Saturating Horn's Inequalities If I have a matrix product of the form: $C = AB$ where $A = UDU^*$ With A and B square, Hermitian and positive semidefinite, D diagonal, U a unitary and * representing the conjugate transpose, then Horn's inequalities give: $c_{i+j-1} \leq d_i b_j \ $ Where $b_i$, $c_i$ and $d_i$ are the eigenvalues of B, C and D respectively, $b_1 > b_2 > ... > b_n$ and similarly for c and d. Is it generally possible to saturate these inequalities (i.e. replace the $\leq$ with an = for the minimum value of the RHS) by the correct choice of U?

No. If all inequalities are ties, you get an overdetermined system for $c_1,\ldots,c_n$ ($n(n+1)/2$ equations and $n$ unknowns), which is not solvable in general. In fact, the two sets of equalities \begin{align*} d_1 b_j &= c_j;\quad j=1,2,...,n,\\\ d_2 b_j &= c_{j+1};\quad j=1,2,\ldots,n-1. \end{align*} imply that $\dfrac{d_2b_j}{d_1b_{j+1}} = 1$. So $B$ and $D$ have to satisfy necessary conditions like this in order that all inequalities can possibly become saturated.