Krdytkyu

Let $T:mathbb{R}^3rightarrow mathbb{R}^3$ be the reflection across the plane $x+2y+3z=0$, find the matrix of this linear operator $T$ in respect to the basis $B=left{v_1,v_2,v_3right}$, where we have:

$v_1=begin{bmatrix}1\ 1\ -1end{bmatrix}$ $v_2=begin{bmatrix}-1\ 2\ -1end{bmatrix}$ $v_3=begin{bmatrix}1\ 2\ 3end{bmatrix}$

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:

$T=begin{bmatrix}1&-1&-1\ 1&2&-2\ -1&-1&-3end{bmatrix}$

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?

The person who prepared you this question has made life very easy for you. Verify following facts:

1. $v_1 in $ the plane (its coordinates verify the equation of the plane)


2. $v_2$ also lies in this plane.


3. $v_3 perp v_1$ (calculate the dot product)


4. $v_3 perp v_2$

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 $begin{pmatrix} 1 & 0 & 0\ 0 & 1 & 0 \ 0& 0& -1end{pmatrix}$

There is more to it. You can find a canonical basis (make an eigenvalue decomposition)
$${bf T = S}^{-1}{bf DS}$$

where $${bf D} = begin{bmatrix}1&0&0\0&1&0\0&0&-1end{bmatrix}$$
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

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


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


3. reassemble our vector.

That is basically what multiplying with ${bf S}^{-1}bf DS$ would mean step-by-step.

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 $Brightarrow B_{std}$.

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 $R'=TRT^{-1}=begin{bmatrix}
frac{6}{7} & frac{-2}{7} & frac{-3}{7}\
&&&\
frac{-2}{7} & frac{3}{7} & frac{-6}{7}\
&&&\
frac{-3}{7} & frac{-6}{7} & frac{-2}{7}
end{bmatrix}$

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:
$$vec r_n=frac{vec rbulletvec n}{vec nbullet vec n}vec n$$
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
$$vec r'=vec r-2vec r_n$$
or
$$vec r'=vec r-2frac{vec rbulletvec n}{vec nbullet vec n}vec n$$
If we substitute the numerical value of the plane normal we get:
$$vec r'=vec r-2frac{x+2y+3z}{14}(1,2,3)$$
or
$$left{begin{array}{c}x'=x-frac{x+2y+3z}{7}\y'=y-2frac{x+2y+3z}{7}\z'=z-3frac{x+2y+3z}{7}end{array}right.$$
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:
$$R=begin{bmatrix}
frac{6}{7}& frac{-2}{7} & frac{-3}{7}\
& & &\
frac{-2}{7} & frac{3}{7} & frac{-6}{7}\
& & &\
frac{-3}{7} & frac{-6}{7} & frac{-2}{7}
end{bmatrix}$$