Standard Fourier series. The linear transformation (instead of a matrix) is $ T = \frac{-d^2}{dx^2}. $ The vector space is smooth functions of $x$ with period $2\pi.$ And we get $ T (\cos n x ) = n^2 \cos n x,$ so $n^2$ is an eigenvalue of $T.$ However, $ T (\sin n x ) = n^2 \sin n x$ as well, so we get two different eigenvectors for that eigenvalues. How different are they? Sticking with real functions, we have an inner product $\langle, \rangle$ on pairs of periodic functions given by $ \langle f,g \rangle = \int_0^{2 \pi} \; f(x) g(x) dx. $ And the pair of functions we gave are orthogonal under the given inner product. Also other pairs, such as $\cos nx, \cos mx$ give $0$ when $m \neq n,$ same for sines, same for sin and cosine, one $m$ the other $n.$ Finally, integration by parts tells us that $T$ is self-adjoint with respect to the inner product, as \int_0^{2 \pi} \; u'(x) v'(x) dx = \langle u,Tv \rangle = \langle Tu,v \rangle.