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In my previous problem, I made a typo. Now I restate it as a new problem.

Let $ \begin{bmatrix} A& B \\ B^* &C \end{bmatrix}$ be positive semidefinite, $A,C$ are of size $n\times n$. Is it true that $$\quad \sum\limits_{i=1}^k\lambda_i\begin{bmatrix} A& B \\ B^* &C \end{bmatrix}\le \sum\limits_{i=1}^k\left(\lambda_i(A)+\lambda_i(C)\right)\quad, $$ where $1\le k\le n$? Here, $\lambda_i(\cdot)$ means the $i$th largest eigenvalue of $\cdot\quad$

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I guess that $A$ and $C$ are Hermitian matrices. If this is the case, the minmax theorem might help to prove the statement. –  Fabian Jul 12 '11 at 19:28
    
Yes, $A,C$ are Hermitian, but using minmax theorem I can proof the case $k=1$, and no more... –  Sunni Jul 13 '11 at 20:34

2 Answers 2

up vote 0 down vote accepted

This question is answered by T. Ito at http://mathoverflow.net/questions/70689/ask-some-matrix-eigenvalue-inequalities

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It holds for $n = 1$. Solving the quadratic equation (with trace and determinant) for the largest eigenvalue of the lhs matrix, the desired inequality reads $$ \frac{a+c+\sqrt{(a-c)^2 + 4|b|^2}}{2} \leq a + c.$$ Some algebra shows that (since $a+c \geq 0$) this is equivalent to $$ |b|^2 \leq ac. $$ This follows (more or less) from Jacobi's criterion.

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