Can the eigenvectors of a linear operator in an infinite-dimensional space span the space and be linearly dependent at the same time?

Question

Consider a vector space $V$ over the complex field which is infinite-dimensional with a Euclidean inner-product. Let $L$ be a linear operator on $V$. Say a subset of eigenvectors of $L$ forms a complete orthonormal basis $B$ for $V$. Can there exist a (normalised) vector $v$ that is an eigenvector of $L$ but does not belong to $B$?

If all the eigenvalues of $L$ are real then would this hold?

Answer 1

Yes. Consider $\mathbb{1} : V \to V$ given by $\mathbb{1}(v) = v$. Then every $v \in V$ is an eigenvector of $\mathbb{1}$ with eigenvalue $1$. Choose some orthonormal basis $B$ of $V$. The elements of $B$ will all be eigenvectors of $\mathbb{1}$. On the other hand, $v_1 + v_2$ is an eigenvector of $\mathbb{1}$, but $v_1 + v_2 \not\in B$ by linear independence.