I have the following problem.
Given is a $A\in \mathbb{R}^{2 \times 2}$ with two eigenvectors $u,v \neq 0$.
Does matrix $A$ have orthogonal eigenvectors?
Normally I would say no, because if you want a matrix with orthogonal eigenvectors then $A$ has to be symmetric and the eigenvalues have to be unequal.
Problem I cannot find a counterexample, so is there anything special with a ${2 \times 2}$ matrix?
No, but there is something special with the fact that the matrix is symmetric. Then, it follows from the spectral theorem that there is an orthonormal basis such that each element of the basis is an eigenvalue.