Let $T\in\mathscr{B(\mathcal{H})}$ where $\mathcal{H}$ is an infinite dimensional seperable Hilbert Space and $k\in\mathbb{N}\cup\{\infty\}$. Now we define k-rank numerical range of $T$ denoted by $\Lambda_k(T)$ is defined as $$\Lambda_k(T):=\{\lambda\in\mathbb{C}: PTP=\lambda P, \text{ for some orthogonal projection } P \text{ of rank } k\}$$ or equivalently we can write $$\lambda\in\Lambda_k(T) \text{ iff there exists an orthonormal set } \{f_j\}_{j=1}^k \text{ s.t. } \langle Tf_j,f_r\rangle=\lambda\delta_{j,r} \text{ for } j,r\in\{1,2\ldots, k\}$$ where $\delta_{j,r}$ is Kronecker delta. Clearly $\Lambda_1(T)=W(T)$ i.e. $\Lambda_k(T)$ is a genaralization of numerical range $W(T)$.


Question: Is the following set equality be true $$\Lambda_k(T+T^*)=\{\lambda+\bar{\lambda}:\lambda\in\Lambda_k(T)\} $$


Comments: I can see the proof of this set equality for $\Lambda_1(T)=W(T)$. One part in the question i.e. $\{\lambda+\bar{\lambda}:\lambda\in\Lambda_k(T)\}\subseteq \Lambda_k(T+T^*)$ is easy to see but for other part I could neither to prove nor to give counter example.

Any Hints or comments is highly appreciated.

  • It's possible that $T+T^*=0$ for a non-zero $T$. – Narutaka OZAWA Dec 6 at 12:29
  • @NarutakaOZAWA that's true but it will not serve as counter example. – Piku Dec 6 at 15:55
  • It does. For example, $T=\mathrm{diag}(0,\sqrt{-1})$ has $\lambda_2(T+T^*)=0$. – Narutaka OZAWA Dec 6 at 22:53
  • Ohh yes here $\Lambda_2(T+T^*)=\{0\}$ but $\Lambda_2(T)=\emptyset$. Thank you @NarutakaOZAWA. – Piku Dec 7 at 5:04

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