How many eigenstates can a Hermitian matrix have? I was learning about the form $H|a\rangle=n|a\rangle$ and one thing wasn't explained clearly. I know that a hermitian matrix can have more than one eigenstate, but my lecture only showed it having two. Can a Hermitian matrix have multiple eigenstates, or just two? 
 A: Summary of comments and some additions:


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*The number of linearly independent eigenvectors of a Hermitian matrix is equal to the dimension of the matrix. As such one can even get infinitely many linearly independent eigenvectors for an infinite dimensional matrix. They are frequently encountered as operators in Quantum Mechanics, e.g. the Hamiltonian of a real system is usually an infinite dimensional matrix. One can however have simple systems where it is only a 2x2 matrix (2-level system, frequently used in quantum optics and hence important).

*The number of eigenvectors is equal to the dimension of a Hermitian matrix if there is no degeneracy. No degeneracy here meaning that there are no repeated eigenvalues.

*For degenerate Hermitian matrices there are always infinitely many normalised eigenvectors. This is since the two (or more) basis vectors with the degenerate eigenvalue span a whole subspace of eigenvectors. This fact can be important, e.g. in the statement of the compatibility theorem.

