Eigenspace and eigenvector inside a Hilbert space Given $\{v_n\}_{n=1}^\infty$ an orthonormal sequence in a Hilbert space.
Let $\{\lambda_n \}_{n=1}^\infty$ a sequence of numbers and $F:H \to H$ defined by $Fx=\sum_{n=1}^\infty \lambda_n \langle x ,v_n \rangle v_n$.
Show that $v_n$ is an eigenvector with eigenvalue $\lambda_n$.
I went by:
$$Fv_n=\sum_{m=1}^\infty \lambda_m \langle v_n , v_m \rangle v_m = \lambda_n \langle v_n , v_n \rangle v_n=\lambda_n \| v_n \|_2^2 v_n=\lambda_n v_n$$
How do I show for each $n$, what is the eigenspace of $\lambda_n$?
what is the eigenspace of  $\lambda=0$?
Does $F$ have eigenvalues outside of $\lambda_n$ and, possibly, $\lambda=0$?
 A: Pedestrian way..
Eigenvector:
$${\sum}_k(\lambda-\lambda_k)\alpha_kv_k=\lambda({\sum}_k\alpha_kv_k)-{\sum}_k\alpha_k(\lambda_kv_k)=\\=F({\sum}_k\alpha_kv_k)-{\sum}_k\alpha_k(Fv_k)
=F({\sum}_k\alpha_kv_k-{\sum}_k\alpha_kv_k)=F(0)=0$$
Nullvector:
$${\sum}_k|\lambda-\lambda_k|^2|\alpha_k|^2=\|{\sum}_k(\lambda-\lambda_k)\alpha_kv_k\|^2=\|0\|=0$$
The rest follows!
A: Without loss of generality, assume that $\lambda_k \ne 0$ for all $k$. Otherwise, discard $v_k$ from your orthonormal set, and discard the term $\lambda_k (x,v_k)v_k$ from the defining sum for $F$. Nothing is changed by making such changes. And assume that $\{ v_k \}$ is an orthonormal set by discarding $0$ vectors and renormalizing.
The largest domain $\mathcal{D}(F)$ on which your definition of $F$ makes sense is
$$
        \mathcal{D}(F) = \left\{ x \in H : \sum_{n=1}^{\infty}|\lambda_n|^2|(x,v_n)|^2 < \infty \right\}.
$$
This domain includes the linear space $\mathcal{M}$ spanned by finite linear combinations of the $v_n$. And the domain includes $\mathcal{M}^{\perp}$ because $Fx=0$ is trivially convergent for all $x\in\mathcal{M}^{\perp}$. Therefore $\mathcal{D}(F)$ is dense in $H$ because $\mathcal{M}\oplus\mathcal{M}^{\perp}$ is dense in $\overline{\mathcal{M}}\oplus\mathcal{M}^{\perp}=H$, where $\overline{\mathcal{M}}$ is the closure of $\mathcal{M}$.
Because of the assumption that $\lambda_k \ne 0$ for all $k$, one has
\begin{align}
         \mathcal{N}(F)&=\{ x\in\mathcal{D}(F) : Fx=0 \} \\
      & =\{ x : \sum_{n=1}^{\infty}|\lambda_n|^2|(x,v_n)|^2 = 0 \} \\
      & = \{ x : (x,v_n) = 0,\;\; n=1,2,3,\ldots\,\} \\
      & = \mathcal{M}^{\perp}.
\end{align}
Every $\lambda_k$ is an eigenvalue because $Fv_k = \lambda_k v_k$, which follows from the definition of $F$ and the fact that $\{ v_k \}$ is an orthonormal set. $F$ has eigenvalue $0$ iff $\{ v_n \}_{n=1}^{\infty}$ is not a complete orthonormal set.
Every $v_k$ is in $\mathcal{D}(F)$ because $\sum_{n=1}^{\infty}|\lambda_n|^2|(v_k,v_n)|^2$ is a trivially convergent sum. And $Fv_k = \lambda_k v_k$, which proves that every $\lambda_k$ is an eigenvalue of $F$. By the above, $0$ may also be an eigenvalue. If $Fx=\lambda x$ for $\lambda\ne 0$ and $\lambda\ne \lambda_k$, then $x\in\mathcal{D}(F)$ and $x = \frac{1}{\lambda}Fx$ must be in $\mathcal{M}$, which gives
$$
               x = \sum_{n=1}^{\infty}(x,v_n)v_n.
$$
Hence,
$$
     0 = Fx-\lambda x = \sum_{n=1}^{\infty}(\lambda_n-\lambda)(x,v_n)v_n \\
   \implies (\lambda_n-\lambda)(x,v_n) =0 \mbox{ for all n.}
$$
Because $\lambda\ne\lambda_n$ is assumed to hold for all $n$, then $(x,v_n)=0$ for all $n$, which gives $Fx=0$. However $Fx=\lambda x$ and $\lambda\ne 0$ were asssumed, which gives $x=0$. Hence, the only eigenvalues are $\{ \lambda_n \}$, and $0$ if $\{ v_n \}$ is not complete.
