The matrix
$$ \begin{bmatrix} {0} & {-1} \\ {-1} & {0} \end{bmatrix} $$
has the following eigenvalues and eigenvectors
$$ \begin{aligned} \lambda_{1} = -1, &\qquad \nu_{1}=( 1,1) \\ \lambda_{2} = 1, &\qquad \nu_{2}=(-1,1) \end{aligned} $$
This makes sense to me, however, there is an additional eigenvector pair I would expect (with same eigenvalues as specified above). Namely, the base vectors $\nu_{x}=(1\,0)$ $\nu_{y}=(0\,1)$ . I have learned that eigenvectors are all vectors that do not get knocked of their span when the linear transformation occurs. Isn't that the case for the base vectors in this '180 degree' rotation?