Reflections in Euclidean plane Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be the counterclockwise rotation of $\frac{\pi}{2}$ and $S: \mathbb{R}^2 \to \mathbb{R}^2$ be the reflection w.r.t. the line $x+3y=0$. There exists a reflection $R$ such that $T^{-1}ST=R$? Is there a canonical way to find which is the line w.r.t. we are reflection through $R$?
 A: One straightforward way is to just compute $R$ and then find an eigenvector for it whose eigenvalue even is 1.
This can be improved a little by noticing that if $v$ is an eigenvector for S, then $T^{-1}v$ is an eigenvector for $R$ with the same eigenvalue. You can just take the line of reflection for $S$ and transform it. 
A: suppose $T^{-1}ST $ fixes a line $l.$ then $$T^{-1}ST(l) = l \implies ST(l) = T(l). $$  that is $T(l)$ is fixed by $S$ and we know that only lines fixed by $S$ is the line $3x+y = 0.$  therefore $T(l)$ is $y-3x = 0$ and $$  T^{-1}ST \text{ is a reflection on the line } y - 3x = 0.$$
A: Let boldfaces represent matrices for corresponding linear transformations, presented in the standard basis.
$\textbf{T}^{-1}\textbf{ST}$ represents the same linear transformation $S$, only presented in the basis $\mathcal B$ whose coordinates in the standard basis are given as the columns of $\textbf T$. However you want to see what transformation $\textbf{T}^{-1}\textbf{ST}$ represents in the standard basis. It is of course a reflection about a line whose equation is $x_1+3y_1=0$, but "in the other basis" $\mathcal B$. That is, in this equation $x_1,y_1$ represent the coordinates of the position vector in the basis $\mathcal B$. Note here that the basis is $\mathcal B = \{(0,1),(-1,0)\}$ so to revert back to the standard basis, we have $x_1=y$ ($x_1$ is the length of the component of the position vector along $(0,1)$, which is the same as $y$) and similarly $y_1=-x$ so the equation of the line about which the transformation is a reflection is $y-3x=0$.
