The matrix representation of a reflection operator across the plane $x+2y+3z=0$ Let $T:\mathbb{R}^3\rightarrow \mathbb{R}^3$ be the reflection across the plane $x+2y+3z=0$, find the matrix of this linear operator $T$ in respect to the basis $B=\left\{v_1,v_2,v_3\right\}$, where we have:
$v_1=\begin{bmatrix}1\\ 1\\ -1\end{bmatrix}$  $v_2=\begin{bmatrix}-1\\ 2\\ -1\end{bmatrix}$ $v_3=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$
First of what does it mean that $T$ is a reflection across the given plane? What will hapen to let's say the vector $(1,0,0)$ if it's passed through $T$? In my textbook, there was a short example that mentioned a reflection operator, which transformed a given vector $(a,b,c)$ to the vector $(a,b,-c)$ So that would be the reflection across the standard plane $x+y+z=0$, but here, will it act as if given a vector $(a,b,c)$ you get out a vector $(a,2b,-3c)$?
EDIT:
after reading the comments, I got the matrix representation as:
$T=\begin{bmatrix}1&-1&-1\\ 1&2&-2\\ -1&-1&-3\end{bmatrix}$
Now how can I find the representation of this operator in respect to the standard basis vectors $(e_1,e_2,e_3)$?
I read about finding the change of basis matrix so I calculated $T^{-1}$:
$T^{-1}=\begin{bmatrix}\frac{4}{7}&\frac{1}{7}&-\frac{2}{7}\\ -\frac{5}{14}&\frac{2}{7}&-\frac{1}{14}\\ -\frac{1}{14}&-\frac{1}{7}&-\frac{3}{14}\end{bmatrix}$
Is this it or is there more to it?
 A: There is more to it. You can find a canonical basis (make an eigenvalue decomposition) 
$${\bf T = S}^{-1}{\bf DS}$$
where $${\bf D} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&-1\end{bmatrix}$$
and the two leftmost columns of $\bf S$ are in the plane, and the rightmost is perpendicular to it. The eigenvalues $1$ means "preserve" vector, which is what happen to the components which lie in the plane. -1 means to flip the direction 180 degrees which is what should happen to the part of the vectors pointing right out of the plane.
In other words, if we 


*

*write the vector to be as a linear combination of two vectors in the plane and one perpendicular to it.

*we can just flip the one perpendicular (multiply with -1)

*reassemble our vector. 


That is basically what multiplying with ${\bf S}^{-1}\bf DS$ would mean step-by-step.
A: The person who prepared you this question has made life very easy for you. Verify following facts:


*

*$v_1 \in $ the plane (its coordinates verify the equation of the plane)

*$v_2$  also lies in this plane.

*$v_3 \perp v_1$ (calculate the dot product)

*$v_3 \perp v_2$


So the reflection maps: $\begin{cases}
 v_1 \mapsto v_1 \\ v_2 \mapsto v_2 \\ v_3 \mapsto -v_3 \end{cases}$
And the matrix w.r.t. this basis is  $\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0& 0& -1\end{pmatrix}$
A: As noted above, vectors $v_1,v_2 \in p$ (plane),and $v_3$ is $\perp$ to $v_1,v_2$.
Since $v_1,v_2$ are in the plane their reflectection will be the same vector $v_1,v_2$ and since $v_3$ is the normal vector for this plane it's reflection will be the same vector just in the opposite direction ( $-v_3$).
So your reflection matrix in the base $B=\{v_1,v_2,v_3\}$ will look like this:
$R=\begin{bmatrix}
1 & 0 & 0\\ 
0 & 1 & 0\\ 
0 & 0 & -1
\end{bmatrix}_B$
To get the matrix representation in the standard base you can use the change of basis matrix $T$.
Notice that $T^{-1}$ is the matrix which will change $B_{std} \rightarrow B$ and $T$ will change $B\rightarrow B_{std}$.
The matrices will look like this:
$T^{-1}=\begin{bmatrix}
 \frac{4}{7}& \frac{1}{7} & \frac{-2}{7}\\ 
& & &\\
\frac{-5}{14} & \frac{2}{7} & \frac{-1}{14}\\ 
& & &\\
\frac{1}{14} & \frac{1}{7} & \frac{3}{14}
\end{bmatrix}
, T=\begin{bmatrix}
1 & -1 & 1\\ 
1 & 2 & 2 \\ 
-1 & -1 & 3 
\end{bmatrix}$
And finally $R'=TRT^{-1}=\begin{bmatrix}
 \frac{6}{7} & \frac{-2}{7} & \frac{-3}{7}\\ 
&&&\\
\frac{-2}{7} & \frac{3}{7} & \frac{-6}{7}\\ 
&&&\\
\frac{-3}{7} & \frac{-6}{7} & \frac{-2}{7}
\end{bmatrix}$
A: Let us call the plane normal as vector $\vec n=(1,2,3)$ and let the incident vector be $\vec r=(x,y,z)$ and the reflected vector be $\vec r'=(x',y',z')$. When the vector $\vec r$ is reflected upon the plane, its projection on the plane remains unchanged and only its normal component with respect to the plane will be reversed to become the reflected vector $\vec r'$. 
The normal component of $\vec r$ with respect to the plane is:
$$\vec r_n=\frac{\vec r\bullet\vec n}{\vec n\bullet \vec n}\vec n$$
And the projection of $\vec r$ on the plane is:
$$\vec r_p=\vec r-\vec r_n$$
So the reflected vector is:
$$\vec r'=-\vec r_n+\vec r_p$$
or:
$$\vec r'=-\vec r_n+\vec r-\vec r_n$$
or
$$\vec r'=\vec r-2\vec r_n$$
or
$$\vec r'=\vec r-2\frac{\vec r\bullet\vec n}{\vec n\bullet \vec n}\vec n$$
If we substitute the numerical value of the plane normal we get:
$$\vec r'=\vec r-2\frac{x+2y+3z}{14}(1,2,3)$$
or 
$$\left\{\begin{array}{c}x'=x-\frac{x+2y+3z}{7}\\y'=y-2\frac{x+2y+3z}{7}\\z'=z-3\frac{x+2y+3z}{7}\end{array}\right.$$
or 
$$\left\{\begin{array}{c}x'=\frac{6x-2y-3z}{7}\\y'=\frac{-2x+3y-6z}{7}\\z'=\frac{-3x-6y-2z}{7}\end{array}\right.$$
So the matrix reperesentation is:
$$R=\begin{bmatrix}
 \frac{6}{7}& \frac{-2}{7} & \frac{-3}{7}\\ 
& & &\\
\frac{-2}{7} & \frac{3}{7} & \frac{-6}{7}\\ 
& & &\\
\frac{-3}{7} & \frac{-6}{7} & \frac{-2}{7}
\end{bmatrix}$$
