Given a $xy$ plane, if we have line like $10x+6y=0$.
To find the eigenvalues and eigenvectors of its reflection.
Could we say for reflection, we always have eigenvalues $0$ and $-1$?
And by matrix $\begin{pmatrix}0&-1\\-1&0\\\end{pmatrix}$, we have two eigenvectors.
Best Answer
A somewhat slightly long hint:
On $\mathbb{R}^n$, every reflection $T$ about an $(n-1)$-dimensional hyperplane $P$ has only two distinct eigenvalues, namely, $\lambda=-1$ with multiplicity $1$ and $\lambda=1$ with multiplicity $n-1$ (so, your claim is wrong; $0$ cannot be an eigenvalue of a reflection). The $1$-dimensional eigenspace associated with $\lambda=-1$ is the line normal to the hyperplane. That is, $Tv=-v$ for every $v\perp P$ (that's why $T$ is called a reflection: $v$ is reflected to the opposite direction). The $(n-1)$-dimensional eigenspace associated with $\lambda=1$ is the hyperplane itself, i.e. $Tv=v$ for all $v\in P$ (hence $T$ is a reflection about the hyperplane $P$: every vector lying on $P$ is left unchanged).
In your specific case, the "hyperplane" that the operator reflects about is the line $10x+6y=0$. You may find the eigenvectors using the above description of eigenspaces.