Consider a hermitian operator. So
a) in a space of infinite dimension its eigenvectors are a base.
b) in a finite-dimensional space the matrix that represents the hermitian operator is always diagonalizable.
c) 2 eigenvectors corresponding to different eigenvalues are collinear.
d) in a finite N-dimensional space there are N linearly dependent eigenvectors
I have to justify what is true and why it is true, and why the others are false. My teacher said that the true answer was b).
I learned in Quantum Mechanics that a Hermitian operator has always real eigenvalues. The operator is diagonalizable and the values of the diagonal are its eigenvalues.
An observable is a Hermitian operator whose eigenvectors constitute an orthonormal basis for the space E, even if it is of infinite dimension.
In my opinion both a) and b) are correct. I do not understand why a) is wrong.
What would be the answer if in the statement, instead of considering a Hermitian operator, we consider an observable?