Let $H$ be a separable Hilbert space and let $T\in B(H)$, such that $\displaystyle \sum_{j=1}^\infty\langle T\xi_j,\eta_j\rangle$ converges for any choice of orthonormal bases $\{\xi_j\}$, $\{\eta_j\}$. Does this imply that $T$ is trace-class?

I think it is, but I couldn't really write a proof.


2 Answers 2


If $T$ is not trace class, for any orthonormal basis $\xi_j$ of your Hilbert space $H$, $\sum_j \langle |T| \xi_j, \xi_j \rangle$ diverges. In particular, there are infinitely many $\xi_j$ such that $\langle |T| \xi_j, \xi_j \rangle > 0$. By the polar decomposition, there is a partial isometry $V$ such that $T = V |T|$, where $|T| = (T^* T)^{1/2}$. This is an isometry of closed subspaces $A$ to $B$, where $B$ contains $\text{Ran}(T)$ and $A$ contains $\text{Ran}(|T|)$. Since $|T|$ is self-adjoint, $|T|v = 0$ for any $v$ orthogonal to $A$. So start with an orthonormal basis $\alpha_j$ of $A$. Corresponding to this is $\beta_j = V \alpha_j$, an orthonormal basis of $B$. We have $$\sum_j \langle T \alpha_j, \beta_j \rangle = \sum_j \langle |T| \alpha_j, V^* \beta_j \rangle = \sum_j \langle |T| \alpha_j, \alpha_j \rangle = \infty$$ The only trouble is that we might not be able to simultaneously extend both $\alpha_j$ and $\beta_j$ to orthogonal bases of the whole space, because one of $A$ and $B$ might have finite codimension while the other has infinite codimension. To fix this problem, split the index set into two infinite subsets $K$ and $L$ such that we still have $\sum_{j \in K} \langle |T| \alpha_j, \alpha_j \rangle = \infty$. Since the closed spans of $\{\alpha_j: j \in K\}$ and $\{\beta_j: j \in K\}$ both have infinite codimension, we can extend both of these to orthonormal bases $\xi_j$ and $\eta_j$.

Note that $$\sum_j \left| \langle T \xi_j, \eta_j \rangle \right| \ge \sum_{j \in K} \langle |T| \alpha_j, \alpha_j \rangle = \infty$$ so $\sum_j \langle T \xi_j, \eta_j \rangle$ does not converge absolutely. It might converge conditionally, but we can always rearrange a conditionally convergent series to be divergent.


Robert Isreal's answer is correct, but at least to my way of thinking, not as clear as it would be possible. So, let me provide a proof which targets the hidden details:

(Note that $H$ doesn't need to be separable. So, I'll remove this assumption.)

  • $T$ is a bounded linear operator on $H$ $\Rightarrow$ There is a unique partial isometry $U:H\to H$ with $$\ker U=\ker T\tag1$$ and $$T=U|T|\tag2$$
  • $(\ker U)^\perp$ is a closed subspace of $H$ $\Rightarrow$ $(\ker U)^\perp$ admits an orthonormal basis $(e_i)_{i\in I}$
  • $U$ is an isometric isomorphism between $(\ker U)^\perp$ and $\operatorname{im}U$ $\Rightarrow$ $$f_i:=Ue_i\;\;\;\text{for }i\in I$$ is an orthonormal basis of the closed subspace $$\operatorname{im}U=(\ker U^\ast)^\perp\tag3$$ of $H$ (the only crucial thing is that $(f_i)_{i\in I}$ is an orthonormal basis of a closed subspace of $H$)
  • Now, $$U^\ast\left.U\right|_{(\ker U)^\perp}=\operatorname{id}_{(\ker U)^\perp}\tag4$$ and hence $$\sum_{i\in I}\langle Te_i,f_i\rangle_H=\sum_{i\in I}\langle U|T|e_i,Ue_i\rangle_H=\sum_{i\in I}\langle|T|e_i,U^\ast Ue_i\rangle_H=\sum_{i\in I}\langle|T|e_i,e_i\rangle_H\tag5$$ by $(2)$
  • $H=(\ker U)^\perp\oplus\ker U$ $\Rightarrow$ $(e_i)_{i\in I}$ can be supplemented to an orthonormal basis $(\tilde e_j)_{j\in J}$ of $H$ by elements of $\ker U$
  • $H=(\ker U^\ast)^\perp\oplus\ker U^\ast$ and $(3)$ $\Rightarrow$ $(f_i)_{i\in I}$ can be supplemented to an orthonormal basis $(\tilde f_k)_{k\in K}$ of $H$
  • By a simple renumbering (and insertion of zeros, if necessary), we may assume $J=K$
  • $(1)$ $\Rightarrow$ $$\ker U=\ker T=\ker|T|\tag6$$ and hence $$\langle T\tilde e_j,\tilde f_j\rangle_H=\langle U|T|\tilde e_j,\tilde f_j\rangle_H=0\;\;\;\text{for all }j\in J\text{ with }\tilde e_j\not\in(e_i)_{i\in I}\tag7$$
  • Thus, $$\sum_{j\in J}\langle T\tilde e_j,\tilde f_j\rangle_H=\sum_{i\in I}\langle Te_i,f_i\rangle_H=\sum_{i\in I}\langle|T|e_i,e_i\rangle_H=\sum_{j\in J}\langle|T|\tilde e_j,\tilde e_j\rangle_H\tag8$$ by $(5)$
  • Suppose $T$ is not nuclear $\Rightarrow$ $$\sum_{j\in J}\langle|T|\tilde e_j,\tilde e_j\rangle_H=\infty\tag9$$
  • $|T|$ is nonnegative $\Rightarrow$ $$\langle Te_i,f_i\rangle_H=\langle |T|e_i,e_i\rangle_H\ge0$$ and hence $\sum_{j\in J}\langle T\tilde e_j,\tilde f_j\rangle_H=\sum_{i\in I}\langle Te_i,f_i\rangle_H$ cannot converge by $(8)$ and $(9)$
  • Thus, we have found an orthonormal bases of $H$ for which your assumption is not satisfied

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