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If G is a symmetric matrix, it admits a decomposition of the form $G = G^+ − G^−$ , where $G^+,G^−$ are positive semidefinite matrix , and which satisfies the following properties:
(i) The nonzero eigenvalues of $G^+$ are the positive eigenvalues of G;
(ii) The nonzero eigenvalues of $G^−$ are the absolute values of the negative eigenvalues of G;
(iii) If $G = A − B$ for another decomposition with positive semidefinite matrices $A, B,$ then we have trace $A ≥ $trace $G^+$ , and trace $B ≥ trace G^-$.

I have no idea to answer the question. Can anyone give me some hints? Thank you in advance!

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  • $\begingroup$ You never actually said what the question was. Are we supposed to prove the statement? $\endgroup$ Commented Dec 5, 2016 at 15:53
  • $\begingroup$ @Amnommomnom Yea. Thank you for pointing out $\endgroup$ Commented Dec 5, 2016 at 20:21

2 Answers 2

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(i,ii) We can write $G = U \Lambda U^*$ where $U$ is unitary and $\Lambda $ is diagonal (the eigenvalues).

Let $\Lambda^+$ be $\Lambda $ with the negative entries replaced by zeros, and $\Lambda^- = \Lambda^+-\Lambda$.

Then $G^+ = U \Lambda^+ U^*$, $G^- = U \Lambda^- U^*$ are a suitable decomposition.

(iii) Note that if $u_k$ is an orthonormal basis then for any $A$ we have $\operatorname{tr} A = \sum_k u_k^*Au_k$.

Let $P = \{ B | B \ge 0, G+B \ge 0 \}$.

Pick $B \in P$ and let $u_k$ be orthonormal eigenvectors of $G$, and let $I$ be the indices corresponding to non positive eigenvalues $\lambda_k$, then we have $\sum_{k \in I} u_k^*(G+B)u_k \ge 0$, and so $\operatorname{tr} B \ge \sum_{k \in I} u_k^*Bu_k \ge -\sum_{k \in I} \lambda_k $. In particular, we see that $G^- \in P$ and $\operatorname{tr} G^- = -\sum_{k \in I} \lambda_k$, hence we have $\operatorname{tr} B \ge \operatorname{tr} G^-$ for all $B \in P$.

If $G = A-B$ with $A,B \ge 0$, then it follows that $\operatorname{tr} B \ge \operatorname{tr} G^-$.

Since $\operatorname{tr} G = \operatorname{tr} A - \operatorname{tr} B = \operatorname{tr} G^+ - \operatorname{tr} G^-$, we see that $\operatorname{tr} A = \operatorname{tr} G^+ + \operatorname{tr} B - \operatorname{tr} G^- \ge \operatorname{tr} G^+$.

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Hint: Start with the case in which $G$ is diagonal. Generalize this answer using the spectral theorem.

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  • $\begingroup$ Part (iii) seems a bit tricky, though $\endgroup$ Commented Dec 5, 2016 at 15:52

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