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| visits | member for | 2 years, 1 month |
| seen | Feb 19 at 23:46 | |
| stats | profile views | 24 |
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May 12 |
awarded | Teacher |
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Apr 16 |
awarded | Commentator |
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Apr 16 |
comment |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared @cardinal: Claim 1 was a nice observation! And you did a good job in rederiving the majorization relation of Claim 2! With respect to the use/not use of the absolute value, the equation appears in the statement of a theorem that assumes "arbitrary complex-valued" matrices :) It does not matter that later you apply the theorem to a special case. |
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Apr 16 |
comment |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared @cardinal 1) thanks for the clarification of Claim 1; let me say that your explanation in the comment is much better than the formulation of the proof of the claim in the answer :) 2) as announced in the comments to the question, I had also found a proof; you can see it in the edited text of the question [and now as answer] 3) I found that Claim 2 corresponds to what is also known as "Schur's theorem" (see Exercise II.1.12 of Bathia, "Matrix Analysis") 4) for your statement of von Neumann's theorem the absolute value can not be avoided as it regards matrices with complex entries. Thanks again. |
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Apr 16 |
answered | Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared |
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Apr 16 |
awarded | Supporter |
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Apr 16 |
comment |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared Thanks! I have to check the steps you provide, but I believe they are correct. I did not know the theorem of von Neumann (BTW, the statement probably misses an absolute values on the the left-hand side), and is good that you pointed it out! I do not have my matrix analysis books with me, but I think I saw Claim 2 somewhere in those books today. For Claim 1, I admit I do not see it now, but it is probably because I am tired. Thanks for putting such effort in solving the problem! |
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Apr 16 |
revised |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared added 2206 characters in body |
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Apr 15 |
comment |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared OK, I think I got the proof :) I will post it as soon as possible. |
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Apr 15 |
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Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared A trivial observation: the conjecture is obviously true for $2\times 2$ matrices. Indeed, as soon as we sum the squares of the two singular values we have already reached the maximum. On the other hand, by definition the square of the largest singular value is larger than the largest absolute value squared of every element of the matrix. |
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Apr 15 |
comment |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared In Horn&Johnson, "Topics in Matrix Analysis", problem 21 of chapter 3.3, a related majorization relation is proven (Eq. (3.3.36)) , but it involves only the diagonal elements of the matrix rather than all the entries of the matrix. What I ask seems much stronger... |
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Apr 15 |
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Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared @cardinal: thanks for the suggestions and for the hope you give me :) I did not have much time to skim through Horn and Johnson today, but I am hoping the answer is there (somewhere :) ) |
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Apr 15 |
comment |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared @cardinal: thanks! Looks good to me. |
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Apr 15 |
awarded | Editor |
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Apr 15 |
comment |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared @cardinal: no problem :) Thanks for your consideration of the question! |
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Apr 15 |
revised |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared Edited to make more clear the relation between the sum of the entries of the two considered vectors --- they are both equal to 1: this is way I talk of majorization rather than weak maorization. |
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Apr 15 |
comment |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared @joriki: the notion of majorization is such that involves ordering the entries of the vectors in descending order; for more details please see the first lines of the Wikipedia entry @cardinal: thanks, but yours is not a countexample. For one thing, the matrix you propose does not satisfies the hypothesis $\textrm{Tr}(A^\dagger A)=1$. If properly rescaled by multiplying it by $1/\sqrt{2}$, it satisfies the conjecture, as it has only one singular value equal to 1. |
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Apr 15 |
awarded | Student |
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Apr 15 |
revised |
Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared edited tags |
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Apr 15 |
asked | Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared |
