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For $A \in \mathbb{R}^{n\times n}$, I am wondering if the following inequality holds, $$\text{tr}(A)^2 - \text{tr}(A^2) \leq ||A||_*^2 - ||A||_F^2$$ This is equivalent to the following inequality on the elementary symmetric polynomial, $$\sum_{i<j}\lambda_i\lambda_j \leq \sum_{i<j} \sigma_i \sigma_j$$ where $\lambda_i$ and $\sigma_i$ denote the eigenvalues and singular values of $A$.

This question arises originally from a discussion here. I am still looking for a proof or counterexample to this inequality.

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  • $\begingroup$ Are you assuming $A$ to have positive, or at least real eigenvalues? $\endgroup$
    – darij grinberg
    Commented Feb 20, 2019 at 2:54
  • $\begingroup$ @darijgrinberg No. $A$ is any real valued square matrix. $\endgroup$
    – user1952770
    Commented Feb 20, 2019 at 2:55
  • $\begingroup$ Oh, I forgot that the symmetric polynomials are real anyway :) $\endgroup$
    – darij grinberg
    Commented Feb 20, 2019 at 2:55
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    $\begingroup$ I think your inequality is just the inequality from math.stackexchange.com/questions/381808/… , but applied to the compound matrix $\wedge^2 A$ (of size $\dbinom{n}{2} \times \dbinom{n}{2}$) instead of $A$. Can you check? $\endgroup$
    – darij grinberg
    Commented Feb 20, 2019 at 2:58
  • $\begingroup$ What is $\|A\|_*$ ? $\endgroup$
    – Christian Remling
    Commented Feb 20, 2019 at 16:15

1 Answer 1

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As suggested by Darij in the comments to the OP, the compound matrix $\wedge^2 A$ has eigenvalues $\lambda_i \lambda_j$ and singular values $\sigma_i \sigma_j$. This result can be found, for example in Theorem 2.16 of

Zhan, Xingzhi, Matrix theory, Graduate Studies in Mathematics 147. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-9491-0/hbk). x, 253 p. (2013). ZBL1275.15001.

see also. Applying the inequality from this post proves the desired inequality.

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