# An eigenvalue problem for positive semidefinite matrix

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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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.
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.