I'm looking at the following problem:
Let $H$ be a Hilbertspace over $\mathbb{K}, T\neq 0$ be a compact, self-adjoint operator with orthonormalsystem $(e_i)_{i\in J}$ of eigenvectors to eigenvalues $(\lambda_i)_{i\in J}\subset \mathbb{K}\backslash\{0\}, J\subset \mathbb{N}$. Then:
$( x,Tx)_H>0\quad \forall x\in H\backslash\{0\}$
$\Leftrightarrow\lambda_i\geq 0\quad \forall i\in J$ and $(e_i)_{i\in J}$ is a Schauder-basis.
I already got "$\Leftarrow$", so what's left to show is "$\Rightarrow$". For that part, I already got that $(x,Tx)\geq 0\quad \forall x\in H\backslash\{0\}\Leftrightarrow\lambda_i\geq0 \quad \forall i\in J$, but I don't know how to show now that $(e_i)_{i\in J}$ is a Schauder-basis. Can someone help me?
Thanks in advance.
By spectral decomposition, we have for all $x$ $$ Tx = \sum_{i\in J} \lambda_i \langle x_i,e_i\rangle e_i $$ and there exists uniquely determined $x_0\in \ker(T)$ such that $$ x= x_0 + \sum_{i\in J} \langle x_i,e_i\rangle e_i $$ Now, if $\langle x,Tx\rangle>0$ for all $x\ne0$, this implies $\ker(T)=\{0\}$, and $$ x= \sum_{i\in J} \langle x_i,e_i\rangle e_i $$ for all $x$, hence $(e_i)$ is a Schauder basis. (In this case $H$ is separable. In non-separable space, $\ker(T)$ has to be non-separable, and $\langle x,Tx\rangle>0$ for all $x\ne0$ is impossible.)