Question

Let P be a $2\times 2$ real orthogonal matrix and $\overrightarrow{x}$ is a real vector $[x_1,x_2{]}^T$ with length $||\overrightarrow{x}||=(X_1^2+X_2^2{)}^{\frac{1}{2}}$.
Then, which one of the following statements is correct?

• A. $||P\overrightarrow{x}||\le ||\overrightarrow{x}||$ where at least on vector satisfies $||P\overrightarrow{x}||<||\overrightarrow{x}||$
• B. $||P\overrightarrow{x}||=||\overrightarrow{x}||$ for all vectors $\overrightarrow{x}$
• C. $||P\overrightarrow{x}||\ge ||\overrightarrow{x}||$ where at least one vector satisfies $||P\overrightarrow{x}||>||\overrightarrow{x}||$
• D. No relationship can be established between $||\overrightarrow{x}||$ and $||P\overrightarrow{x}||$

Answer 1

The correct option is B $||P\overrightarrow{x}||=||\overrightarrow{x}||$ for all vectors $\overrightarrow{x}$

Let P an orthogonal matrix

$P=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]$
So $P\overrightarrow{X}=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$
$=\left[\begin{array}{c}{x}_{1}cos\theta +x_{2}sin\theta \\ -x_{1}sin\theta +x_{2}cos\theta \end{array}\right]$

$||P\overrightarrow{x}||=\sqrt{(x_{1}cos\theta +x_{2}sin\theta {)}^2+(-x_{1}sin\theta +x_{2}cos\theta {)}^2}$
$=\begin{array}{c}\sqrt{x_1^{2}co{s}^2\theta +x_2^{2}si{n}^2\theta +2x_1{x}_{2}cos\theta sin\theta +x_1^{2}si{n}^2\theta +x_2^{2}si{n}^2\theta -2x_1{x}_{2}sin\theta cos\theta }\end{array}$
$=\sqrt{x_1^2+x_2^2}=||\overrightarrow{x}||$
$\Rightarrow ||P\overrightarrow{x}||=||\overrightarrow{x}||forany\overrightarrow{x}$