Do eigenvectors with pairwise distinct eigenvalues of a bounded, linear, nonnegative, symmetric operator on a Hilbert space build an orthogonal basis? Let $H$ be a Hilbert space and $Q$ be a bounded, linear, nonnegative and symmetric operator on $H$ with finite trace.

By the Hilbert–Schmidt theorem, there is an orthonormal basis $(e_n)_{n\in\mathbb N}$ of $H$ with $$Qe_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N\tag 1$$ for some non-increasing $(\lambda_n)_{n\in\mathbb N}\subseteq[0,\infty)$ with $\lambda_n\stackrel{n\to\infty}\to 0$.

I'm new to this topic. If $\dim H=d<\infty$, e.g. $H=\mathbb R^d$, and we can find $0<\lambda_1<\cdots<\lambda_n$ with $(1)$ for some $e_1,\ldots,e_d\in H$, then $e_1,\ldots,e_d$ are pairwise orthogonal and hence induce an orthonormal basis of $H$.

Does this fact generalize to the case $\dim H=\infty$?

I assume things are different here. From the Hilbert-Schmidt theorem we see, that the "eigenvalues" don't need to be pairwise distinct in order for the "eigenvectors" being an orthonormal basis.
 A: Regardless of dimension, if $Q$ is symmetric on a complex inner product space, and $Qx =\lambda x$ for some $x \ne 0$, then $\lambda$ is real because
$$
          (\lambda-\overline{\lambda})\|x\|^2=(Qx,x)-(x,Qx)=0.
$$
And, if $Qx_1=\lambda_1 x_1$, $Qx_2 = \lambda_2 x_2$ with $\lambda_1\ne \lambda_2$, then $x_1 \perp x_2$ because
$$
            (\lambda_1-\lambda_2)(x_1,x_2)=(Qx_1,x_2)-(x_1,Qx_2)=0.
$$
So those things carry over from finite-dimensional spaces, and
$$
             \mathcal{N}(Q-\lambda_1 I)\perp\mathcal{N}(Q-\lambda_2 I).
$$
If $Q$ is continuous, $\mathcal{N}(Q-\lambda I)= (Q-\lambda I)^{-1}\{0\}$ is a closed subspace; therefore if the underlying space is a Hilbert space, then $\mathcal{N}(Q-\lambda I)$ is a Hilbert space, and you can choose an orthonormal basis of $\mathcal{N}(Q-\lambda I)$. Then, if you combine orthonormal bases for different $\mathcal{N}(Q-\lambda_j I)$, the result is still an orthonormal subset because of the above orthogonality condition. And you can do that with any number of the $\lambda_j$. Of course, $\mathcal{N}(Q-\lambda I)$ need not be finite-dimensional.
If $Q$ is bounded and positive, i.e., $(Qx,x) \ge 0$), with spectrum consisting of isolated points, except for possibly $0$, then each non-zero $\lambda$ in the spectrum is an eigenvalue, and the eigenvalues can be ordered
$$
  \lambda_1 > \lambda_2 > \lambda_3 > \cdots > \lambda_n > \cdots > 0.
$$
Then, if you choose orthonormal bases $\mathscr{E}_{k}$ of $\mathcal{N}(Q-\lambda_kI)$, the union of these is an orthonormal basis of the underlying Hilbert space, provided you include $\mathscr{E}_0=\mathcal{N}(Q)$ if this space is non-trivial. Completeness is not obvious, but it does hold in this case. So you end up with an orthonormal basis of eigenvectors, which is what you wanted. The dimensions of the eigenspaces are allowed to be finite or infinite in the above discussion, which generalizes the results for compact operators by assuming a condition on the spectrum only.
