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# Question

Reference - Question2

Because it is just rotation, the negative rotation is described by positive rotation.

Therefore, if $y^2 = x^2$ is shown to be established, regardless of whether its root is positive or negative, is the proof completed?

Below is a reply from Jaemin Shin, PhD.

You need know the definition of $\left\|\cdot\right\|$, called the norm, first. There are several ways to define the norm but in general in $R^{2}$ it is defines as follows (called 2-norm); for $X = \mathbf{(x_{1}, x_{2})}^\top \in R^2$

$\left\|X\right\| = \sqrt{\left\|x_{1}\right\|^2 + \left\|x_{2}\right\|^2}$

or it is equivalent to

$\left\|X\right\| = (\mathbf{X}^{\top}X)^{\frac{1}{2}}$

Here $\mathbf{X}^\top$ is the transpose of the vector.  Note that

$\mathbf{(AB)}^\top = \mathbf{B}^\top\mathbf{A}^\top$

and

$\mathbf{R(\theta)}^\top = R(\theta)^{-1} = R(-\theta)$

Thus,

$\left\|Y\right\|^{2} = \left\|RX\right\|^{2} = \mathbf{(RX)}^{\top}(RX)=(\mathbf{X}^{\top}\mathbf{R}^{\top})(RX)$

Using the associative law

$(\mathbf{X}^{\top}\mathbf{R}^{\top})(RX) = \mathbf{X}^{\top}(\mathbf{R}^{\top}R)X = \mathbf{X}^{\top}(R^{-1}R)X = \mathbf{X}^{\top}X = \left\|X\right\|^2$

Since the norm is non-negative, it follows that

$\left\|Y\right\| = \left\|X\right\|.$

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