Linear transformations and change of basis Let $T:\in \Bbb R^3  \rightarrow \Bbb R^3$ be the linear transformation given by reflecting across the plane $-x_1+x_2+x_3=0$ (...) 
If $S=\{x:-x_1+x_2+x_3=0\} \implies S=\gen[(1,1,0),(1,0,1)]$
But how can I get the reflection matrix across $(1,1,0)$ and $(1,0,1)$?
Thanks!

You can find the projection matrix P onto S using $P=A(A^{T}A)^{-1}A^{T}$ where $A=\begin{bmatrix}1&1\\1&0\\0&1\end{bmatrix}$, 
which gives $P=\begin{bmatrix}\frac{2}{3}&\frac{1}{3}&\frac{1}{3}\\\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\\\frac{1}{3}&-\frac{1}{3}&\frac{2}{3}\end{bmatrix}$. 
Another way to do this is to find an orthonormal basis $\{q_1,q_2\}$ for S using the Gram-Schmidt process, and then use $P=QQ^{t}$ where Q is the matrix with columns $q_1$ and $q_2$.
Then the reflection matrix for S is given by $R=2P-I=\begin{bmatrix} \frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\ \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ \frac{2}{3} & -\frac{2}{3} & \frac{1}{3} \end{bmatrix}$.
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Another approach would be to use the fact that R takes the vectors (1,1,0) and (1,0,1) to themselves, and takes the vector (-1,1,1) to (1,-1,-1), so
$\;\;\;\;$R$\begin{bmatrix}1 & 1 &-1\\1 & 0 &1\\0 & 1 & 1\end{bmatrix}=\begin{bmatrix}1 & 1 & 1\\1 & 0 & -1\\0 & 1 & -1\end{bmatrix}$ and $R=\begin{bmatrix}1 & 1 & 1\\1 & 0 & -1\\0 & 1 & -1\end{bmatrix}\begin{bmatrix}1 & 1 &-1\\1 & 0 &1\\0 & 1 & 1\end{bmatrix}^{-1}$.
