Let $\mathbb{S}^{dn} \ni X \succeq 0$ with $d,n \in \mathbb{N}$, where $X \succeq 0$ indicates that $X$ is positive semidefinite. Now partition $X$ into the block form \begin{gather} \begin{pmatrix} X_{11} & X_{12} & \cdots & X_{1d} \\ * & X_{22} & \cdots & X_{2d} \\ \vdots & \vdots & \ddots & \vdots \\ * & * & \cdots & X_{dd} \\ \end{pmatrix}, \; \; X_{ii} \in \mathbb{S}^{n} \;\; \text{and} \;\; X_{i,j} \in \mathbb{R}^{n \times n} \;\; \text{with}\;\; i \neq j \end{gather} where the notation $*$ represents the symmetric blocks. Does the matrix inequality \begin{equation} \mathbb{S}^{dn} \ni I_d \otimes \left( \sum_{i=1}^d X_{ii} \right) = \begin{pmatrix} \sum_{i=1}^d X_{ii} & 0_{n\times n} & \cdots & 0_{n\times n} \\ 0_{n\times n} & \sum_{i=1}^d X_{ii} & \cdots & 0_{n\times n} \\ \vdots & \vdots & \ddots & \vdots \\ 0_{n\times n} & 0_{n\times n} & \cdots & \sum_{i=1}^d X_{ii} \\ \end{pmatrix} \succeq X \end{equation} hold for any $\mathbb{S}^{dn} \ni X \succeq 0$?

I believe the above matrix inequality might be considered as an extension of the trace inequality $\mathsf{tr}(X)I_n \succeq X$ with $\mathbb{S}^{n} \ni X \succeq 0$. That is why I believe the inequality is true, however I am not able to give a proof for it.

A followed-up question:

Thank you for the suggestions, indeed, the above inequality is not ture. However, I would like to have a follow-up question which is in fact the original property I intended to prove.

Given matrices $X := \begin{bmatrix} X_1 & X_2 & \cdots & X_d \end{bmatrix} \in \mathbb{R}^{ n \times d \rho n} $ and $\mathbb{S}^{n} \ni U \succ 0$ with $n;d; \rho \in \mathbb{N}$, where the notation $\succ$ standards for the positive definite relation. Let $ \widehat{X} := \begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_d \end{bmatrix} \in \mathbb{R}^{dn \times \rho n}$, can we prove \begin{equation} I_d \otimes \left( \widehat{X}^\top \left( I_d \otimes U \right) \widehat{X} \right) \succeq X^\top UX \end{equation} for any given $X \in \mathbb{R}^{ n \times d \rho n}$ and $U \succ 0$?

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    $\begingroup$ Your question is exactly equivalent to whether or not the reduction criterion from quantum information theory holds for all quantum states. It indeed holds for all separable states, but it can fail for entangled states (for example, Nik Weaver's answer is the maximally-entangled Bell state). $\endgroup$ Commented Jan 30, 2018 at 13:21
  • $\begingroup$ @NathanielJohnston Thanks for your comments, are there general mathematics results on this criterion without using the framework of quantum information theory? I have made a follow-up question. Thanks again for your comments. $\endgroup$
    – Johannes
    Commented Jan 30, 2018 at 21:46

1 Answer 1


No, this is false. Consider $\left[\matrix{1&0&0&1\cr 0&0&0&0\cr 0&0&0&0\cr 1&0&0&1}\right]$, regarded as a $2 \times 2$ matrix of $2\times 2$ blocks. This is a rank 1 matrix with operator norm 2 but the diagonal sum you describe is the $4\times 4$ identity, whose norm is 1. So the latter cannot be larger than the former.

  • $\begingroup$ Thank you so much for your comment. Indeed, that inequality is not true. I have made a follow-up question. $\endgroup$
    – Johannes
    Commented Jan 30, 2018 at 21:48

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