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Are there any matrices for which the Gaussian method yields wrong/ or most inaccurate results? I've implemented a full choice algorythm, where i switch rows and columns so that the current element is biggest.

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    $\begingroup$ If you mean Gaussian Elimination (en.wikipedia.org/wiki/Gaussian_elimination), then it can proven to always reduce a matrix to a reduced row echelon form. $\endgroup$
    – Dan Rust
    Apr 8, 2013 at 17:49
  • $\begingroup$ Gaussian elimination with complete pivoting is quite stable. $\endgroup$ Apr 8, 2013 at 17:51
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    $\begingroup$ The main issue with Gaussian Elimination is 'element growth'. In practice, this is not an issue. In fact, just partial pivoting is pretty good, in practice. $\endgroup$
    – copper.hat
    Apr 8, 2013 at 18:00
  • $\begingroup$ With respect to @copper.hat's comment, you will want to read the book by Golub and Van Loan (which I linked to in my previous comment). $\endgroup$ Apr 8, 2013 at 18:14
  • $\begingroup$ @J.M.: Useful link. I need to get a newer edition. I just read about rank pivoting in link. $\endgroup$
    – copper.hat
    Apr 8, 2013 at 18:17

2 Answers 2

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If you are looking for matrices which will give inaccurate results, when solving numerically, you might want to look at the Hilbert matrix which has a condition number that grows exponentially with its size.

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  • $\begingroup$ right condition number for matrix should be close to $1$,+1 for Hilbert matrix $\endgroup$ Apr 8, 2013 at 17:57
  • $\begingroup$ what about the last column? Since the Hilbert Matrix is square, should i fill that with random values? $\endgroup$ Apr 8, 2013 at 18:08
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    $\begingroup$ The funny thing is that since the Hilbert matrix is symmetric positive definite, you don't even need to pivot! In any event, see this paper for things on the decomposition of the Hilbert matrix. $\endgroup$ Apr 8, 2013 at 18:13
  • $\begingroup$ kalandarson.blogspot.com/2012/03/… $\endgroup$ Apr 8, 2013 at 18:17
  • $\begingroup$ @coolbartek Sure, just try different values. $\endgroup$ Apr 8, 2013 at 18:24
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as wikipedia says it always works,but sometimes it needs minor changes, so you can read this document http://www.personal.psu.edu/bmw5075/360notes.pdf

here is given advantages and disadvantages of this method ,their comparison,please see it.

EDITED:

for matrix ,on last page is given formulas and this statement

Matrix Norms Matrix norms are natural extensions of vector norms, but are not as clearly defined. They have to be tested over the whole area of the matrix:

we can's simply apply on matrix vector's norm definition,just test on whole matrix

use wikipedia http://en.wikipedia.org/wiki/Matrix_norm

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  • $\begingroup$ looks really nice, but could you explain the condition with the norms a bit more? Should i check for norm 1 or norm infinity? $\endgroup$ Apr 8, 2013 at 18:02
  • $\begingroup$ please see my edited version $\endgroup$ Apr 8, 2013 at 18:06

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