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?
Best Answer
The person who prepared you this question has made life very easy for you. Verify following facts:
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 $\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0& 0& -1\end{pmatrix}$