Question

Find the QR factorization of the matrices$\left[\begin{array}{cc}2& 4\\ 3& 4\\ 0& 2\\ 6& 13\end{array}\right]$.

Short answer

The QR factorization of the matrix is$\left[\begin{array}{cc}2& 4\\ 3& 4\\ 0& 2\\ 6& 13\end{array}\right]=\left[\begin{array}{cc}2/7& 0\\ 3/7& -2/3\\ 0& 2/3\\ 6/7& 1/3\end{array}\right]\left[\begin{array}{cc}7& 14\\ 0& 3\end{array}\right]$.

Step-by-step solution

Determine column u→1 and entries r11 of R.

Consider the matrix $M=\left[\begin{array}{cc}2& 4\\ 3& 4\\ 0& 2\\ 6& 13\end{array}\right]$where localid="1659956780727" ${\overrightarrow{v}}_1=\left[\begin{array}{c}2\\ 3\\ 0\\ 6\end{array}\right]$and ${\overrightarrow{v}}_2=\left[\begin{array}{c}4\\ 4\\ 2\\ 13\end{array}\right]$.

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

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

Simplify the equation$r_{11}=\left|\left|\overrightarrow{v_1}\right|\right|$as follows.

$$\begin{gathered} r_{11}=||{\overrightarrow{v}}_1|| \\ {r}_{11}=||\left[\begin{array}{c}2\\ 3\\ 0\\ 6\end{array}\right]|| \\ {r}_{11}=\sqrt{2^2+3^2+0^2+6^2} \\ {r}_{11}=7 \end{gathered}$$

Substitute the values 7 for$r_{11}$and$\left[\begin{array}{c}2\\ 3\\ 0\\ 6\end{array}\right]$for ${\overrightarrow{v}}_1$in the equation ${\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1$as follows.

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1 \\ {\overrightarrow{u}}_1=\frac{1}{7}\left[\begin{array}{c}2\\ 3\\ 0\\ 6\end{array}\right] \\ {\overrightarrow{u}}_1=\left[\begin{array}{c}2/7\\ 3/7\\ 0\\ 6/7\end{array}\right] \end{gathered}$$

Therefore, the values${\overrightarrow{u}}_1=\left[\begin{array}{c}2/7\\ 3/7\\ 0\\ 6/7\end{array}\right]$and $r_{11}=7$.

Determine column v→2⊥ and entries r12 of R.

As $r_{12}={\overrightarrow{u}}_1-\overrightarrow{v_2}$, substitute the values $\left[\begin{array}{c}4\\ 4\\ 2\\ 13\end{array}\right]$for${\overrightarrow{v}}_2$and$\left[\begin{array}{cccc}\frac{2}{7}& \frac{3}{7}& 0& \frac{6}{7}\end{array}\right]$for${\overrightarrow{u}}_1$in the equation$r_{11}={\overrightarrow{u}}_1-\overrightarrow{v_2}$as follows.

$$\begin{gathered} r_{12}={\overrightarrow{u}}_1-\overrightarrow{v_2} \\ {r}_{12}=\left[\begin{array}{cccc}\frac{2}{7}& \frac{3}{7}& 0& \frac{6}{7}\end{array}\right].\left[\begin{array}{c}4\\ 4\\ 2\\ 13\end{array}\right] \\ {r}_{12}=\begin{array}{cccc}\frac{8}{7}& \frac{12}{7}& 0& \frac{78}{7}\end{array} \\ {r}_{12}=14 \end{gathered}$$

Substitute the values$\left[\begin{array}{c}4\\ 4\\ 2\\ 13\end{array}\right]$for ${\overrightarrow{v}}_2$, 14 for $r_{12}$and $\left[\begin{array}{c}2/7\\ 3/7\\ 0\\ 6/7\end{array}\right]$for ${\overrightarrow{u}}_1$in the equation ${\overrightarrow{v}}_2^{\perp }={\overrightarrow{v}}_2-r_{12}{\overrightarrow{u}}_1$as follows.

$$\begin{gathered} {\overrightarrow{v}}_2^{\perp }={\overrightarrow{v}}_2-r_{12}{\overrightarrow{u}}_1 \\ {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}4\\ 4\\ 2\\ 13\end{array}\right]-14\left[\begin{array}{c}2/7\\ 3/7\\ 0\\ 6/7\end{array}\right] \\ {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}4\\ 4\\ 2\\ 13\end{array}\right]-\left[\begin{array}{c}4\\ 6\\ 0\\ 12\end{array}\right] \\ {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}0\\ -2\\ 2\\ 1\end{array}\right] \end{gathered}$$

Therefore, the values${\overrightarrow{v}}_z^{\perp }=\left[\begin{array}{c}0\\ -2\\ 2\\ 1\end{array}\right]$and$r_{12}=14$.

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.

$$\begin{gathered} r_{22}=||{\overrightarrow{v}}_2^{\perp }|| \\ {r}_{22}=||\left[\begin{array}{c}0\\ -2\\ 2\\ 1\end{array}\right]|| \\ {r}_{22}=\sqrt{{\left(0\right)}^2+{\left(-2\right)}^2+{\left(2\right)}^2+{\left(1\right)}^2} \\ {r}_{22}=3 \end{gathered}$$

Substitute the values 3 for $r_{22}$and $\left[\begin{array}{c}0\\ -2\\ 2\\ 1\end{array}\right]$for ${\overrightarrow{v}}_2^{\perp }$in the equation $\overrightarrow{u_2}=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp }$as follows.

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp } \\ {\overrightarrow{u}}_2=\frac{1}{3}\left[\begin{array}{c}0\\ -2\\ 2\\ 1\end{array}\right] \\ {\overrightarrow{u}}_2=\left[\begin{array}{c}0\\ -2/3\\ 2/3\\ 1/3\end{array}\right] \end{gathered}$$

The values ${\overrightarrow{u}}_2=\left[\begin{array}{c}0\\ -2/3\\ 2/3\\ 1/3\end{array}\right]$and $r_{22}=3$.

Therefore, the matrices $Q=\left[\begin{array}{cc}2/7& 0\\ 3/7& -2/3\\ 0& 2/3\\ 6/7& 1/3\end{array}\right]$and $R=\left[\begin{array}{cc}7& 14\\ 0& 3\end{array}\right]$.

Hence, the QR factorization of the matrix is$\left[\begin{array}{cc}2& 4\\ 3& 4\\ 0& 2\\ 6& 13\end{array}\right]=\left[\begin{array}{cc}2/7& 0\\ 3/7& -2/3\\ 0& 2/3\\ 6/7& 1/3\end{array}\right]\left[\begin{array}{cc}7& 14\\ 0& 3\end{array}\right]$.