Picture of the problem I'm trying to solve
I did my best to mathjax the question: Consider the linear transformation T:R2 to R2 that first rotates a vector with pi/4 radians clockwise and then projects onto the x2 axis
(a) Find $$ T\begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} $$
Hello! I am confused on how to solve this problem - specifically (a). If someone could teach me step by step how to solve (a) I would be very grateful! I want to try solving (b) and (c) by myself after I learn how to do (a).
What I have tried to do so far on (a) is first I did- $$ \left[ \begin{array}{cc|c} \cos(\pi/4)&-\sin(\pi/4)\\ \sin(\pi/4)&\cos(\pi/4) \end{array} \right] $$
solved that matrix and whatever matrix I got as an answer I multiplied it by the x2 matrix
$$ \left[ \begin{array}{cc|c} -1&0\\ 0&1 \end{array} \right] $$
and whatever was the result of that I would multiply by matrix
$$ \left[ \begin{array}{cc|c} 1\\ 1 \end{array} \right] $$
and that was my result for (a) but i have no idea if what i just did is totally wrong or what steps I need to take to get to the right answer?
Thank you!
For part (a), let's go over step by step. This will help for part (b) as well.
The vector $\begin{bmatrix}1\\1\end{bmatrix}$ has a length of $\sqrt{2}$, rotated $\pi/4$ counterclockwise from $x_1$ axis. When you rotate it clockwise by $\pi/4$, it will just point along $x_1$. You can write this in a matrix form: $$\begin{bmatrix}\cos(\pi/4)&\sin(\pi/4)\\ -\sin(\pi/4)&\cos(\pi/4)\end{bmatrix} \begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}\sqrt 2/2+\sqrt 2/2\\-\sqrt2/2+\sqrt2/2\end{bmatrix}=\begin{bmatrix}\sqrt2\\0\end{bmatrix}$$ The next step is to project this vector onto the $x_2$ axis. What that means is that the second component is unchanged, and the first component become $0$. Then the final answer for (a) is $$T\left(\begin{bmatrix}1\\1\end{bmatrix}\right)=\begin{bmatrix}0\\0\end{bmatrix}$$ The projection matrix onto the $x_2$ axis can be written as $$\begin{bmatrix}0&0\\0&1\end{bmatrix}$$ since $$\begin{bmatrix}0&0\\0&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}0\\x_2\end{bmatrix}$$ When we apply this to the $\begin{bmatrix}\sqrt2\\0\end{bmatrix}$ vector we obtain $\begin{bmatrix}0\\0\end{bmatrix}$.
To summarize, we applied first the rotation (we multiplied the vector on the right with the rotation matrix), then we projected onto $x_2$ (we multiplied on the right the previous result by the projection matrix). So we write $$T\left(\begin{bmatrix}1\\1\end{bmatrix}\right)=\begin{bmatrix}0&0\\0&1\end{bmatrix}\left(\begin{bmatrix}\cos(\pi/4)&\sin(\pi/4)\\ -\sin(\pi/4)&\cos(\pi/4)\end{bmatrix} \begin{bmatrix}1\\1\end{bmatrix}\right)$$
Therefore $$T=\begin{bmatrix}0&0\\0&1\end{bmatrix}\begin{bmatrix}\cos(\pi/4)&\sin(\pi/4)\\ -\sin(\pi/4)&\cos(\pi/4)\end{bmatrix}=\begin{bmatrix}0&0\\-\sqrt2/2&\sqrt2/2\end{bmatrix}$$