Let $Q_1$: $R^2$ → $R^2$ be the linear transformation that reflects vectors about the line $y$ $=$ $x$. Determine where the zero vector, $\begin{pmatrix}-1\\ 0\end{pmatrix}$, and $\begin{pmatrix}5\\ 1\end{pmatrix}$ are mapped to by $Q_1$.
Let $R−π/4$ denote the clockwise rotation by 45 degrees, let $Q_0$ denote the reflection about the x-axis, and let $R_{π/4}$ denote the counterclockwise rotation by 45 degrees. Prove that $Q_1 = R_{π/4} \circ Q_0\circ R_{−π/4}$.
I managed to figure out the first part. The zero vector would map to zero, $\begin{pmatrix}-1\\ 0\end{pmatrix}$ would map to $\begin{pmatrix}0\\ -1\end{pmatrix}$, and $\begin{pmatrix}5\\ 1\end{pmatrix}$ would map to $\begin{pmatrix}1\\ 5\end{pmatrix}$.
But I am not sure about the second part. I know I have to derive the matrices for each of those transformations and multiply them in that order, but I am not sure how to find them.
Any help would be highly appreciated!
Find a basis $B$, for $\mathbb{R}^2$, if for all $x \in B$, if $T_1(x)=T_2(x)$, then $T_1=T_2$.
For example, it could have pick it to be the standard unit basis.
Alterantively, let $B= \{(1,1)^T, (1,-1)^T\}.$
$$Q_1(1,1)^T=(1,1)^T$$
$$R_{\frac{\pi}4}Q_0R_{\frac{-\pi}4}(1,1)^T=R_{\frac{\pi}4}(\sqrt2,0)=(1,1)^T$$
$$Q_1(1,-1)^T=(-1,1)^T$$
$$R_{\frac{\pi}4}Q_0R_{\frac{-\pi}4}(1,-1)^T=R_{\frac{\pi}4}(0,\sqrt2)=(-1,1)^T$$