Find the matrix of ortogonal reflection

Question

Let $e_1, e_2, e_3$ be an orthonormal basis for $R^3$ and consider the plane with equation $x_1 + 2x_2 - 2x_3 = 0$. Find the matrix of orthogonal reflection in that plane with respect to the given basis.

So, first I know that an orthogonal reflection satisfies: 1. F is a reflection 2. F is symmetric

So if I could find a transformation matrix, I would easily be able to verify that it is a reflection by the shape of the matrix and by asserting that $A^2 = I$. However, I don't know how to find such a matrix.

Answer 1

Hint:

I suppose that ${e_i}$ is the standard basis. In this case you can find the matrix from a geometrical construction.

If $\vec u$ is the vector orthogonal to the plane, the projection of a vector $x=[x_1,x_2,x_3]^T$ on $\vec u$ is: $$ \vec v= \frac {\langle \vec x, \vec u \rangle}{|u|^2}\vec u $$ and the reflection of $\vec x$ in the plane gives a vector

$$\vec x'=\vec x - 2 \vec v$$

In your case $\vec u=[1,2,-2]^T$. can you do from this?

Answer 2

If $\vec{n}$ is the normal vector of a given plane, then for any other vector $\vec{v}$ we have that $\vec N=\dfrac{\langle \vec v,\vec n\rangle}{\langle \vec n,\vec n\rangle}\vec n$ is the orthogonal projection of $\vec v$ on $\vec n$, so $\vec N$ is another perpendicular vector to the plane.

Let $\vec u=\vec v-\vec N$ and note that $\vec u$ is perpendicular to $\vec n$, because $$\langle\vec u,\vec n\rangle=\langle\vec v,\vec n\rangle-\langle\vec N,\vec n\rangle=\langle\vec v,\vec n\rangle-\dfrac{\langle\vec v,\vec n\rangle\langle\vec n,\vec n\rangle}{\langle\vec n,\vec n\rangle}=\langle\vec v,\vec n\rangle-\langle\vec v,\vec n\rangle=0$$

Then we can express the vector $\vec v$ as $$\vec v=\underbrace{\left(\vec v-\dfrac{\langle \vec v,\vec n\rangle}{\langle \vec n,\vec n\rangle}\vec n\right)}_{\text{in the plane}}+\underbrace{\left(\dfrac{\langle \vec v,\vec n\rangle}{\langle \vec n,\vec n\rangle}\vec n\right)}_{\text{perpendicular to the plane}}$$

The orthogonal reflection of $\vec v$ with respect to the plane can be found by just reversing the direction of the normal component. Then the orthogonal reflection is given by $$\vec w=\left(\vec v-\dfrac{\langle \vec v,\vec n\rangle}{\langle \vec n,\vec n\rangle}\vec n\right)-\dfrac{\langle \vec v,\vec n\rangle}{\langle \vec n,\vec n\rangle}\vec n=\vec v-2\dfrac{\langle \vec v,\vec n\rangle}{\langle \vec n,\vec n\rangle}\vec n$$

In your example we have $\vec v=(x,y,z)$ and $\vec n=(1,2,-2)$. Then, \begin{align*} R(x,y,z)&=(x,y,z)-\dfrac{2(x+2y-2z)}{9}(1,2,-2)\\&=\dfrac19(7x-4y+4z,\,-4 x + y + 8 z,\,4 x + 8 y + z) \end{align*} and the matrix (in the standard basis) that you are looking for is $$A=\begin{bmatrix}\dfrac79&-\dfrac49&\dfrac49\\-\dfrac49&\dfrac19&\dfrac89\\\dfrac49&\dfrac89&\dfrac19\end{bmatrix}$$

Indeed $A$ is symmetric and $A^2=I$, as you expected !