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In $\mathbb{R}$, given $a,b>0$, you have $a+b>b\iff \frac{1}{a+b} < \frac{1}{b} \iff (a+b)^{-1} - b^{-1}<0$. Is this true for positive definite $A,B\in\mathbb{R}^{n\times n}$ of arbitrary $n$?

Orlando
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2 Answers2

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Yes. The same logic almost works, albeit with more justification. First, $A+B$ is positive definite also (which is easily checked). Then this result applied to $A+B$ and $B$ gives that $B^{-1} - (A+B)^{-1}$ is positive definite, as required.

B. Mehta
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  • Thank you! I was pretty sure it was true because I sampled a ton of positive definite matrices, and it always worked. – Orlando Dec 23 '17 at 01:32
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You can simultaneously diagonalise $ A $ and $ B $. So, you get $A = diag(1,1,1,.....,1) $ and $ B = diag(\lambda_1, \lambda_2,.....,\lambda_n) $ .This is possible as at least one matrix is positive definite. As both are positive definite the $ \lambda$ are positive. If you do the calculation you get the result $ diag(\mu_1, \mu_2,.....,\mu_n) $ where $ \mu_i = \frac {1} {1+\lambda_i} - \frac {1} {\lambda_i}. $ This is a negative definite matrix.

kieran
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    What do you mean by "simultaneously diagonalise"? $A$ and $B$ do not commute, so they can not be simultaneously diagonalized. Even when they can be, where do you get the diagonal elements of $A$ to be all $1$'s? – Hans Dec 23 '17 at 02:01
  • This is the best answer. They can be simultaneously diagonalized, see: http://fourier.eng.hmc.edu/e161/lectures/algebra/node7.html – Nick Alger Dec 23 '17 at 02:13
  • @Hans: The commuting result is orthogonally diagonalize. – max_zorn Dec 23 '17 at 02:49
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    @NickAlger: I know what he is trying to do. But that is not the correct nomenclature. Diagonalization is defined specifically as the eigenvalue decomposition problem. Not any matrix operation resulting in a diagonal matrix is qualified as diagonalization. $A^{-\frac12}$ does NOT diagonalize $A$, despite $A^{-\frac12}AA^{-\frac12}=I$. – Hans Dec 23 '17 at 05:16
  • @max_zorn: See my response to NickAlger above. – Hans Dec 23 '17 at 05:17
  • @Hans it is the solution to the generalized eigenvalue problem $Ax=\lambda Bx$. Using the term "simultaneous diagonalization" to refer to this is absolutely standard. See Horn and Johnson page 465, where they discuss the different usages of the phrase "simultaneous diagonalization". – Nick Alger Dec 23 '17 at 07:01
  • @NickAlger: I think you are looking at the first edition of Horn and Johnson. There they discuss simultaneous diagonalization BY * join congruence vs. simultaneous diagonalization BY similarity. So there needs to be differentiation when mentioned, which is lacking in this answer. Moreover, in the second edition, despite still keeping the name in the section title, the mention of simultaneous diagonalization completely disappears from the text of that section. Every time the subject is discussed, it is either explicitly called * congruent or the explicit matrix expression is written out. – Hans Dec 23 '17 at 10:04
  • @NickAlger: Every time diagonalization is mentioned, it is only referred to as diagonalization by similarity. I think they probably realized the great confusion this sloppy usage of the term would bring and ditched the usage of diagonalization by congruence altogether. Besides, writing out explicitly what one means does not take more than 10 letters. The words "simultaneous diagonalization" brings no meaning but confusion. for people who do not already know the technique. For people who already in the know, nothing new needs to be said. So there is no information present in this answer. – Hans Dec 23 '17 at 10:11
  • @Hans Look, all I know is that I've read the term in books and heard people use it in real life, routinely, to refer to this concept. You should consider un-downvoting kieran (assuming it was you who originally downvoted them) – Nick Alger Dec 23 '17 at 20:55
  • @NickAlger: Yes, you are saying that. Your two examples you used as your rationale have been have been refuted. So your claim does not stand, so far. You can come back with a better argument or accept my claim. Do yuo not think it is a bit odd that you should have a request without reason that STANDS? On the other hand, we can circumvent this whole argument by simply asking kieran to augment his answer to write out explicitly the mathematical derivation. As it stands now, I have argued why this answer does not give any information. You have ignored this particular argument of mine. – Hans Dec 23 '17 at 23:40
  • @NickAlger: It is not so hard to edit the answer and write an explicit mathematical derivation so that people who do not know learn something new, is it? Is this not the purpose of this site, not to show off what one knows but to get and give help for things people do not know? Maybe you can help with the edit, if kieran thinks it is beneath him to say things explicitly and in detail. Do you not think this is a better resolution? – Hans Dec 23 '17 at 23:44