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Prove the angle of $v \in \mathbb R^2 $ after rotation by $R_{\theta}$ (rotation matrix) change $\theta$-degrees.

We have $R_{\theta} =\begin{bmatrix}\cos(\theta) & \sin(\theta)\\-\sin(\theta) & \cos(\theta)\end{bmatrix}$ is a rotation matrix in $\mathbb R^2$.

It is proven graphically that $R_{\theta}$ rotates the vector $v$ by $\theta$-degrees. However I'm interested in a rigorous proof.

Let $v=\begin{bmatrix}x \\y \end{bmatrix}$ and suppose the angle of $v$ is $\phi$ (with respect to the $x$-axis).

Then $\begin{bmatrix}\cos(\theta) & \sin(\theta)\\-\sin(\theta) & \cos(\theta)\end{bmatrix}\begin{bmatrix}x \\y \end{bmatrix}=\begin{bmatrix}x\cos(\theta) + y\sin(\theta)\\-x\sin(\theta) +y\cos(\theta)\end{bmatrix} = v^{'}$

We can easily verify that$ ||v^{'}||=||v||$, so the length of $v$ has been preserved.

How does I verify that the angle of $v^{'}$ is $\phi + \theta$ ?

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    $\begingroup$ Instead of $x,y$ write $x=r\cos\phi, y=r\sin\phi$. Then use some trig identities. $\endgroup$ – David Peterson Feb 27 '14 at 20:11
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You can look at it as multiplication in the complex plane $\mathbb{C}$.

For $c \in \mathbb{C},\ c = a + bi$ assign matrix $\begin{pmatrix}a & b\\-b & a\end{pmatrix}$.

If we interpret $x-yi$ as $\begin{pmatrix}x\\y\end{pmatrix}$, we have

$\begin{pmatrix}a & b\\-b & a\end{pmatrix}\begin{pmatrix}x \\y \end{pmatrix} = \begin{pmatrix}ax+by \\ ay - bx\end{pmatrix}$ which corresponds to $(ax + by) - (ay - bx)i$ which is exactly the result of complex multiplication $(a + bi)(x - yi)$.

If $|c|=1$ we can write $c = cos\ \theta + i\ sin\ \theta$ and you see the multiplication by $c$ is exactly the multiplication by $R_\theta$.

But multiplication by $c$ is rotation counterclockwise by $\theta$, which results in rotation clockwise by $\theta$ of your vector since we flipped the $y$ coordinate.

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