Question

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

16.$\left[\begin{array}{cc}6& 2\\ 3& -6\\ 2& 3\end{array}\right]$.

Short answer

The QR factorization of the matrix is $\left[\begin{array}{cc}6& 2\\ 3& -6\\ 2& 3\end{array}\right]=\frac{1}{7}\left[\begin{array}{cc}6& 2\\ 3& -6\\ 2& 3\end{array}\right]\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$.

Step-by-step solution

Determine column  and entries  of .

Consider the matrix $M=\left[\begin{array}{cc}6& 2\\ 3& -6\\ 2& 3\end{array}\right]$where ${\overrightarrow{v}}_1=\left[\begin{array}{c}6\\ -3\\ 2\end{array}\right]$and ${\overrightarrow{v}}_2=\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$.

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.

$$\begin{gathered} r_{11}=\left|\left|{\overrightarrow{v}}_1\right|\right| \\ {r}_{11}=\left|\left|\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]\right|\right| \\ {r}_{11}=\sqrt{6^2+3^2+2^2} \\ {r}_{11}=7 \end{gathered}$$

Substitute the values 7 for $r_{11}$and $\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]$for in the equation ${\overleftrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1$as follows.

localid="1659440988937" $$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1 \\ {\overrightarrow{u}}_1=\frac{1}{7}\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right] \\ {\overrightarrow{u}}_1=\left[\begin{array}{c}6/7\\ 3/7\\ 2/7\end{array}\right] \end{gathered}$$

Therefore, the values ${\overrightarrow{u}}_1=\left[\begin{array}{c}6/7\\ 3/7\\ 2/7\end{array}\right]$and $r_{11}=7$.

Determine column v2⊥ and r12entries  of .

As $r_{12}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2$, substitute the values $\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$for ${\overrightarrow{v}}_2$and $\left[\begin{array}{ccc}\frac{6}{7}& \frac{3}{7}& \frac{2}{7}\end{array}\right]$for ${\overrightarrow{u}}_1$in the equation $r_{11}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2$as follows:

$$\begin{gathered} r_{12}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2 \\ {r}_{12}={\left[\begin{array}{ccc}\frac{6}{7}& \frac{3}{7}& \frac{2}{7}\end{array}\right]}^T-\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right] \\ {r}_{12}=\frac{12}{7}-\frac{18}{7}+\frac{6}{7} \\ {r}_{12}=0 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$for ${\overrightarrow{v}}_2$, 0 for $r_{12}$and $\left[\begin{array}{c}6/7\\ 3/7\\ 2/7\end{array}\right]$for ${\overrightarrow{u}}_1$in the equation ${\overrightarrow{v}}_2^{\perp }={\overrightarrow{v}}_2-r_{12}{\overrightarrow{u}}_1$as follows:

$$\begin{gathered} {\overrightarrow{v}}_2^{\perp }={\overrightarrow{v}}_2-r_{12}{\overrightarrow{u}}_1 \\ {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]-0\left[\begin{array}{c}6/7\\ 3/7\\ 2/7\end{array}\right] \\ {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right] \end{gathered}$$

Therefore, the values ${\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$and $r_{12}=0$.

Determine column u→2and entries r22 of R

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

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

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

$$\begin{gathered} r_{22}=\left|\left|{\overrightarrow{v}}_2^{\perp }\right|\right| \\ {r}_{22}=\left|\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]\right| \\ {r}_{22}=\sqrt{{\left(2\right)}^2+{\left(-6\right)}^2+{\left(3\right)}^2} \\ {r}_{22}=7 \end{gathered}$$

Substitute the values 7 for $r_{22}$and $\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$for${\overrightarrow{v}}_2^{\perp }$ in the equation ${\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp }$as follows:

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp } \\ {\overrightarrow{u}}_2=\frac{1}{7}\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right] \\ {\overrightarrow{u}}_2=\left[\begin{array}{c}2/7\\ -6/7\\ 3/7\end{array}\right] \end{gathered}$$

The values ${\overrightarrow{u}}_2=\left[\begin{array}{c}2/7\\ -6/7\\ 3/7\end{array}\right]$and $r_{22}=7$.

Therefore, the matrices $Q=\left[\begin{array}{cc}6/7& 2/7\\ 3/7& -6/7\\ 2/7& 3/7\end{array}\right]$and $R=\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$.

Hence, the QR factorization of the given matrix is, $\left[\begin{array}{cc}6& 2\\ 3& -6\\ 2& 3\end{array}\right]=\frac{1}{7}\left[\begin{array}{cc}6& 2\\ 3& -6\\ 2& 3\end{array}\right]\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$.