eigenvalue/singular value inequality, block matrix


If $\begin{bmatrix}A & X \\ X^* & B\end{bmatrix}$ is positive semidefinite with each block $n\times n$, we prove that $$2s_j\Big(\Phi(X)\Big)\le s_j\Big(\Phi(A+B)\Big), \qquad j=1, \ldots, n,$$ where $\Phi: X\mapsto X+(\tr X)I$ and $s_j(\cdot)$ means the $j$-th largest singular value. This confirms a conjecture of the author in [Linear Algebra Appl. 459 (2014) 404-410].



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