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Let $A$ be an $n-$square complex matrix. Show that if A is Hermitian and $\mathrm{trace}(A)\geq\mathrm{Re}\,\,\mathrm{trace}(AU)$ for all unitary $U\in\mathbb{C}^{n\times n}$, then A is positive semidefinite.

Is that possible to show this assertion? Any help is appreciated.

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Since $A$ is Hermitian all eigenvalues $\lambda_k$ of $A$ are real and can be decomposed as $$A=V\Lambda V^*$$ with $V$ unitary and $\Lambda$ a diagonal matrix with elements $\lambda_k$. Select a unitary matrix of the form $$U=VDV^*$$ with $D$ a diagonal unitary matrix with elements $d_k$ in the diagonal (obviously $|d_k|=1$). Then, $$AU=V(\Lambda D) V^*$$ with eigenvalues the elements $\lambda_kd_k$ of the diagonal matrix $\Lambda D$. Thus, $${trace}(AU)=\sum_{k=1}^n{\lambda_kd_k}$$ Let $\lambda_j$ the minimum eigenvalue of $A$. Then, if we select $d_k=1$ for all $k\neq j$ and $d_j=i$ the inequality $$trace(A)\geq Re[trace(AU)]$$ yields $$\sum_{k=1}^n\lambda_k\geq Re[\sum_{k=1,k\neq j}^n{\lambda_k}+i\lambda_j]=\sum_{k=1,k\neq j}^n{\lambda_k}$$ i.e. $\lambda_j\geq 0$ and the matrix $A$ is positive semidefinite.

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  • $\begingroup$ Thank you so much, this is a perfect answer to this question. $\endgroup$ – like_math Nov 25 '17 at 21:12
  • $\begingroup$ @like_math You are welcome! $\endgroup$ – RTJ Nov 25 '17 at 21:14

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