Show that the following matrix has no real eignvalues, and thus no eigenvectors. Interpret your result geometrically.
[0 -1]
[-1 0]
Are you sure you have written the matrix correctly?
Charikar is right. The matrix has real eigenvalues and eigenvectors. The matrix is a reflection - the eigenvectors are the bisectors of the axes.
Your question would make sense if the matrix had been
[0 -1]
[1 0]
as that would have been a rotation, which does not have real eigenvectors.
If $\displaystyle det(A-\lambda I) = f(\lambda)$ for some function f, then the eigenvalues of A are the roots of $\displaystyle f(\lambda)$. If it only has solutions in $\displaystyle F = \mathbb{C} - \mathbb{R}$ then its eigenvalues are not real, therefore it has no eigenvectors as your question described.
In this case, $\displaystyle f(\lambda) = \lambda^2 + 1$.
Can you find the roots of f?