Eigenvalues, Hermitian operators and observables in quantum mechanics 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?
 A: First of all, you are correct that b) is correct, and c) is wrong.
d) is a bit weird - if $\vec v$ is an eigenvector, then so are $2 \vec v$, $3 \vec v$, ..., $N \vec v$ - those are $N$ linearly dependent eigenvectors. However, the intention of the questioner is obviously for you to realize that the "correct" statement would be "there are $N$ linearly independent eigenvectors".
The problem with a) is that operators can be a bit weird in infinite-dimensional spaces and sometimes it is not a good idea to think about them as matrices. The maybe easiest counterexample to a) is the position operator (on $L^2(\mathbb R)$)
$$ (\hat x \psi)(x) = x \psi(x) . $$
$\hat x$ is Hermitian (with the right domain), but it does not have any eigenvectors:
$x\psi_\lambda(x) = \lambda\psi_\lambda(x)$ implies that $\psi_\lambda(x) = 0$ whenever $x \neq \lambda$, so $\psi_\lambda(x) = 0$ almost everywhere.
In Physics lectures, you will work with "generalized eigenvalues" $\lambda \in \mathbb R$ that correspond to "generalized eigenfunctions" $\psi_\lambda(x) = \delta(x - \lambda)$, but note that $\psi_\lambda$ is not in the Hilbert space $L^2(\mathbb R)$.
Mathematicians will tell you that the spectrum of $\hat x$ is not discrete, but continuous.
In any case, the claim that "the eigenvectors are a basis of the Hilbert space" is wrong.
