Rotation of a matrix through $\cfrac{-\pi}{4}$

Question

So this is my question.

$T:R2→R2$ rotates points (about the origin) through $\cfrac{-\pi}{4}$ (clock-wise)

So what I do know that I need is that the origin is represented as $$ T = \begin{bmatrix} cos \theta &-sin\theta \\ sin \theta & cos \theta \\ \end{bmatrix} $$

Then I know that rotation is $\cfrac{-\pi}{4}$.

I was also given the hint that $$T(e_1) = \begin{bmatrix} 1/\sqrt2 & 1/\sqrt2 \\ -1/\sqrt2 & 1/\sqrt2 \\ \end{bmatrix} $$

I'm not exactly sure how to proceed with this to get $$T(e_1)$$ I'm thinking I need to do something like

$$ T = \begin{bmatrix} cos (-\pi/4) &-sin(-\pi/4) \\ sin (-\pi/4) & cos (-\pi/4) \\ \end{bmatrix} $$

Am I headed in the right direction?

Answer 1

Yes, You are in the good direction.

Answer 2

So just in case, anyone is wondering. I came to the $ T(e_1) $ answer by using output of $ cos(-\pi/4)$, $-sin(-\pi/4)$, $sin(-\pi/4)$, and $cos(-\pi/4)$ because $cos(-\pi/4)$ is $1/\sqrt2$, $-sin(-\pi/4)$ is $1/\sqrt2$, $sin(-\pi/4)$ is $-(1/\sqrt2)$, and $cos(-\pi/4)$ is $1/\sqrt2$. This concludes to $ T(e_1) = \begin{bmatrix} 1/\sqrt2 & 1/\sqrt2 \\ -1/\sqrt2 & 1/\sqrt2 \\ \end{bmatrix} $