Can an eigenbasis be divided into bases of the eigenspaces?

Question

I am considering an operator $A$ defined on a subspace of some Hilbert space $H$ and I am given a Hilbert-basis $(v_n)_{n\in\mathbb N}$ such that all $v_n$ are eigenvectors of $A$: $$Av_n=\lambda v_n$$ I am wondering whether the basis can be divided into bases of the eigenspaces, possibly under additional assumptions (e.g. $A$ compact and symmetric). The equivalance relation is obvious: $$v_i\sim v_j:\Leftrightarrow\lambda_i=\lambda_j$$ But how do I know that each eigenspace is spanned by an equivalence class? The answers to this question suggest that the issue is subtle.