I have read the chapter in my textbook on Linear Transformations, but I'm still at a bit of a loss on a particular question.
Find the standard matrix representation of the following linear transformations, $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$
A) Rotation by 45 degrees counterclockwise followed by reflection in the line $y=-x$
B) Projection in the line $y=\frac{x}{2}$ followed by rotation by 60 degrees clockwise.
I attempted part A, and these are my results.
$R$ is the rotation, and $F$ is the reflection
$R_{45ccw}=\begin{bmatrix}
0 & -\frac{\sqrt{2}}{2}\\
1 & \frac{\sqrt{2}}{2}
\end{bmatrix}$ and $F_{y=-x}=\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}$
Then $[F_{y=-x}][R_{45ccw}]=\begin{bmatrix}
-1 & -\frac{\sqrt{2}}{2}\\
0 & -\frac{\sqrt{2}}{2}
\end{bmatrix}$
I'm not sure if I followed the process correctly or not. If I did, I'm not quite sure how to get started on part B of the problem.
Best Answer
For part A) your procedure is correct, but your matrices are not. For a $45$-degree rotation, it should be $$\left(\begin{array}{cc}\cos(\pi/4)&-\sin(\pi/4)\\\sin(\pi/4)&\cos(\pi/4)\end{array}\right) = \frac{\sqrt{2}}{2}\left(\begin{array}{cc}1&-1\\1 &1\end{array}\right).$$ For instance we know this rotation should take the vector $(1,0)^T$ to $(\sqrt{2}/2,\sqrt{2}/2)^T$ and you can check that this is the case.
For a reflection over the line $y=-x,$ it is $$ \left(\begin{array}{cc}0 & -1\\-1&0\end{array}\right)$$ which you can see is plausible by checking that it takes the vector $(1,1)^T$ to $(-1,-1)^T$ and $(1,-1)^T$ to $(1,-1)^T.$ Can you see by drawing a picture that this reflection should take $(x,y)^T$ to $(-y,-x)^T$? Another guideline is that rotations always have determinant $1$ and reflections have determinant $-1$.
For part B), the rotation can be done using the same formula as above but with $\pi/4$ replaced by $-\pi/3.$ For the projection, start by figuring out what it must do to some test vectors. For instance it must take $(1,1/2)^T$ to $(1,1/2)^T.$ What must it do to, say $(1,0)$? You need to figure out how to project it onto the line $y=x/2$ which is a matter of drawing some triangles. How about $(1,-2)^T$? It should go to $(0,0)^T.$ Do you see why? After you do this for a few test vectors, you should be able to narrow down on what the form of the matrix needs to be. It should have determinant zero and eigenvalues $1$ and $0.$