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Let $\mathcal{H}$ be an infinite dimensional separable Hilbert Space and $T\in\mathscr{B(\mathcal{H})}$. Suppose $$\mu_k(T):=\left\{ \begin{pmatrix} \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_k \end{pmatrix}\in\mathbb{C}^k: PTP= \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 &\cdots & 0 \\ \vdots & \vdots & \ddots\\ 0 & 0 & \cdots & \lambda_k \\ \end{pmatrix}, \text{ for some orthogonal projection $P$ of rank } k \right\}$$ Is $\mu_k(T)$ convex in $\mathbb{C}^k$?

Comments: I can only see that $\mu_k(T)$ is non-empty but I could not construct desired projection of rank k to show convexity of $\mu_k(T).$
Any hint/comment regarding convexity of $\mu_k(T)$ is highly appreciated. Thanks in advance.

Piku
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  • Well, I don't know, but the spectrum needs not be convex. These sets look like a kind of spectrum so probably they are not convex as well. – Giuseppe Negro Jan 22 '19 at 09:38
  • If I understand your notation correctly, then $\mu_1(T)$ is the numerical range of $T$. Have you tried to immitate the convexity proof for the numerical range? Some references can be found in the comments to this question: https://math.stackexchange.com/questions/171729/why-is-the-numerical-range-of-an-operator-convex – MaoWao Jan 22 '19 at 14:12
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    It's not clear how your set is defined. What's a matrix in $B(H)$? Do you want $PTP=\sum_j\lambda Q_j$ for fixed pairwise orthogonal rank-one projections, or for any such family of projections? – Martin Argerami Jan 23 '19 at 01:52
  • @MartinArgerami As $\mathcal{H}$ is a separable Hilbert space, we can think an operator $T$ on $\mathscr{B(\mathcal{H})}$ as a matrix of infinite order where (i,j)-th entry of that matrix is $\langle Te_i,e_j\rangle$ and $(e_i)_{i\in\mathbb{N}}$ being an orthonormal basis of $\mathcal{H}$. However one can think also the way you mentioned in the comment that $PTP=\sum_j\lambda_jQ_j$ where $Q_j$ are pairwise orthogonal rank one projection but $Q_j$ are dependent on $P$ i.e. for different $P$ we will get different $Q_j$'s. – Piku Jan 23 '19 at 05:25
  • @MaoWao In the proof of convexity of $W(T)$, it was boiled down to show numerical range of any 2-by-2 matrix is convex which was just computation. In the same line if we want to show $\mu_k(T)$ is convex, it is sufficient of proof $\mu_k(S)$ is convex where $S$ is compression of $T$ on union of $Ran(P_1)$ and $Ran(P_2)$, $P_1$ and $P_2$ are orthogonal projection obtained corresponding to two points of $\mu_k(T)$. After that I couldn't proceed to show $\mu_k(S)$ is convex like numerical range case. – Piku Jan 23 '19 at 06:44
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    @GiuseppeNegro: it looks like a numerical range, and those are usually convex. – Martin Argerami Jan 23 '19 at 19:40
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    Something seems off in your definition; if $P$ is a rank-$k$ projection (acting on $\mathcal H$) then $PTP\in\mathcal B(\mathcal H)$ but the right-hand side is a matrix. I assume you want $P$ to be some sort of compression (?), i.e. for the standard basis $(\hat e_i){i=1}^k$ in $\mathbb C^k$ and an arbitrary orthonormal system $(e_i){i=1}^k$ in $\mathcal H$ define $$P:\mathbb C^k\to\mathcal H\quad \hat e_i\mapsto e_i$$ for all $i=1,\ldots,k$ (plus linear extension onto all of $\mathbb C^k$). This way $P^\dagger TP\in\mathbb C^{k\times k}$ with fixed basis. – Frederik vom Ende Jan 24 '19 at 15:01
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    (2/2) In that case, the object in question is close to (but slightly more general than) the rank-$k$ numerical range. Maybe this connection alone is of use to you already? – Frederik vom Ende Jan 24 '19 at 15:06

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