ and $r_{11}=5$.

Determine column v→2⊥ and entries r12 of R

As $r_{12}={\overrightarrow{u}}_1·{\overrightarrow{v}}_2$, substitute the values$\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right]$ for${\overrightarrow{v}}_2$ and${\left[\frac{4}{5}0\frac{3}{5}\right]}^T$ for${\overrightarrow{u}}_1$ in the equation$r_{11}={\overrightarrow{u}}_1·{\overrightarrow{v}}_2$ as follows:

$$\begin{gathered} r_{12}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2 \\ {r}_{12}={\left[\frac{4}{5}0\frac{3}{5}\right]}^T.\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right] \\ {r}_{12}=20+0-15 \\ {r}_{12}=5 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right]$for ${\overrightarrow{v}}_2$, 35 for$r_{12}$ and$\left[\begin{array}{c}4/5\\ 0\\ 3/5\end{array}\right]$ for${\overrightarrow{u}}_1$ in the equation${\overrightarrow{v}}_2^{\perp }=\overrightarrow{v_2}-r_{12}{\overrightarrow{u}}_1$ as follows:

$$\begin{gathered} {\overrightarrow{v}}_2^{\perp }={\overrightarrow{v}}_2-r_{12}{\overrightarrow{u}}_1 \\ {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right]-5\left[\begin{array}{c}4/5\\ 0\\ 3/5\end{array}\right] \\ {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right]-\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right] \\ {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right] \end{gathered}$$

Therefore, the values${\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right]$ and $r_{12}=5$.

Determine column u→2 and entries r22 of R

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

$\begin{array}{l}{r}_{22}=\left|\left|{\overrightarrow{v}}_2^{\perp }\right|\right|\\ {\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp }\end{array}$

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

$$\begin{gathered} r_{22}=\left|\left|{\overrightarrow{v}}_2^{\perp }\right|\right| \\ {r}_{22}=\left|\left|\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right]\right|\right| \\ {r}_{22}=\sqrt{{\left(21\right)}^2+{\left(0\right)}^2+{(-28)}^2} \\ {r}_{22}=35 \end{gathered}$$

Substitute the values 35 for$r_{22}$and$\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right]$for${\overrightarrow{v}}_2^{\perp }$in the equation${\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp }$as follows:

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp } \\ {\overrightarrow{u}}_2=\frac{1}{35}\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right] \\ {\overrightarrow{u}}_2=\left[\begin{array}{c}21/35\\ 0\\ -28/35\end{array}\right] \end{gathered}$$

Therefore, the values${\overrightarrow{u}}_2=\left[\begin{array}{c}21/35\\ 0\\ -28/35\end{array}\right]$and $r_{22}=35$.

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

As $r_{13}={\overrightarrow{u}}_1·{\overrightarrow{v}}_3$, substitute the values$\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right]$ for${\overrightarrow{v}}_3$ and for${\left[\frac{4}{5}0\frac{3}{5}\right]}^T$ in${\overrightarrow{u}}_1$ the equation$r_{13}={\overrightarrow{u}}_1·{\overrightarrow{v}}_3$ as follows:

role="math" localid="1659959616037" $$\begin{gathered} r_{13}={\overrightarrow{u}}_1.{\overrightarrow{v}}_3 \\ {r}_{13}={\left[\frac{4}{5}0\frac{3}{5}\right]}^T.\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right] \\ {r}_{13}=0 \end{gathered}$$

substitute the values$\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right]$ for${\overrightarrow{v}}_3$ and$\left[\frac{21}{35}0-\frac{28}{35}\right]$ for${\overrightarrow{u}}_2$ in the equation$r_{23}={\overrightarrow{u}}_2·{\overrightarrow{v}}_3$ as follows:

$$\begin{gathered} r_{23}={\overrightarrow{u}}_2.{\overrightarrow{v}}_3 \\ {r}_{23}={\left[\frac{21}{35}0\frac{28}{35}\right]}^T.\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right] \\ {r}_{23}=0 \end{gathered}$$

Substitute the values$\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right]$ for ${\overrightarrow{v}}_3$,0 for $r_{13}$,0 for $r_{23}$,$\left[\begin{array}{c}4/5\\ 0\\ 3/5\end{array}\right]$ for$\overrightarrow{u_1}$ and$\left[\begin{array}{c}21/35\\ 0\\ -28/35\end{array}\right]$ for${\overrightarrow{u}}_2$ in the equation${\overrightarrow{v}}_3^{\perp }={\overrightarrow{v}}_3-r_{13}{\overrightarrow{u}}_1-r_{23}{\overrightarrow{u}}_2$ as follows:

$$\begin{gathered} {\overrightarrow{v}}_2^{\perp }={\overrightarrow{v}}_3-r_{13}{\overrightarrow{u}}_1-r_{23}{\overrightarrow{u}}_2 \\ {\overrightarrow{v}}_3^{\perp }=\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right]-0\left[\begin{array}{c}4/5\\ 0\\ 3/5\end{array}\right]+0\left[\begin{array}{c}21/35\\ 0\\ -28/35\end{array}\right] \\ {\overrightarrow{v}}_3^{\perp }=\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right] \end{gathered}$$

Therefore, the values ${\overrightarrow{v}}_3^{\perp }=\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right]$,$r_{13}=0$ and $r_{23}=0$.

Determine column u→3 and entries r33 of R

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

$\begin{array}{l}{r}_{33}=\left|\left|{\overrightarrow{v}}_3^{\perp }\right|\right|\\ {\overrightarrow{u}}_3=\frac{1}{r_{33}}{\overrightarrow{v}}_3^{\perp }\end{array}$

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

$$\begin{gathered} r_{33}=\left|\left|{\overrightarrow{v}}_3^{\perp }\right|\right| \\ {r}_{33}=\left|\left|\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right]\right|\right| \\ {r}_{33}=\sqrt{{\left(0\right)}^2+{\left(-2\right)}^2+{\left(0\right)}^2} \\ {r}_{33}=2 \end{gathered}$$

Substitute the values 2 for$r_{33}$ and$\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right]$ for${\overrightarrow{v}}_3^{\perp }$ in the equation${\overrightarrow{u}}_3=\frac{1}{r_{33}}{\overrightarrow{v}}_3^{\perp }$ as follows:

role="math" localid="1659960743539" $$\begin{gathered} {\overrightarrow{u}}_3=\frac{1}{r_{33}}{\overrightarrow{v}}_3^{\perp } \\ {\overrightarrow{u}}_3=\frac{1}{2}\left[\begin{array}{c}0\\ -2\\ 0\end{array}\right] \\ {\overrightarrow{u}}_3=\left[\begin{array}{c}0\\ -1\\ 0\end{array}\right] \end{gathered}$$

Therefore, the matrices $\mathrm{Q}=\left[\begin{array}{cc}4/5& 21/35\\ 0& 0\\ 3/5& -28/35\end{array}\begin{array}{c}0\\ -1\\ 0\end{array}\right]\mathrm{and}\mathrm{R}=\left[\begin{array}{ccc}5& 5& 0\\ 0& 35& 0\\ 0& 0& 2\end{array}\right]$.

Hence, the QR factorization of the matrix is .

$\left[\begin{array}{ccc}4& 25& 0\\ 0& 0& -2\\ 3& -25& 0\end{array}\right]=\left[\begin{array}{ccc}4/5& 21/35& 0\\ 0& 0& -1\\ 3/5& -28/35& 0\end{array}\right]\left[\begin{array}{ccc}5& 5& 0\\ 0& 35& 0\\ 0& 0& 2\end{array}\right]$