# Rotation of vector by rotation matrix

Assume the following expression

$$\begin{bmatrix} a_1^* \\ a_2^* \end{bmatrix} = \begin{bmatrix} \cos(45) & - \sin(45) \\ \sin(45) & \cos(45) \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}$$

Which is a 45 degree rotation clockwise.

Show where $$P=(2, 3)$$ is moved to as a result of this rotation.

If I multiply

$$\frac{\sqrt{2}}{2} \begin{bmatrix} 1 & - 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix}$$

I get

$$\frac{\sqrt{2}}{2} \begin{bmatrix} -1 \\ 5 \end{bmatrix}$$

Which I'm confused about as this would put it into the second quadrant, which doesn't seem to make sense for a clockwise rotation.

In the following image $$e_1, e_2$$ are the original axis and $$f_1, f_2$$ are the new axis after rotating.

How to graphically demonstrate where $$P$$ ends up?

• Your matrix is a counter-clockwise rotation. You need the minus sign in the bottom left $\sin(45)$ for clockwise. – ryan221b May 19 at 18:54