Question

Find the QR factorization of the matrices$\left[\begin{array}{cc}4& 5\\ 0& 2\\ 0& 14\\ 3& 10\end{array}\right]$.

Short answer

The QR factorization of the matrix is$\left[\begin{array}{cc}4& 5\\ 0& 2\\ 0& 14\\ 3& 10\end{array}\right]=\left[\begin{array}{cc}4/5& -1/5\\ 0& 2/15\\ 0& 14/15\\ 3/5& 4/15\end{array}\right]\left[\begin{array}{cc}5& 10\\ 0& 15\end{array}\right]$ .

Step-by-step solution

Determine column u→1 and entries r11 of R.

Consider the matrix$M=\left[\begin{array}{cc}4& 5\\ 0& 2\\ 0& 14\\ 3& 10\end{array}\right]\mathrm{where}{\overrightarrow{\mathrm{v}}}_1=\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right]\mathrm{and}{\overrightarrow{\mathrm{v}}}_2=\left[\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\right]$

By the theorem of QR method, the value of and is defined as follows.

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

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

role="math" localid="1659961806612" $$\begin{gathered} r_{11}=||{\overrightarrow{v}}_1|| \\ {r}_{11}=||\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right]|| \\ {r}_{11}=\sqrt{4^2+0^2+0^2+3^2} \\ {r}_{11}=5 \end{gathered}$$

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

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

Therefore, the values${\overrightarrow{u}}_1=\left[\begin{array}{c}4/5\\ 0\\ 0\\ 3/5\end{array}\right]$ 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 $\left[\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\right]$for ${\overrightarrow{v}}_2$and $\left[\begin{array}{cccc}\frac{4}{5}& 0& 0& \frac{3}{5}\end{array}\right]$for in the equation $r_{11}={\overrightarrow{u}}_1.\overrightarrow{v_2}$as follows.

$r_{12}={\overrightarrow{u}}_1.\overrightarrow{v_2}$

$r_{12}=\left[\begin{array}{cccc}\frac{4}{5}& 0& 0& \frac{3}{5}\end{array}\right].\left[\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\right]$

$$\begin{gathered} r_{12}=4+0+0+6 \\ {r}_{12}=10 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\right]$for ${\overrightarrow{v}}_2$, 10 for $r_{12}$and $\left[\begin{array}{c}4/5\\ 0\\ 0\\ 3/5\end{array}\right]$for ${\overrightarrow{u}}_1$in the equation as follows.

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

$$\begin{gathered} {\overrightarrow{V}}_2^{\perp }=\overrightarrow{v_2}-r_{12}{\overrightarrow{u}}_1 \\ {\overrightarrow{V}}_2^{\perp }=\left[\begin{array}{c}\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\end{array}\right]-10\left[\begin{array}{c}4/5\\ 0\\ 0\\ 3/5\end{array}\right] \\ {\overrightarrow{V}}_2^{\perp }=\left[\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\right]-\left[\begin{array}{c}8\\ 0\\ 0\\ 6\end{array}\right] \\ {\overrightarrow{V}}_2^{\perp }=\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right] \end{gathered}$$

Therefore, the values${\overrightarrow{V}}_2^{\perp }=\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right]$ and $r_{12}=10$.

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.

$$\begin{gathered} r_{22}=||{\overrightarrow{v}}_2^{\perp }|| \\ {r}_{22}=||\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right]|| \\ {r}_{22}=\sqrt{{\left(-3\right)}^2+{\left(2\right)}^2+{\left(14\right)}^2+{\left(4\right)}^2} \\ {r}_{22}=15 \end{gathered}$$

Substitute the values 15 for $r_{22}$and $\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right]$for ${\overrightarrow{v}}_2^{\perp }$in the equation ${\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp }$as follows.

$$\begin{gathered} \overrightarrow{u}=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp } \\ \overrightarrow{u}=\frac{1}{15}\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right] \\ {\overrightarrow{u}}_2=\left[\begin{array}{c}-1/5\\ 2/15\\ 14/15\\ 4/15\end{array}\right] \end{gathered}$$

The values ${\overrightarrow{u}}_2=\left[\begin{array}{c}-1/5\\ 2/15\\ 14/15\\ 4/15\end{array}\right]$and $r_{22}=15$.

Therefore, the matrices role="math" localid="1659963313418" $Q=\left[\begin{array}{cc}4/5& -1/5\\ 0& 2/15\\ 0& 14/15\\ 3/5& 4/15\end{array}\right]\mathrm{and}R=\left[\begin{array}{cc}5& 10\\ 0& 15\end{array}\right]$

Hence, the QR factorization of the matrix is$\left[\begin{array}{cc}4& 5\\ 0& 2\\ 0& 14\\ 3& 10\end{array}\right]=\left[\begin{array}{cc}4/5& -1/5\\ 0& 2/15\\ 0& 14/15\\ 3/5& 4/15\end{array}\right]\left[\begin{array}{cc}5& 10\\ 0& 15\end{array}\right]$ .