# Eigenvalues and eigenvectors of reflection matrix

I'm working on the following problem:

Let A be a reflection matrix, such that, $$a_{ij}=\delta_{ij}-2n_{i}n_{j}$$, about a plane perpendicular to $$\vec{n}$$, $$\vec{n}$$ being the unitary vector. Find its eigenvalues and eigenvectors algebraically.

My first thought was on using $$A\vec{v}=\lambda\vec{v} \Rightarrow (A-\lambda I)\vec{v}=0$$, but I got an absolute huge equation, given that the matrix was $$\begin{bmatrix} 1-2n_1^2-\lambda & -2n_1n_2 & -2n_1n_3 \\ -2n_2n_1 & 1-2n_2^2-\lambda & n_2n_3 \\ -2n_3n_1 & n_3n_2 & 1-2n_3^2-\lambda \\ \end{bmatrix}$$ So, $$det(A)$$ has given a lot of terms to manipulate. I was thinking if there is a different approach to this problem, maybe using determinant properties or writing the matrix on a different basis, but could not develop any further.
Any tips?

• Since $A^2 = I$, the only eigenvalues can be $1$ and $-1$. – Ben Grossmann Mar 11 '19 at 17:53
• Or a bunch of other complex numbers. – Vasily Mitch Mar 11 '19 at 17:58
• @VasilyMitch Which other complex numbers ? As far as I know, the only roots of $x^2=1$ are $x=\pm 1$. – gandalf61 Mar 12 '19 at 9:21
• Sorry, my bad. I didn't read your comment correctly. – Vasily Mitch Mar 12 '19 at 10:28

If a vector is parallel to $$n_i$$, then $$a_{ij}(\alpha n_j) = \alpha(n_i-2n_in_jn_j) = -\alpha n_j.$$

So $$n_i$$ is a first eigenvector with eigenvalue $$-1$$.

All vectors $$v_i$$ from subspace $$R^n/n_i$$ are orthogonal to $$n_i$$. Then, $$a_{ij}v_j = v_i - 2n_in_jv_j = v_i.$$ So they are all are eigenvectors with eigenvalue $$1$$. You can choose any $$n-1$$ to form an orthogonal basis.

Hint: think geometrically.

What does $$A$$ map $$\vec n$$ itself to ?

A vector $$\vec m$$ such that $$\vec m . \vec n = 0$$ is the in the plane of the reflection. So what does $$A$$ map $$\vec m$$ to ?

• I forgot to add that the problem requires an algebraic solution. – Lincon Ribeiro Mar 11 '19 at 18:05

Hint This problem illustrates well the follow principle: Use a basis adapted to the geometry of the transformation. In this case, pick a basis $$({\bf e}_1, \ldots, {\bf e}_{n - 1})$$ of $${\bf n}^{\perp} := \{{\bf x} \in \Bbb R^n : {\bf x} \cdot {\bf n} = 0\}$$ and compute the matrix representation of $$A$$ with respect to the basis $$({\bf n}, {\bf e}_1, \ldots, {\bf e}_n) .$$

• I am not sure if I understand how to do it. Suppose I have a basis with an axis paralell to $\vec{n}$. So, the unitary vectors could be: (1,0,0),(0,1,0) and $\vec{n}=(0,0,1)$. If I compute the matrix representation on those, I will get the same $A$. Does it make sense? – Lincon Ribeiro Mar 11 '19 at 20:01
• There's no reason to restrict to an orthogonal/unitary basis---the eigenvalues and eigenvectors are independent of the choice of basis. Using the definition of $A$, what is $A {\bf n}$? What is $A {\bf e}_i$? – Travis Willse Mar 11 '19 at 20:50