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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?

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  • $\begingroup$ Since $A^2 = I$, the only eigenvalues can be $1$ and $-1$. $\endgroup$ Mar 11, 2019 at 17:53
  • $\begingroup$ Or a bunch of other complex numbers. $\endgroup$ Mar 11, 2019 at 17:58
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    $\begingroup$ @VasilyMitch Which other complex numbers ? As far as I know, the only roots of $x^2=1$ are $x=\pm 1$. $\endgroup$
    – gandalf61
    Mar 12, 2019 at 9:21
  • $\begingroup$ Sorry, my bad. I didn't read your comment correctly. $\endgroup$ Mar 12, 2019 at 10:28

3 Answers 3

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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 ?

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  • $\begingroup$ I forgot to add that the problem requires an algebraic solution. $\endgroup$ Mar 11, 2019 at 18:05
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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.

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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 $$({\bf n}, {\bf e}_1, \ldots, {\bf e}_n) .$$

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  • $\begingroup$ 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? $\endgroup$ Mar 11, 2019 at 20:01
  • $\begingroup$ 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 $A {\bf e}_i$? $\endgroup$ Mar 11, 2019 at 20:50

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