In $\textbf{R}^{2}$, let $L$ be the line $y = mx$, where $m \neq 0$. Find an expression for $T(x,y)$, where
(a) $T$ is the reflection of $\textbf{R}^{2}$ about $L$.
(b) $T$ is the projection on $L$ along the line perpendicular to $L$.
MY ATTEMPT
(a) Let us consider the basis $\mathcal{B}' = \{(1,m),(-m,1)\}$. Then we must have $T(1,m) = (1,m)$ and $T(-m,1) = (m,-1)$.
Thus the matrix representation of $T$ in the basis $\mathcal{B}'$ is given by \begin{align*} [T]_{\mathcal{B}'} = \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix} \end{align*}
On the other hand, the change of basis matrix from $\mathcal{B}'$ to $\mathcal{B} = \{(1,0),(0,1)\}$ is given by \begin{align*} Q = \begin{bmatrix} 1 & -m\\ m & 1 \end{bmatrix} \Rightarrow Q^{-1} = \frac{1}{m^{2}+1} \begin{bmatrix} 1 & m\\ -m & 1 \end{bmatrix} \end{align*} Finally, we have that $[T]_{\mathcal{B}} = Q[T]_{\mathcal{B}'}Q^{-1}$: \begin{align*} [T]_{\mathcal{B}} =\frac{1}{m^{2}+1} \begin{bmatrix} 1-m^{2} & 1 + m\\ 2m & m^{2}-1 \end{bmatrix} \end{align*}
(b) In this case, the basis is the same: $\mathcal{B}' = \{(1,m),(-m,1)\}$. But $T(1,m) = (1,m)$ and $T(-m,1) = 0$. Thus we get that
\begin{align*} [T]_{\mathcal{B}'} = \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} \end{align*} from whence we get that $[T]_{\mathcal{B}} = Q[T]_{\mathcal{B}'}Q^{-1}$: \begin{align*} [T]_{\mathcal{B}} = \frac{1}{m^{2}+1} \begin{bmatrix} 1 & m\\ m & m^{2} \end{bmatrix} \end{align*}
Is my approach to solve this problem correct? Could someone propose another way to solve it?