# Vanishing Trace

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?

• True in finite dimension... – Igor Rivin Jan 19 '11 at 2:18

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), 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.

• Mikael, you only get that $0$ is in the closure of the convex hull of the $\lambda_n$'s. IIRC, the numerical range need not be closed. For those who don't know: in finite dimensions Mikael's argument works and is the standard proof that an $n$ by $n$ matrix that has trace zero is unitarily equivalent to a matrix that has zero diagonal. Of course, here simple induction is used. – Bill Johnson Jan 19 '11 at 23:10
• Bill: unless I miss something, I really believe that my argument proves that the condition $Tr(a)=0$ implies that the numerical range of $a$ is closed. See my edit. – Mikael de la Salle Jan 20 '11 at 8:28
• Mikael, I'm confused with a couple of things in your answer. First you seem to imply that an operator with zero trace is trace-class (and thus compact), which is not the case. Second, in your maximality argument in the second paragraph you say that the compression of a trace-zero operator by a projection is still trace-zero, which again I think is not true. – Martin Argerami Jan 21 '11 at 19:40
• @Martin: How do you define the trace of an operator which is not of trace class? For you second remark, what I meant is that if the compression to a closed subspace $K$ of a trace-zero operator has trace zero, then so has its compression to the orthogonal of $K$. This is just because of the equality $Tr(PaP+(1-P)a(1-P))=Tr(a)$, valid for any projection $P$. – Mikael de la Salle Jan 21 '11 at 22:02
• Thanks a lot, Mikael! I should start thinking before writing... – Martin Argerami Jan 22 '11 at 2:05

The Igor's answer also works if $a$ is compact self-adjoint and $\mathcal{H}$ is isomorphic to $\ell^2(\mathbb{N})$ .

Because of the spectral theorem and $\text{Tr}(a)=0$, for any $v\in \mathcal{H}$ we have :

$$\langle v,av\rangle =\sum_{i=1}^{\infty}\lambda_i v^2_{r_i} -\sum_{i=1}^{\infty}\beta_i v^2_{s_i}\qquad (1)$$

where $(\lambda_i)_{i\in\mathbb{N}}\neq (0,0,\ldots)$ and $(\beta_i)_{i\in\mathbb{N}}\neq (0,0,\ldots)$ and for all $i$ we have $\lambda_i,\beta_i\geq 0$. Here $\lambda$'s and $\beta$'s form the spectrum of $a$.

Note that $\ell^2(\mathbb{N})=V_+\oplus V_{-}$, where the spaces $V_{+}=\bigoplus_{i=1}^{\infty}e_{r_i}$ and $V_{-}=\bigoplus_{i=1}^{\infty}e_{s_i}$

We can rewrite (1) as

$|||\pi_{1}(v)|||$- $||\pi_{2}(v)||=0,$

where the norms appearing above are defined on the subspaces $V_{+}$ and $V_{-}$ by two positive bilinear forms associated to the $\lambda$'s and $\beta$'s respectively and $\pi_1$ and $\pi_2$ are projections on the subspaces $V_{+}$ and $V_{-}$ .

Taking an orthonormal sets in both spaces in the unit ball, we can find the vectors you want. I guess this idea can be generalized using the functional calculus.

• I am interested specifically in non-self-adjoint operators $a \in \mathcal B(\mathcal H)$. Thank you for your idea of obtaining positive bilinear forms from $a$. – Andre Jan 19 '11 at 9:46