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

Here, $\lambda_i(\cdot)$ means the $i$th largest eigenvalue of $\cdot\quad$

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2 Answers 2

up vote 7 down vote accepted

Since the sum of all eigenvalues is equal to the trace of a matrix, we can say the former actually equals the latter sum, so long as you change the indices of the left sum to run over all values from 1 to $2n$. If you're interested in keeping the indexing as it is, then the sum on the left will be smaller than or equal to what it would have been running over all eigenvalues (since all eigenvalues of a positive semi-definite matrix are nonnegative), thus your inequality is correct.

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I am struggling with a somewhat related problem here math.stackexchange.com/questions/144890/… –  Dilawar May 14 '12 at 9:58

Note that, using your notation, $\quad \sum\limits_{i=1}^{2n}\lambda_i(\begin{bmatrix} A& B \\ B^* &C \end{bmatrix}) = \text{tr}( \begin{bmatrix} A& B \\ B^* &C \end{bmatrix}) = \text{tr} (A) + \text{tr} (C) = \sum\limits_{i=1}^n\left(\lambda_i(A)+\lambda_i(C)\right),$

so your inequality can actually be made stronger. But your inequality holds, since all the eigenvalues of a positive semidefinite matrix satisfy $\lambda_i \ge 0$, so if the sum on the LHS only goes up to $n$ it will of course be lessened.

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Somehow I missed @anon's answer to this, which is exactly the same and was posted before mine - please delete this –  barf Jul 12 '11 at 3:51
    
You can delete it yourself by clicking on the delete link on the lower left of your answer. But why should you? There is no harm in having a slightly different take or exposition on a question! –  t.b. Jul 12 '11 at 16:31

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