k-rank numerical range of an operator.

Question

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 $$\Lambda_k(T+T^*)=\{\lambda+\bar{\lambda}:\lambda\in\Lambda_k(T)\} $$ Note: To prove it, you can assume $T$ is normal in case if you need.

---

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.

Answer 1

If $\lambda\in \Lambda_k(T)$ then there exists some orthogonal projective operator $P$ of rank $k$ such that

$PTP=\lambda P$

So

$PT^*P=P^*T^*P^*=P^*(PT)^*=$

$((PT)P)^*=(\lambda P)^*=\lambda^* P^*=\lambda^* P$

Then

$P(T+T^*)P=(\lambda+\lambda^*)P$

Answer 2

It is not true. For example if we take $T=\text{ diag }(0,i)$, then $\Lambda_2(T+T^*)=\{0\}$ but $\Lambda_2(T)=\emptyset$. We can see easily that $\{\lambda+\bar\lambda:\lambda\in\Lambda_k(T)\}\subseteq\Lambda_k(T+T^*)$ is true.