Question

Find the QR factorization of the matrices$\left[\begin{array}{cc}5& 3\\ 4& 6\\ 2& 7\\ 2& -2\end{array}\right]$.

Short answer

The QR factorization of the matrix is $\left[\begin{array}{cc}5& 3\\ 4& 6\\ 2& 7\\ 2& -2\end{array}\right]=\left[\begin{array}{cc}\frac{5}{7}& -\frac{2}{7}\\ \frac{4}{7}& \frac{2}{7}\\ \frac{2}{7}& \frac{5}{7}\\ \frac{2}{7}& \frac{-4}{7}\end{array}\right]\left[\begin{array}{cc}7& 7\\ 0& 7\end{array}\right]$.

Step-by-step solution

Determine column  u→1 and entries  r11 of R.

Consider the matrix $M=\left[\begin{array}{cc}5& 3\\ 4& 6\\ 2& 7\\ 2& -2\end{array}\right]\mathrm{where}{\overrightarrow{v}}_1=\mathrm{M}=\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right]\mathrm{and}{\overrightarrow{\mathit{v}}}_2=\left[\begin{array}{c}3\\ 6\\ 27\\ -2\end{array}\right]$

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

$$\begin{gathered} {r_1}_1=||{\overrightarrow{v}}_1|| \\ {\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1 \end{gathered}$$

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

$$\begin{gathered} {\mathrm{r}}_{11}=||{\overrightarrow{\mathrm{v}}}_1|| \\ {\mathrm{r}}_{11}=||\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right]|| \\ {\mathrm{r}}_{11}=\sqrt{5^2+4^2+2^2+2^2} \\ {\mathrm{r}}_{11}=7 \end{gathered}$$

Substitute the values 7 for $r_{11}$and $\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right]$for ${\overrightarrow{v}}_1$in the equation $\overrightarrow{u}=\frac{1}{r_{11}}{\overrightarrow{v}}_1$as follows,

$$\begin{gathered} \overrightarrow{u}=\frac{1}{r_{11}}{\overrightarrow{v}}_1 \\ \overrightarrow{u}=\frac{1}{7}\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right] \\ \overrightarrow{u}=\left[\begin{array}{c}\frac{5}{7}\\ \frac{4}{7}\\ \frac{2}{7}\\ \frac{2}{7}\end{array}\right] \end{gathered}$$

Therefore, the values$\overrightarrow{u}=\left[\begin{array}{c}\frac{5}{7}\\ \frac{4}{7}\\ \frac{2}{7}\\ \frac{2}{7}\end{array}\right]$ 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 $\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]$for ${\overrightarrow{v}}_2$and $\left[\begin{array}{cccc}\frac{5}{7}& \frac{4}{7}& \frac{2}{7}& \frac{2}{7}\end{array}\right]$for in the equation $r_{11}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2$as follows.

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

$$\begin{gathered} r_{12}=\left[\begin{array}{cccc}\frac{5}{7}& \frac{4}{7}& \frac{2}{7}& \frac{2}{7}\end{array}\right].\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right] \\ {r}_{12}=\begin{array}{cccc}\frac{15}{7}& \frac{24}{7}& \frac{14}{7}& \frac{4}{7}\end{array} \\ {r}_{12}=7 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]$for ${\overrightarrow{v}}_2$, 7 for role="math" localid="1659942303914" ${\overrightarrow{r}}_{12}$and $\left[\begin{array}{c}\frac{5}{7}\\ \frac{4}{7}\\ \frac{2}{7}\\ \frac{2}{7}\end{array}\right]$for ${\overrightarrow{u}}_1$in the equation

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

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

role="math" localid="1659942632253" $$\begin{gathered} {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]-7\left[\begin{array}{c}\frac{5}{7}\\ \frac{4}{7}\\ \frac{2}{7}\\ \frac{2}{7}\end{array}\right] \\ {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]-7\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right] \\ {\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}-2\\ 2\\ 5\\ -4\end{array}\right] \end{gathered}$$

Therefore, teh values${\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}-2\\ 2\\ 5\\ -4\end{array}\right]\mathrm{and}{\mathrm{r}}_{12}=7$

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,

$$\begin{gathered} r_{22}=||{\overrightarrow{v}}_2^{\perp }|| \\ {r}_{22}=||\left[\begin{array}{c}-2\\ 2\\ 5\\ -4\end{array}\right]|| \end{gathered}$$

$$\begin{gathered} r_{22}=\sqrt{{\left(-2\right)}^2+2^2+5^2+{\left(-4\right)}^2} \\ {r}_{22}=7 \end{gathered}$$

Substitute the values 7 for $r_{22}$and $\left[\begin{array}{c}-2\\ 2\\ 5\\ 4\end{array}\right]$for ${\overrightarrow{v}}_2^{\perp }$in the equation ${\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp }$as follows,

${\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp }$

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{7}\left[\begin{array}{c}-2\\ 2\\ 5\\ -4\end{array}\right] \\ {\overrightarrow{u}}_2=\left[\begin{array}{c}\frac{-2}{7}\\ \frac{2}{7}\\ \frac{5}{7}\\ \frac{-4}{7}\end{array}\right] \end{gathered}$$

Therefore, the values ${\overrightarrow{u}}_2=\left[\begin{array}{c}\frac{-2}{7}\\ \frac{2}{7}\\ \frac{5}{7}\\ \frac{-4}{7}\end{array}\right]\mathrm{and}{\mathrm{r}}_{22}=7$

Therefore, the matrices $Q=\left[\begin{array}{cc}\frac{5}{7}& -\frac{2}{7}\\ \frac{4}{7}& \frac{2}{7}\\ \frac{2}{7}& \frac{5}{7}\\ \frac{2}{7}& -\frac{4}{7}\end{array}\right]$and $R=\left[\begin{array}{cc}7& 7\\ 0& 7\end{array}\right]$.

Hence, the QR factorization of the matrix is$\left[\begin{array}{cc}5& 3\\ 4& 6\\ 2& 7\\ 2& -2\end{array}\right]=\left[\begin{array}{cc}\frac{5}{7}& -\frac{2}{7}\\ \frac{4}{7}& \frac{2}{7}\\ \frac{2}{7}& \frac{5}{7}\\ \frac{2}{7}& \frac{-4}{7}\end{array}\right]\left[\begin{array}{cc}7& 7\\ 0& 7\end{array}\right]$ .