babeeb0oL

2021-02-26

Let
b)Compute $‖Ux‖$ (it should be the same as $‖x‖$)

Lacey-May Snyder

Step 1
Given matrices are,

Step 2 Now the multiplication of the matrices is,
$Ux=\left[\begin{array}{cc}\frac{\sqrt{2}}{2}& \frac{\sqrt{3}}{3}\\ -\frac{\sqrt{2}}{2}& \frac{\sqrt{3}}{3}\\ 0& -\frac{\sqrt{3}}{3}\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}\frac{\sqrt{2}}{2}+\frac{2\sqrt{3}}{3}\\ -\frac{\sqrt{2}}{2}+\frac{2\sqrt{3}}{3}\\ -\frac{2\sqrt{3}}{3}\end{array}\right]$
Step 3
Norm of the matrix is,
$‖Ux‖=‖\begin{array}{c}\left[\begin{array}{c}\frac{\sqrt{2}}{2}+\frac{2\sqrt{3}}{3}\\ -\frac{\sqrt{2}}{2}+\frac{2\sqrt{3}}{3}\\ -\frac{2\sqrt{3}}{3}\end{array}\right]\end{array}‖=\sqrt{{\left(\frac{\sqrt{2}}{2}+\frac{2\sqrt{3}}{3}\right)}^{2}+{\left(-\frac{\sqrt{2}}{2}+\frac{2\sqrt{3}}{3}\right)}^{2}+{\left(-\frac{2\sqrt{3}}{3}\right)}^{2}}$
$=\sqrt{\frac{2}{4}+\frac{12}{9}+2\frac{\sqrt{2}}{2}\frac{2\sqrt{3}}{2}+\frac{2}{4}+\frac{12}{9}-2\frac{\sqrt{2}}{2}\frac{2\sqrt{3}}{2}+\frac{12}{9}}$
$=\sqrt{\frac{1}{2}+\frac{4}{3}+\frac{1}{2}+\frac{4}{3}+\frac{4}{3}}=\sqrt{1+\frac{12}{3}}=\sqrt{5}$
Step 4
Therefore, norm of the matrix is,
$‖Ux‖=\sqrt{5}$

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