Let $\mathcal H$ be a Hilbert space, and let $a \in \mathcal B(\mathcal H)$ satisfy $\mathrm{Tr}(a)=0$. If $a$ is self-adjoint, then we can find a vector $\xi \in \mathcal H$ such that $\langle \xi | a \xi \rangle=0$. Is this true in general? Is it always possible to find an orthonormal basis of such vectors?



Vanishing Trace
The answers to both your questions are yes. (ie for any trace class operator $a$ on a Hilbert space $H$ with $Tr(a)=0$, there exists an orthonormal basis of $H$ consisting of vectors $\xi$ such that $\langle a \xi,\xi\rangle=0$).


First note that the fact that there exists an orthonormal basis of vectors $\xi$ such that $\langle a \xi,\xi\rangle=0$ is immediate (by something like "transfinite induction") from the fact that for any $a$ with trace zero, there exists one non-zero vector $\xi$ with $\langle a \xi,\xi\rangle=0$. Indeed, Zorn's Lemma implies that there exists a maximal closed subspace $H$ of $K$ such that there is an orthonormal basis of $K$ consisting of vectors such that $\langle a \xi,\xi\rangle=0$. If $K$ were different from $H$, consider $P$ the orthonormal projection on the orthogonal of $K$. The restriction of $Pa$ to the orthogonal of $K$ is still of trace class and zero trace, so there exists a unit vector in the orthogonal of $K$ such that $\langle a \xi,\xi\rangle=0$, which contradicts the maximality of $K$.


Now to the first point. You want to prove that is $Tr(a)=0$, zero belongs to the numerical range $W(a)$ of $a$, which is the set of $\langle a \xi,\xi\rangle$ for $\xi$ in the unit sphere of $H$. $W(a)$ is a convex subset of $\mathbb C$ for any operator $a$. This is usually stated for matrices (see [this page](http://en.wikipedia.org/wiki/Numerical_range)), but since everything happens in two dimensions this remains true in general.


But the assumption $Tr(a)=0$ implies that there is a sequence $(\lambda\_n)\_n$ in $W(a)$ such that $\sum\_n \lambda\_n=0$. This clearly implies that $0$ belongs to the convex hull of the $\lambda\_n$'s, which is contained in $W(a)$.


**Edit:** To answer Bill's objection, here is a proof that if $\sum\_n \lambda\_n=0$, then $0$ belongs to the convex hull of the $\lambda\_n$'s. (I agree that for a general compact operator, $0$ does not always belong to $W(a)$ but only to its closure, but my point was that here the assumption $Tr(a)=0$ makes it different).


Assume, by contradiction, that $0$ does not belong to the convex hull of the $\lambda\_n$'s. By (some form of) Hahn-Banach, the $\lambda\_n$'s all belong to some closed half-plain, say for simplicity $Im(\lambda\_n)\geq 0$ for all $n$. The equality $\sum\_n Im(\lambda\_n)=0$ then implies that the $\lambda\_n$'s are all real. They cannot be all positive (or all negative), otherwise $\sum\_n \lambda\_n \ne 0$. This is a contradiction.


