3 added 129 characters in body

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_{V_+}(v)|{+}-|\pi{V_-}(v)|_{-}=0,$$||\pi_{2}(v)||=0, where the norms appearing above are defined by on the subspaces V_{+} and V_{-} by two positive bilinear forms obtained by associated to the \lambda's and \beta's respectively and \pi_1 and \pi_2 are projections on the subspaces V_{+} and V_{-} . Taking an orthonormal sets in both spaces in the unit ball, we can find the vectors you want. I guess this idea can be generalized using the functional calculus. 2 added 16 characters in body; added 18 characters in body The Igor's answer also works if a is compact self-adjoint case and \mathcal{H} is isomorphic to \ell^2(\mathbb{N}) . Because of the spectral theorem and \text{Tr}(a)=0, for any v\in \mathcal{H} we have :$$ \sum_{i=1}^{\infty}\lambda_i 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) $$where (\lambda_i)_{i\in\mathbb{N}}\neq (0,0,\ldots) and (\beta_i)_{i\in\mathbb{N}}\neq (0,0,\ldots) and for all i we have \lambda_i,\beta_i\geq 0. Here \lambda's and \beta's form the spectrum of a. Note that \ell^2(\mathbb{N})=V_+\oplus V_{-}, where the spaces V_{+}=\bigoplus_{i=1}^{\infty}e_{r_i} and V_{-}=\bigoplus_{i=1}^{\infty}e_{s_i} We can rewrite (1) as$$ |v_1|_{+}-|v_2|_{-}=0,$$\pi_{V_+}(v)|{+}-|\pi{V_-}(v)|_{-}=0,$$ where the norms are defined by the positive bilinear forms obtained by the$\lambda$'s and$\beta$'s respectively. 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. 1 The Igor's answer also works if$a$is compact self-adjoint case 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 : $$\sum_{i=1}^{\infty}\lambda_i v^2_{r_i} -\sum_{i=1}^{\infty}\beta_i v^2_{s_i}\qquad (1)$$ 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 $$|v_1|_{+}-|v_2|_{-}=0,$$ where the norms are defined by the positive bilinear forms obtained by the$\lambda$'s and$\beta\$'s respectively.

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.