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Dec 31, 2011 at 3:59 vote accept Victor Liu
Dec 28, 2011 at 10:05 comment added Alexander Chervov @Federico, You mean - probably the paper mentioned above deals with LL^t ? Then it is Okay- we need squares. Just in the question LDL^t was mentioned which is sometimes called Cholesky or Cholesky without square roots also. In this case you do not need square roots. Any way puzzle seems to be resolved :) What about my questions ? What the reason can be to have eigs = D ?
Dec 28, 2011 at 8:57 comment added Federico Poloni In Cholesky, $D=I$ and the diagonal scaling is "included" in the $L$ factor, which need not have ones on its main diagonals. So that's where you get those square roots from.
Dec 28, 2011 at 8:23 comment added Alexander Chervov I've heard similar, but there were some conditions, may be small eigs, or like that... But "squares" seems to me misprint - take A=D then they are equal. Any way... What can be the reason for relation in case "L" is not very small ? Is there some intuitive explanation ?
Dec 28, 2011 at 1:05 history answered Igor Rivin CC BY-SA 3.0