I need to calculate the eigenvalues, eigenvectors and matrix of a linear operator
$$ A:\Bbb R^3 \to\Bbb R^3$$
which is a reflection on the plane$ y=-x/2$.
I know that reflector transforms $(x,y,z)$ into $(x,y,-z)$ but I don't know how to deal with it when given a specific axes of symmetry.
Is it possible to know the eigenvalues and eigenvectors without calculating the operator matrix?
Best Answer
Let $v_1=(-2,1,0), v_2=(-2,1,1)$ and $v_3=(1,2,0)$ so $B=(v_1,v_2,v_3)$ is a basis of $\Bbb R^3$ and the two former vectors span the plane $x+2y=0$ and the last vector is orthogonal to it. The matrix of the reflection $T$ in the basis $B$ is
$$[T]_B=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}$$ and clearly that $1,1$ and $-1$ are the eigenvalues associated to the eigenvectors $v_1,v_2$ and $v_3$ respectively. now let $P$ the change matrix from the standard basis $B_c$ to $B$ then $$P=(v_1\;v_2\;v_3)$$ and then the matrix of $T$ in the standard basis is $$[T]_{B_c}=P[T]_BP^{-1}=\frac15\begin{pmatrix}3&-4&0\\-4&-3&0\\0&0&5\end{pmatrix}$$