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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?

enter image description here

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  • 2
    $\begingroup$ Your matrix is a counter-clockwise rotation. You need the minus sign in the bottom left $\sin(45)$ for clockwise. $\endgroup$ – ryan221b May 19 at 18:54

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