Reflections over a plane. Suppose $E$ is a three-dimensional Euclidean vector space.
The reflection plane $F$ through the origin can be defined by a vector $v$ which is orthogonal to the plane. The reflection of a point $x$ about this plane is the linear isometry of $E$.
$$s_Fx = x-2\frac{\langle v,x \rangle}{|v|^2}v$$
Now I want to show that $-s_F$ is the reflection in a line orthogonal to that plane, or similarly the rotation of $180^\circ$ around that axis.
I can easily do this if $F$ is the plane orthogonal to a base vector of E. Since then for a unit vector $e_i$ orthogonal to $F$ and belonging to the basis of E: $s_Fx = x - 2\langle e_i, x \rangle e_i = x-2x_i$ and thus $-s_Fx = -x + 2x_i$. And thus the different kind of reflections can be easily shown.
Now I was wondering if I could choose a similar approach to an arbitrary $F$ one that isn't necessarily orthogonal to one of the base vectors of $E$. I guess taking any change of basis to a basis triple including the the orthogonal vector could let me give a similar argument.
 A: Yes, you are free to change to another orthonormal basis with $f_1,f_2\in F$.
Geometrically, observe that $x\mapsto -x$ is the composition of reflections through the planes $(e_1,e_2),\ (e_1,e_3),\ (e_2,e_3)$ (in any order) for any orthogonal basis $e_1,e_2,e_3$, and the composition of any two of these is the reflection through their intersection. 
Thus, with the basis $f_1,f_2,f_3$ above, we get
$$-s_F=\left(s_{(f_2,f_3)}\circ s_{(f_1,f_3)}\circ s_{(f_1,f_2)}\right)\circ s_F=s_{(f_2,f_3)}\circ s_{(f_1,f_3)}=s_{f_3}\,. $$
A: If I understand well, you are looking for a matrix expression for an "oblique" symmetry wrt to the given plane (with basis $x_1,x_2$) parallel to the given vector $x_3$ (pictured in red on the following figure), this vector being in general non normal to the plane.

We know that the orthogonal reflection with respect to horizontal plane $xOy$ is 
$$S=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}$$
Now, let us "adapt" this transformation to the new base $x_1,x_2,x_3$. More explicitly, let us define :
$$B=[x_1|x_2|x_3]$$ 
where the $x_i$s are here meant as the coordinates of vectors $x_i$ wrt to the canonical basis.
The looked for oblique symmetry matrix is therefore merely a change of basis operation :
$$S=BSB^{-1}$$
Otherwise said, the eigenvalue-eigenvector decomposition of the symmetry matrix...
