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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.
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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.
show/hide this revision's text 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.