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!


1 Answer 1


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 $z$.


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