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!
 A: Hint: The normal vector of the plane is $\vec n=(-1,1,1)$. Given any vector $\vec x=(x,y,z), $ the projection of $\vec x$ onto $\vec n$ is given by:
$$ Projection _{\, \vec x}\, \vec n = \frac{\vec x · \vec n}{\vec n · \vec n} \vec n $$
And the reflection of $\vec x$ across the plane is:
$$ Reflection (\vec x)=\vec x - 2\frac{\vec x · \vec n}{\vec n · \vec n} \vec n$$
Which is, in your case:
$$=(x,y,z)-2\frac{(x,y,z) · (-1,1,1)}{(-1,1,1)·(-1,1,1)}(-1,1,1)=(x,y,z)+(x-y-z) \left( -\frac{2}{3},\frac{2}{3},\frac{2}{3}\right) = \frac{1}{3}(x+2y+2z,2x+y-2z,2x-2y+z) $$
Now that you have the transformed of $\vec x$ you can find the matrix. Quite messy but I hope you understood
A: 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 $R=\begin{bmatrix}1 & 1 & 1\\1 & 0 & -1\\0 & 1 & -1\end{bmatrix}\begin{bmatrix}1 & 1 &-1\\1 & 0 &1\\0 & 1 & 1\end{bmatrix}^{-1}$.
