When is the numerical range of a nilpotent matrix circular?
The problem formulated in the title is examined. A \(2\times 2\) nilpotent matrix always has circular numerical range. The situation when \(3\times 3\) nilpotent matrices have circular numerical ranges is completely described by \textit{D. S. Keeler, L. Rodman} and \textit{I. M. Spitkovsky} [Linear Algebra Appl. 252, 115--139 (1997; Zbl 0876.15020)]. The \(4\times 4\) case is covered by \textit{H.-L. Gau} [Taiwanese J. Math. 10, No.~1, 117--128 (2006; Zbl 1099.15022)] as follows: A necessary and sufficient condition that the numerical range of a nilpotent matrix \(M\) of size at most 4 be circular is that the traces \(\text{tr} M^*M^2\) and \(\text{tr} M^*M^3\) be all null. For matrices of size 5, the situation becomes more complicated. The authors show that the analog of the so-called trace condition appearing in the result by Gau does not hold in the \(5\times 5\) case, counterexamples being given. The main result of the paper completely characterizes the situation when a \(5\times 5\) nilpotent matrix has circular numerical range. Also, the case when a \(5\times 5\) nilpotent matrix satisfying, respectively non-satisfying, the trace condition has circular numerical range is discussed thoroughly.
scientific article; zbMATH DE number 5688870