Question

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

17.$\left[\begin{array}{cc}4& 25\\ 0& 0\\ 3& -25\end{array}\right]$

Short answer

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

Step-by-step solution

Determine column u→1 and entries r11 of R

Consider the matrix$M=\left[\begin{array}{cc}4& 25\\ 0& 0\\ 3& -25\end{array}\right]$where${\overrightarrow{v}}_1=\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right]$and localid="1659954267209" ${\overrightarrow{v}}_2=\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right]$.

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 equation${\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1$ as follows:

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

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

Determine column v→2⊥ and r12 entries  of R

As $r_{11}={\overrightarrow{u}}_1·\overrightarrow{v}2$, 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 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 ${\overrightarrow{v}}_2$, 5 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,role="math" localid="1659954948498" ${\overrightarrow{v}}_2^{\perp }=\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right]$ and $r_{12}=5$.

Determine column u→2 and entries r22 of R

The value of and 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 equationrole="math" localid="1659956549783" $r_{22}=\left|\left|{\overrightarrow{v}}_2^{\perp }\right|\right|$ as follows:

role="math" localid="1659956569123" $$\begin{gathered} r_{22}=\left|\left|{\overrightarrow{v}}_2^{\perp }\right|\right| \\ {r}_{22}=\left|\left|\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right]\right|\right| \\ {r}_{22}=\sqrt{{\left(21\right)}^2+{\left(0\right)}^2+{(-28)}^2} \\ {r}_{22}=35 \end{gathered}$$

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:

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp } \\ {\overrightarrow{u}}_2=\frac{1}{35}\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right] \\ {\overrightarrow{u}}_2=\left[\begin{array}{c}21/35\\ 0\\ -28/35\end{array}\right] \\ {\overrightarrow{u}}_2=\left[\begin{array}{c}3/5\\ 0\\ -4/5\end{array}\right] \end{gathered}$$

The values${\overrightarrow{u}}_2=\left[\begin{array}{c}3/5\\ 0\\ -4/5\end{array}\right]$ and $r_{22}=35$.

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

Hence, the QR factorization of the given matrix is, $\left[\begin{array}{cc}4& 25\\ 0& 0\\ 3& -25\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}4& 3\\ 0& 0\\ 3& -4\end{array}\right]\left[\begin{array}{cc}5& 5\\ 0& 35\end{array}\right]$.