Qbit pure vs mixed state space

[PeterDonis]

A vector is an eigenvector for an operator regardless of the picked basis.

Yes, but it is one of the basis vectors in only one basis.

it could be for example in the form $$\ket{0}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

Yes, it would be a basis vector in this basis, but not in any other basis.