Linear Algebra 9g: Which Columns Are Pivot Columns in Gaussian Elimination
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- čas přidán 27. 10. 2014
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Thank you!
Thought I had this previously and was wrong. Key for me is that the linearly dependent columns ( col 2 and 4 ) are made up of the two independent columns ( col 1 and 3) that come before it.
Was confusing to me how you knew that column 2 was 3 times column 1. Could see this was true, but it took a minute for me to see it was made up of the preceding independent column.
Same for column 4 being made up of the two preceding independent columns. Was not clear to me why column 2 was left out of column 4 until realized that 2 was dependent on col 1.
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All of this made sense, but at 8:54 things started to get confusing. "One of this column and one of the third". That kind of made sense as it means whatever the values are for x and z are linearly independent. Even "the second column is 3 times the first" kind of makes sense. We are using a scalar, alpha, because of the dependence, right? The final beta scalar is then trying to account for the second dependent column, but that's a little harder to piece together.
What if i switch column and RHS same time accordingly then do our different arrangement bring any change in which columns will get pivot element?
Should some way be made to emphasize the pivots in positions '1' and '3' ('x' and 'z') in the particular solution vector together with the _complementary_ positions of the other - I suppose 'slack' - variables of '-1' in positions '2' ('y') and '4' ('t') in each of the _null_ vectors? Since those particular entries will always be '-1', perhaps putting a these in brackets (i.e. as '(-1)' ) within their columns, suggesting complements, might be appropriate to bring out the relations?
| 1 | | 3 | | -1 |
| 0 | |(-1)| | 0 |
| 1 | | 0 | | 2 |
| 0 | | 0 | |(-1)|
Not a bad idea. I think some people might find it useful.
Would you explain to me why you get the -1in the second and fourth row?
+Sophia Zhao Yes. When determining the null space, we are looking for linear combinations of columns that equal zero. So if the second column is 3x the first, the corresponding linear combination is [ 3 -1 0 0 ]. And if the last column equals -1x the first column +2x the second column, then the corresponding linear combination is [ -1 0 2 -1 ].
The following video explains why we are looking for the null space in the first place: czcams.com/video/0b9bYQEBAZc/video.html
4:18, why did you eliminate the entry above the pivot, instead of eliminating the first entry of the last column ?
Ultimately, either one is a viable option. However, to stay systematic, you use your chosen *pivots* to do all the eliminating. According to the RREF convention, pivots are chosen as leftmost as possible. In this case, the second pivot is in column 3 so that's where the elimination takes place.
Thank you.