Let
- $U$ be a $\mathbb R$-Hilbert space
- $Q$ be a bounded linear operator on $U$
- $(v_n)_{n\in\mathbb N}$ and $(w_n)_{n\in\mathbb N}$ be orthonormal bases of $\ker(Q)^\perp$ and $\ker Q$ respectively and $$E:=\bigcup_{n\in\mathbb N}\left\{v_n,w_n\right\}$$
- $u\in U$
Since $\ker(Q)^\perp$ is a closed subspace of $U$ and (since $Q$ is continuous) $$\ker Q^{\perp\perp}=\overline{\ker Q}=\ker Q\tag 1\;,$$ we obtain $$U=\ker(Q)^\perp\oplus\ker Q\tag 2$$ and hence $$u=v+w\tag 3$$ for some $v\in\ker(Q)^\perp$ and $w\in\ker Q$. Since $$\langle u,v_n\rangle_U=\langle v,v_n\rangle_U\;\;\;\text{and}\;\;\;\langle u,w_n\rangle_U=\langle w,w_n\rangle_U\;,\tag 4$$ we should obtain $$\sum_{e\in E}\langle u,e\rangle_Ue=\sum_{n\in\mathbb N}\langle v,v_n\rangle_Uv_n+\sum_{n\in\mathbb N}\langle w,w_n\rangle_Uw_n=v+w=u\;.\tag 5$$ Thus, $E$ should be a countable orthonormal basis of $U$.
Is this the way we can supplement $(v_n)_{n\in\mathbb N}$ to an orthonormal basis of $U$ or am I missing something?