Is the kernel finite-dimensional if only finitely many basis-vectors of a Hilbert basis are mapped to zero?

Question

Let $H$ be a Hilbert space, $A$ a self-adjoint operator on a subspace of $H$ and $(v_n)_{n\in\mathbb N}$ a Hilbert basis (all $v_n$ are assumed to be in the domain of $A$). If $Av_n=0$ for at most finitely many $n$, can we already deduce that the kernel of $A$ is finite-dimensional?

Answer 1

No. For instance, let $E\subset H$ be the subspace spanned by the vectors $v_{2n}+v_{2n+1}$ and let $A$ be the orthogonal projection onto $E$. Then $Av_n\neq 0$ for all $n$ but $\ker(A)=E^\perp$ is infinite-dimensional (it contains $v_{2n}-v_{2n+1}$ for all $n$).

Answer 2

Let $A$ be the orthogonal projection onto $$w=\sum_{n=1}^\infty 2^{-n/2}v_n$$ given by $Ax=\langle x,w\rangle w.$ Then $Av_n=2^{-n/2}w\neq 0$ and $\ker A=w^\perp$ is a subspace of codimension $1,$ so it is maximal possible kernel for a nonzero operator.