Eigenvalue and span of vector in Hilbert space Consider the Hilbert space $H=l^{2}( \mathbb{N} )$ and $x=\sum_{j}\frac{e_j}{j} \in H$. Consider an operator $T$ on  $l^{2}( \mathbb{N} )$ such that
\begin{align*}
T(\sum_i <z,e_i>e_i)=(\sum_i <z,e_i> |<x,e_i>|^3e_i)+k<x,z>x. 
\end{align*}
$T$ is a positive operator when restricted to span$(x)$ $^{\perp}$. We also have
\begin{align*}
 \langle T( e_j ) ,e_j \rangle =|\langle x,e_j \rangle|^{3}+k|\langle x,e_j \rangle|^{2}   
,\end{align*}
which is negative for $k<0$ and  $j$ sufficiently large. How do we conclude that  $T$ has only has 1 negative eigenvalue when $k<0$?
I think I might  have some misunderstandings about  span$(x)$. Since span$(x)$ is induced by one vector, it is closed. In the meantime,  $e_i$ can not be in span($x$) $^{\perp}$ as  $\langle e_i,x \rangle\neq 0$ for all $i \in \mathbb{N}$. That would imply that $\{e_i\}_{i=1}^{n}$ is in span($x$). It does not seem right.
 A: It is by no means true that $x \notin M^{\perp}$ implies $ x\in M$ for a closed subspace $M$. ($M^{\perp}$ is not the set-theoretic complement of $M$.) For example, If $M$ is $x-$axis in $\mathbb R^{2}$ then $M^{\perp}$ is the $y-$axis. $(1,1)$ is not in $M^{\perp}$ but it is not in $M$ either.
A: What you have there is the sum of a non-negative (compact) self-adjoint operator $A$ and a self-adjoint operator $F$ with one-dimensional range (and hence a kernel of codimension one). Assume that there are two negative eigenvalues of $A+F$, for example $-1$ and $-2$ (for simplicity). Then there exist corresponding unit eigenvectors $x$ and $y$, that is, $(A+F)x = -x$ and $(A+F)y = -2y$ with $\|x\|=1$ and $\|y\|=1$. [Edit: this is another $x$ as yours] Since $\ker F$ is 1-codimensional and $x,y$ are linearly independent, there exist scalars $a,b$ such that $ax+by\in\ker F$. Hence,
$$
-ax-2by = a(A+F)x + b(A+F)y = A(ax+by)
$$
and therefore
$$
0\le (A(ax+by),ax+by) = (-ax-2by,ax+by) = -a^2 - 2b^2,
$$
since $x$ and $y$ are orthogonal to each other. A contradiction.
