On a recursive inverse eigenvalue problem

Let $s_1,\dots,s_n$ – be arbitrary complex scalars. It is required to construct an $n\times n$ normal matrix $A$ such that $s_i$ is an eigenvalue of the leading principal submatrix $A_i$, $i=1,2,\dots,n$. It is shown that, along with the obvious diagonal solution $\operatorname{diag}(s_1,\dots,s_n)$, this problem always admits a much more interesting nondiagonal solution $A$. As a rule, this solution is a dense matrix; with the diagonal solution, it shares the property that each submatrix $A_i$ is itself a normal matrix, which implies interesting connections between the spectra of the neighboring submatrices $A_i$ and $A_{i+1}$.