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Let $A_1,A_2,\dots,A_M$ be given $N\times N$ hermitian matrices. The numerical range is defined as the set \begin{align} \mathbb{S}=\{(u^HA_1u,\dots,u^HA_Mu)\in \mathbb{R}^M\mid u^Hu=1\} \end{align} By toeplitz hausdorff theorem, $\mathbb{S}$ is convex for

  1. $M=2$, $N\geq 2$, no conditions on $A_i$
  2. $M=3$, $N\geq 3$, no conditions on $A_i$
  3. $A_i$'s are tridiagonal and real in some basis. No conditions on $M$ and $N$. (from Nathaniel Johnston's answer).

I would like to know the other cases where $\mathbb{S}$ is convex. Can some body point me to known references?

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  • $\begingroup$ Can you be more clear about what you mean by "other cases"? If you are only allowed to consider $M$ and $N$, then the conditions you listed are optimal (i.e., the numerical range can be non-convex for all other choices of $M,N$). So do you want conditions on the $A_i$'s instead? $\endgroup$
    – Nathaniel Johnston
    Commented Mar 20, 2014 at 22:02

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The only result of this nature that I'm aware of (beyond the two cases involving $M$ and $N$ that you've listed) is Theorem 3.1 in "C.-K. Li and Y.-T. Poon. Convexity of the joint numerical range. SIAM J. Matrix Analysis Appl. 21 (1999), 668-678", which basically says that the numerical range is convex when each of the $A_i$'s are tridiagonal and real in some basis.

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