Question

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

15.$\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right]$

Short answer

The QR factorization of the matrices $\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right]\mathrm{is}\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right]=\underset{\mathrm{Q}}{\underbrace{\left[\begin{array}{c}2/3\\ 1/3\\ -2/3\end{array}\right]}}\underset{\mathrm{R}}{\underbrace{3}}$

Step-by-step solution

QR factorization

Consider an $\mathit{n}\mathbf{\times }\mathit{m}$matrix Mwith linearly independent columns ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}$. Then there exists an $\mathit{n}\mathbf{\times }\mathit{m}$matrix Qwhose columns ${\overrightarrow{\mathbf{u}}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\overrightarrow{\mathbf{u}}}_{\mathbf{m}}$are orthonormal and an upper triangular matrix Rwith positive diagonal entries such that.

M=QR

Furthermore, $r_{11}=\left|\left|{\overrightarrow{v}}_1\right|\right|,r_{jj}=\left|\left|{{\overrightarrow{v}}_j}^{\perp }\right|\right|,(\mathrm{For}j=2,...m)\mathrm{and}{r}_{ij}={\overrightarrow{u}}_i.{\overrightarrow{v}}_j$.

According to above theorem

Let the given vector is, $M=\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right],\mathrm{where}{\overrightarrow{v}}_1=\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right]$, where .

Consider the terms below to find out the QR factorization.

$$\begin{gathered} r_{11}=\left|\left|{\overrightarrow{v}}_1\right|\right| \\ =\sqrt{2^2+1^2+{\left(-2\right)}^2} \\ =\sqrt{4+1+4} \\ =3 \end{gathered}$$

Now, find ${\overrightarrow{u}}_1$.

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

Thus, the value is$\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right]=\underset{\mathrm{Q}}{\underbrace{\left[\begin{array}{c}2/3\\ 1/3\\ -2/3\end{array}\right]}}\underset{\mathrm{R}}{\underbrace{3}}$

Hence, the QR factorization for M is $\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right]=\underset{\mathrm{Q}}{\underbrace{\left[\begin{array}{c}2/3\\ 1/3\\ -2/3\end{array}\right]}}\underset{\mathrm{R}}{\underbrace{3}}$.