# 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?

• The vectors in the plane will be preserved (multiplied by 1) and the vectors perpendicular to the plane will be mirrored (multiplied by -1). Maybe that can help. – mathreadler Feb 4 '17 at 11:13
• @mathreadler so v1 will map into v1,v2 into v2 and v3 into -v3? – adadaae12313412 Feb 4 '17 at 11:17
• In general, there is a householder matrix that describes reflections of this type. – Andres Mejia Feb 4 '17 at 11:28
• v1 map onto v1, v2 onto v2, v3 onto -v3 – mathreadler Feb 4 '17 at 11:36
• @mathreadler can you review mt question, I edited it – adadaae12313412 Feb 4 '17 at 11:42

## 4 Answers

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

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

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

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.

• Okay, but can you answer this to me? After finding the matrice $T^{-1}$, we can find the matrice representation in respect to the standard basis by calculating $A=TET^{-1}$, where $E$ is the matrice made up of the vectors e1,e2,e3? – adadaae12313412 Feb 4 '17 at 11:58
• if $T$ is defined as you do, it will be like a $S^{-1}$ in my presentation and $E$ will be a diagonal matrix like $\bf D$. – mathreadler Feb 4 '17 at 12:06

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

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: $$R=\begin{bmatrix} \frac{6}{7}& \frac{-2}{7} & \frac{-3}{7}\\ & & &\\ \frac{-2}{7} & \frac{3}{7} & \frac{-6}{7}\\ & & &\\ \frac{-3}{7} & \frac{-6}{7} & \frac{-2}{7} \end{bmatrix}$$