Maximize the first $k$-trace of a positive definite matrix

Asked Jul 15 '21 at 13:51

Active Jul 15 '21 at 14:08

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I have a question that seems simple. But I don't know how to prove it.

Assume $A=(a_{ij})\in\mathbb{R}^{d\times d}$ is a real positive definite matrix. Then I want to prove that $$ \sum^k_{i=1}a_{ii}\leq \sum^k_{i=1}\lambda_{i}, $$ where $\lambda_i$ is the $i$-th largest eigenvalue of $A$.

I am not sure if this is true. If this is true, can anyone give a hint to prove this? If it's not, can anyone tell me a counter example?

edited Jul 15 '21 at 14:08

Zhanxiong

14,040

asked Jul 15 '21 at 13:51

Zhiyan

1,179

There are many ways to prove it. Simple proofs can be obtained, for instances, by using Courant-Fischer minimax principle, von Neumann's trace inequality, or Cauchy's interlacing inequality for bordered matrices. – user1551 Jul 15 '21 at 14:37
Is $A$ symmetric? – Zhanxiong Jul 15 '21 at 14:40
Yes. $A$ is positive definite. So it's symmetric. – Zhiyan Jul 15 '21 at 14:45
What is bordered matrices? Sorry, I can't find it on google. – Zhiyan Jul 15 '21 at 14:51
@superghostt I felt the conclusion is true for just symmetric matrices. The positive definite condition may be superfluous. For the proof, checking user1551's suggestion (just search Courant-Fischer minimax theorem). – Zhanxiong Jul 15 '21 at 14:55
Thank you! I think I got it. Appreciate your help! – Zhiyan Jul 15 '21 at 14:59
1

@superghostt This result is known as the Schur Horn theorem (or is part of this result, in any case). – Ben Grossmann Jul 15 '21 at 16:57
@BenGrossmann Thank you! This theorem can directly imply this result. – Zhiyan Jul 15 '21 at 19:14
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@user8675309 Thank you! This can answer my question! – Zhiyan Jul 15 '21 at 19:15

Maximize the first $k$-trace of a positive definite matrix

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