Regarding the eigenvalues of a symmetric, non-negative bounded linear operator with finite trace on a separable Hilbert space: are eigenvalues unique?

Question

If we consider a symmetric, non-negative bounded linear operator $Q$ with finite trace on a separable Hilbert space $H$, we know from the Hilbert–Schmidt theorem —see, for example, Theorem VI.16 in Reed and Simon's Methods of Modern Mathematical Physics, Volume I: Functional Analysis (Revised and Enlarged Edition), 1980— that there exists an orthonormal basis $\{h_n:n\geq1\}$ of $H$ consisting of eigenvectors of the operator $Q$ with corresponding non-negative eigenvalues $\{\lambda_n:n\geq1\}$.

However, is this decomposition unique (up to relabaling the indices)? Is it true that the earlier sequence of eigenvalues $\{\lambda_n:n\geq1\}$ contains all the eigenvalues of the operator $Q$ and, to be precise, I mean the following: if $\lambda$ is an eigenvalue of the operator $Q$ with corresponding eigenvector $h$, does there exists an $N\in\mathbb{N}$ such that $\lambda = \lambda_N$ and $h = h_N$?

The reason I ask this is because Lidskii's theorem asserts that all eigenvalues of the operator $Q$ sum up to the trace of $Q$, but I wondered whether the above sequence of eigenvalues contained all of them.

Answer 1

As stated, the answer is no for silly and trivial reasons. Namely, the eigenvector associated to an eigenvalue is not unique, so it does not make sense to require $h=h_N$. Indeed, $h$ could be any linear combination of the basis vectors $h_n$ such that $\lambda_n=\lambda$.

If you drop the restriction on $h$ and just require that $\lambda=\lambda_N$ for some $N$, then the answer is yes. To see this, just write $h=\sum c_nh_n$ for scalars $c_n$. We then have $Qh=\sum c_n\lambda_n h_n$. But we also have $Qh=\lambda h=\sum c_n\lambda h_n$. Since the $h_n$ are orthonormal, this is only possible if $c_n\lambda=c_n\lambda_n$ for all $n$. This means either $c_n=0$ for all $n$ (so $h=0$) or else there exists $n$ such that $\lambda=\lambda_n$. Moreover, $c_n$ can only be nonzero for those $n$ such that $\lambda=\lambda_n$, so $h$ must be a linear combination of the basis vectors $h_n$ such that $\lambda=\lambda_n$.

Answer 2

I don't follow your question very well, but here are the facts:

Any positive trace-class operator $Q$ is of the form $$\tag{1} Q=\sum_{j=1}^\infty\lambda_jP_j, $$ where $\{P_j\}$ is a sequence of pairwise orthogonal rank-one projections.

The spectrum of $Q$ consists of $\{\lambda_j\}$ and possibly zero (which might be on the list or not, but it is always in the spectrum in the infinite-dimensional case).

Uniqueness questions about eigenvectors are not usually well-posed. You could have an eigenvalue with multiplicity greater than one: for instance, $\lambda_1=\lambda_2$ in $(1)$; in that case the set of eigenvectors for $\lambda_1$ is (at least) a two-dimensional subspace.

For an easy example, let $$ Q=\begin{bmatrix} 1&0&0\\0&0&0\\0&0&0\end{bmatrix}. $$ The eigenvalues are $0$ and $1$. There is an orthonormal basis of eigenvectors; infinitely many, in fact. For instance, $$\{(1,0,0),(0,1,0),(0,0,1)\}$$ is an orthonormal basis of eigenvectors; but $$\{(i,0,0),(0,1/\sqrt2,1/\sqrt2),(0,-1/\sqrt2,1/\sqrt2)\}$$ is another orthonormal basis of eigenvectors.

Even when all eigenvalues are different, a scalar multiple of an eigenvector is still an eigenvector, so you cannot expect uniqueness.