I want to find the affine map of the reflection through the subspace $U\subseteq\mathbb R^3$ defined by the equation $$2x-y+z=1.$$
I did what I deemed to be sensible and would like to verify whether it's correct or contains errors:
At first I described the subspace with parameters: $$U=\begin{pmatrix}0\\0\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\-2\end{pmatrix}+\mu\begin{pmatrix}0\\1\\1\end{pmatrix},\ \lambda,\mu\in\mathbb R.$$
Now I computed the reflection through the parallel subset $U_0$ through the origin $$U_0=\lambda\begin{pmatrix}1\\0\\-2\end{pmatrix}+\mu\begin{pmatrix}0\\1\\1\end{pmatrix},\ \lambda,\mu\in\mathbb R.$$ using the fact that the two vectors must be eigenvectors corresponding to the eigenvalue $1$ and that their cross product must be an eigenvector corresponding to the eigenvalue $-1$. I ended up with $$A=\frac{1}{3}\begin{pmatrix}-1&2&-2\\2&2&1\\-2&1&2\end{pmatrix}.$$
Now I assume that the reflection $f$ through $U$ is of the form $f(x)=Ax+t$ for some $t\in\mathbb R^3$ and since for $v=(0,0,1)^T\in U$ we have $f(v)=v$ I can just plug it into $f$ which yields $t$ to be $$t=\frac{1}{3}(2,-1,1)^T.$$ To sum it up the reflection $f$ through $U$ must be $$f(x)=Ax+t=\frac{1}{3}\begin{pmatrix}-1&2&-2\\2&2&1\\-2&1&2\end{pmatrix}+\frac{1}{3}\begin{pmatrix}2\\-1\\1\end{pmatrix}.$$ Are my thoughts correct?
Looks good to me. There’s a simpler way to arrive at $A$, however.
Following a standard derivation, observe that a reflection through $U_0$ reverses the component of a vector $\mathbf v$ that’s orthogonal to this subspace. I.e., if you decompose $\mathbf v$ into its components $\mathbf v_\parallel+\mathbf v_\perp$ in $U_0$ and $U_0^\perp$, respectively, then its reflection through $U_0$ is $\mathbf v_\parallel-\mathbf v_\perp$. (This is equivalent to your statement about the eigenvectors of this reflection.) However, $\mathbf v_\parallel = \mathbf v-\mathbf v_\perp$, so the reflection can also be written as $\mathbf v-2\mathbf v_\perp$, reducing the problem to one of computing the matrix for projection onto $U_0^\perp$.
From the equation of $U$ we know that $U_0^\perp$ is spanned by $\mathbf n = (2,-1,1)^T$, so you just need to find the matrix for projection onto this single vector. Using the fact that $\mathbf v^T\mathbf n$ is a scalar, hence commutes with any matrix, and regrouping, $$A\mathbf v = \mathbf v - 2{\mathbf v^T\mathbf n\over\mathbf n^T\mathbf n}\mathbf n = \left(I-2{\mathbf n\mathbf n^T\over\mathbf n^T\mathbf n}\right)\mathbf v.$$