# Transition matrix which rotates counter clockwise

The text says to write $(x,y)$ in $\mathbb R^2$ in terms of basis $$B'=\{(\cos t, \sin t), (-\sin t, \cos t)\}.$$

Since this is the new basis, we find the transition to it by putting it on the left and the standard basis on the right $$\begin{bmatrix} \cos t & -\sin t & 1 & 0 \\ \sin t & \cos t & 0 & 1 \end{bmatrix}.$$ The left matrix (left 2 columns) is a transform which rotates a point $t$ radians counterclockwise.

It reduces to $$\begin{bmatrix} 1 & 0 & \cos t & \sin t \\ 0 & 1 & -\sin t & \cos t \end{bmatrix} .$$

Then it says $(x,y)$ is in $\mathbb R^2$ and to find $(x',y')$ relative to $B'$ we multiply $$\begin{bmatrix}\cos t & \sin t \\ -\sin t & \cos t \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} =\begin{bmatrix} x' \\ y' \end{bmatrix}.$$

But the resulting equation rotates the point $(x,y)$ clockwise. Isn't something backwards?

The basis $B'$ is the result after counterclockwise rotation of the standard basis. For a fixed point in the plane, rotating axes could be viewed as rotating point in the opposite direction when considering the axes relatively still.