# Linear Algebra: Matrices and Linear Equations - Multiplication of Matrices - Question #1

# Question

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?

# Answer

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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