Gaussian elimination (pivoting)

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SUMMARY

The forum discussion centers on the method of Gaussian elimination with pivoting, specifically addressing the correct approach to selecting pivot elements in a matrix. The original poster questions their lecturer's method of maximizing elements across rows instead of columns, as outlined in their textbook. The consensus among respondents is that while both methods can yield correct results, the standard approach is to maximize the leading element in each column before performing elimination. This ensures that the pivoting process is effective and maintains numerical stability.

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  • #31
Mark44 said:
The you look at a33 as a potential pivot.
?? What do you mean ?
 
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  • #32
fonseh said:
?? What do you mean ?
Then you look at a33 as a potential pivot.
 
  • #33
Ray Vickson said:
Generally, we start in column 1 and move the numerically largest value in that column to position (1,1), perhaps by swapping some row i and row 1. Then we do row operations to zero out the remaining elements in column 1.

As I recall, some computer implementations of Gaussian elimination that concern themselves with "numerical methods" begin by switching rows and columns so the element with largest absolute value (picked from the all elements in the entire matrix) is in position 1,1. They follow a similar policy to select subsequent pivots. However, for making theoretical conclusions about the algorithm of Gaussian elimination or prescribing how to do hand-calculations it is simpler to consider an algorithm where less freedom-of-choice is allowed in selecting pivots.

It would be challenging exercise in scholarship to determine if there is a "standard" definition of "Gaussian elimination" as a unique algorithm.
 
  • #34
Stephen Tashi said:
As I recall, some computer implementations of Gaussian elimination that concern themselves with "numerical methods" begin by switching rows and columns so the element with largest absolute value (picked from the all elements in the entire matrix) is in position 1,1. They follow a similar policy to select subsequent pivots. However, for making theoretical conclusions about the algorithm of Gaussian elimination or prescribing how to do hand-calculations it is simpler to consider an algorithm where less freedom-of-choice is allowed in selecting pivots.

It would be challenging exercise in scholarship to determine if there is a "standard" definition of "Gaussian elimination" as a unique algorithm.

In my post I did mention that some implementations just seek a pivot element whose absolute value exceeds some threshold, without bothering to look for the largest. Of course, some implementations do not move rows at all, but just store pointers and lists and the like in order to keep track of which rows have already originated pivots and which are still available for new pivots.

And, of course, with exact, rational arithmetic we do not need to worry at all about the magnitude of the pivot element, just so long as it is nonzero. I believe that "pure" (original) Gaussian elimination is like that; only "floating-point" Gaussian elimination needs to, and does exercise care in the pivot choice.
 
  • #35
Mark44 said:
What is so difficult to understand here? In the visible image of your post, they swapped rows 1 and 2. As I mentioned before, this is one of the three possible row operations. The next operation is to replace row 2 by -.02/500 * R1 + R2. This is another of the three row operations I mentioned. The a2, 1 entry in the second row will be 0 after this operation.
Or the notes in this picture is misleading ? It mentioned that the a11 or a22 is 0 , not when we pivot the a11 , the a21 and a31 is 0 ...
 

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  • #36
fonseh said:
Or the notes in this picture is misleading ? It mentioned that the a11 or a22 is 0 , not when we pivot the a11 , the a21 and a31 is 0 ...

The note you post is 100% clear, so you must be reading it incorrectly. Remember: after swapping rows 1 and ##i##, the old row ##i## becomes the new row 1, and vice-versa. The old ##a_{i1}## becomes the new ##a_{11}##, and the old ##a_{11}## becomes the new ##a_{i1}##.
 
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  • #37
Ray Vickson said:
The note you post is 100% clear, so you must be reading it incorrectly. Remember: after swapping rows 1 and ##i##, the old row ##i## becomes the new row 1, and vice-versa. The old ##a_{i1}## becomes the new ##a_{11}##, and the old ##a_{11}## becomes the new ##a_{i1}##.
It didnt mention about when after swapping row , the old a11 will become 0
 
  • #38
fonseh said:
It didnt mention about when after swapping row , the old a11 will become 0

No, you didn't, but the posted note DID, and that is all that counts here. You are trying to understand what the note is all about, and whether the note is correct. It is correct. (The note itself did not say that the old row i becomes the new row 1, etc., but it takes that as understood by the reader.)
 

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