# A Trace Bound Identity of Matrix Products (known for reals) in the Complex Space

Please refer to this beautiful paper on trace inequalities for matrix products. Theorem $$3$$ of the article (rephrased) states:

For any real $$n\times n$$ matrix $$A$$ and any real symmetric $$B$$ of the same size, let $$\bar{A}=(A+A^T)/2$$. Then \begin{align} \lambda_n(\bar{A})\operatorname{Tr}(B) &- \lambda_n(B)\Big(n \lambda_n(\bar{A}) - \operatorname{Tr}(A) \Big) \\ & \le \ \operatorname{Tr}(AB) \ \le \\ \lambda_1(\bar{A})\operatorname{Tr}(B) &- \lambda_n(B)\Big(n \lambda_1(\bar{A}) - \operatorname{Tr}(A) \Big) \end{align} where $$\lambda_1$$ and $$\lambda_n$$ denote the maximum and minimum eigenvalues respectively.

1. Is this valid over the complex field $$\mathbb{C}^{n\times n}$$ when $$A$$ is positive semidefinite and $$B$$ is Hermitian?
2. If yes, will having $$A$$ and $$B$$ as Hermitian be sufficient to ensure its validity?

The conditions mean that complex matrices $$A$$ and $$B$$ have real eigenvalues (trace also) and that $$A = \bar{A}$$.

While I'm of the thought that this is valid, I need its affirmation or correction. Thanks!

For any Hermitian matrix $$H\in M_n(\mathbb C)$$ and positive semidefinite matrix $$P\in M_n(\mathbb C)$$, by unitarily diagonalising $$H$$, we see that $$\lambda_n(H)\operatorname{Tr}(P)\le\operatorname{Tr}(HP)\le\lambda_1(H)\operatorname{Tr}(P).\tag{1}$$ Therefore, for any complex square matrix $$A$$ and Hermitian matrix $$B$$, if $$\bar{A}$$ denotes the Hermitian part (as opposed to the symmetric part) of $$A$$, we have $$\lambda_n(\bar{A})\operatorname{Tr}\left(B-\lambda_n(B)I\right) \le\operatorname{Tr}\left(\bar{A}(B-\lambda_n(B)I)\right) \le\lambda_1(\bar{A})\operatorname{Tr}\left(B-\lambda_n(B)I\right)\tag{2}$$ or equivalently, \begin{align} &\lambda_n(\bar{A})\operatorname{Tr}(B)-\lambda_n(B)\left(n\lambda_n(\bar{A})-\operatorname{Tr}(\bar{A})\right)\\ \le\ &\operatorname{Tr}(\bar{A}B)\\ \le\ &\lambda_1(\bar{A})\operatorname{Tr}(B)-\lambda_n(B)\left(n\lambda_1(\bar{A})-\operatorname{Tr}(\bar{A})\right).\tag{3} \end{align} $$(2)$$ and $$(3)$$ can also be rewritten as $$\lambda_n(\bar{A})\operatorname{Tr}\left(B-\lambda_n(B)I\right) \le\Re\operatorname{Tr}\left(A(B-\lambda_n(B)I)\right) \le\lambda_1(\bar{A})\operatorname{Tr}\left(B-\lambda_n(B)I\right)\tag{4}$$ and \begin{align} &\lambda_n(\bar{A})\operatorname{Tr}(B)-\lambda_n(B)\left(n\lambda_n(\bar{A})-\Re\operatorname{Tr}(A)\right)\\ \le\ &\Re\operatorname{Tr}(AB)\\ \le\ &\lambda_1(\bar{A})\operatorname{Tr}(B)-\lambda_n(B)\left(n\lambda_1(\bar{A})-\Re\operatorname{Tr}(A)\right)\tag{5} \end{align} where $$\Re z$$ denotes the real part of a complex number $$z$$.