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How can we with linear algebra prove that if two vectors in 2 dimensions form an angle $\phi$ then by multiplying those two vectors with the same rotation matrix the cosine of the formed angle will be preserved?

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up vote 2 down vote accepted

You can maybe use the fact that matrix corresponding to the rotation is orthogonal: $$R^T = R^{-1}$$ and that the norm of a vector after rotation is the same as the original one: $$||R \mathbf{v}|| = ||\mathbf{v}||$$

So, by direct computation, for two vectors $\mathbf{v}$ and $\mathbf{w}$, you have:

$$ \cos(\alpha_{R\mathbf{v},R\mathbf{w}}) = \frac{<R\mathbf{v},R\mathbf{w}>}{||R\mathbf{v}|| \ ||R\mathbf{w}||} = \frac{\mathbf{v}^T R^T R \mathbf{w}}{||\mathbf{v}|| \ ||\mathbf{w}||} = \frac{\mathbf{v}^T \mathbf{w}}{||\mathbf{v}|| \ ||\mathbf{w}||} = \cos(\alpha_{\mathbf{v},\mathbf{w}}) $$

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We have ${\bf a}\cdot{\bf b} = {\bf a}^T{\bf b} = ab\cos\phi$. Let $R$ be the rotation matrix. Then $$\begin{eqnarray*} {\bf a}^T {\bf b} &\rightarrow& (R{\bf a})^T {R\bf b} \\ &=& {\bf a}^T R^T R {\bf b} \\ &=& {\bf a}^T {\bf b}, \end{eqnarray*}$$ since $R$ is orthogonal.

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