# k-rank numerical range of an operator

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