eigenvalue question I think this question isn't that hard, but I am a bit confused.
Define the linear operator $T_k:H\mapsto H$ by 
\begin{align}
T_ku=\sum^\infty_{n=1}\frac{1}{n^3}\langle u,e_n\rangle e_n+k\langle u,z\rangle z,
\end{align}
where $z=\sqrt{6/\pi^2}\sum^\infty_{n=1}e_n/n$. For negative $k$, show $T_k$ has at most one negative eigenvalue.
 A: *

*A computation shows that $T_k$ is self-adjoint if $k$ is a real number (as a sum of such operators), so if $\lambda$ and $\mu$ are two different eigenvalues of $T_k$, the associated eigenvectors for these eigenvalues are orthogonal. 

*If $u$ is an eigenvector for $\lambda<0$, then for each $j\geqslant 1$,
$$\langle u,e_j\rangle+kj^2|z|^{-1}\langle u,z\rangle=\lambda j^3\langle u,e_j\rangle,$$
hence, 
$$\langle u,e_j\rangle=-\frac{kj^2}{1-\lambda j^3},$$
as we can assume $\langle u,z\rangle =1$ and $\lambda\neq j^{-3}$. 

*Now we conclude. Let $\lambda$ and $\mu$ two distinct eigenvalues of $T_k$ and $u,v$ associated eigenvectors. Then
$$0=\langle u,v\rangle =\sum_{j\geq 1}\frac{kj^2}{1-\lambda j^3}\frac{kj^2}{1-\mu j^3},$$
a contradiction. 

*Even if it's not required in the question but in the comments, $A_k$ is a compact operator for each $k$. Let $$T_N:=\sum_{j=1}^N\frac 1{j^3}\langle u,e_j\rangle e_j+k\sum_{j=1}^N\frac 1{j}\langle u,e_j\rangle \frac 1{\lVert z\rVert}z.$$
It's a finite ranked operator, and $\lVert T-T_N\rVert\to 0$.
