Comments on: Hilbert Matrices https://blogs.mathworks.com/cleve/2017/06/07/hilbert-matrices-2/?s_tid=feedtopost Cleve Moler is the author of the first MATLAB, one of the founders of MathWorks, and is currently Chief Mathematician at the company. He writes here about MATLAB, scientific computing and interesting mathematics. Tue, 18 Jul 2017 02:01:44 +0000 hourly 1 https://wordpress.org/?v=6.2.2 By: Cleve Moler https://blogs.mathworks.com/cleve/2017/06/07/hilbert-matrices-2/#comment-6242 Tue, 18 Jul 2017 02:01:44 +0000 https://blogs.mathworks.com/cleve/?p=2528#comment-6242 See: blogs.mathworks.com/cleve/2017/07/17/what-is-the-condition-number-of-a-matrix/

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By: Michele https://blogs.mathworks.com/cleve/2017/06/07/hilbert-matrices-2/#comment-6227 Fri, 16 Jun 2017 05:19:59 +0000 https://blogs.mathworks.com/cleve/?p=2528#comment-6227 This is more than I expected, thank you! I will be waiting.

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By: Cleve Moler https://blogs.mathworks.com/cleve/2017/06/07/hilbert-matrices-2/#comment-6226 Thu, 15 Jun 2017 15:48:39 +0000 https://blogs.mathworks.com/cleve/?p=2528#comment-6226 In reply to Michele.

Good question, Michele. The answer deserves a proper explanation, so I’ll make it the subject of a blog post. I’ve already got something cooking for next week, so it will be a couple of weeks before I get to this topic. Thanks for asking. — Cleve

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By: Michele https://blogs.mathworks.com/cleve/2017/06/07/hilbert-matrices-2/#comment-6225 Thu, 15 Jun 2017 13:41:33 +0000 https://blogs.mathworks.com/cleve/?p=2528#comment-6225 You made the remark that on the Burroughs 205, you did better on inverting the Hilber matrix than the condition number seemed to imply. cond(H) gives an upper estimate of the error, but this upper estimate can be way off. Suppose I have two matrices: the identity matrix and the diagonal matrix with elements 1,2,4,8,16…2^n; their condition number is 1 and 2^n respectively. I don’t see any reason why in the second case I would lose log10(2^n) digits.
Can you comment on when the condition number gives a tight estimate and whether there is a better estimator?

michele

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By: Cleve Moler https://blogs.mathworks.com/cleve/2017/06/07/hilbert-matrices-2/#comment-6221 Sun, 11 Jun 2017 01:49:54 +0000 https://blogs.mathworks.com/cleve/?p=2528#comment-6221 In reply to José-Javier Martínez.

Hi —

Thanks for the comment and the references. Koev’s paper is available from his web site:
https://math.mit.edu/~plamen/files/acctp.pdf

But maybe I didn’t make this point strongly enough — we can’t use these high accuracy algorithms, for two reasons. First, we only start with a floating point approximation to what would be a totally positive matrix. The roundoff errors made in generating the matrix in the first place have a bigger effect on the inverse than those generated during the inversion process. Second, to take advantage of totally positivity, it is necessary to have the representation as a product of exact, rational, bidiagonal matrices.

— Cleve

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By: José-Javier Martínez https://blogs.mathworks.com/cleve/2017/06/07/hilbert-matrices-2/#comment-6218 Sat, 10 Jun 2017 18:30:43 +0000 https://blogs.mathworks.com/cleve/?p=2528#comment-6218 Dear Professor Moler: I would like to recall an additional useful property of the Hilbert matrix, namely its total positivity. Interestingly, this property is also related to the work of John Todd and Olga Taussky, since the reference given by Nick Higham in his “Test Matrix Toolbox for MATLAB” for the total positivity of Cauchy and Hilbert matrices was

-Olga Taussky and Marvin Marcus: Eigenvalues of finite matrices. In John Todd, editor, Survey of Numerical Analysis, pages 279—313. McGraw-Hill, New York, 1962.

The power of this property int the context of computing with high relative accuracy has been shown in

– Plamen Koev: Accurate computations with totally nonnegative matrices. SIAM J. Matrix Anal. Appl. 29(3), pp. 731-751 (2007).

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