Since the pivot columns of a matrix are always linearly independent and span the columns of the matrix, they usually make up the basis of the matrix. However, this had me wondering about the non pivot columns of the matrix. According to column correspondence theory, the non pivot columns are linear combinations of the pivot columns. Does this mean that the non pivot columns could also be a generating set of the columns like how the pivot columns are a generating set of the columns? Then, if the non pivot columns were proven to be linearly independent by finding the reduced row echelon form or by some other means and the number of non pivot columns was equal to the number of pivot columns (therefore, the dimension wouldn't be a problem), could the non pivot columns make up a basis for the column space of a matrix?
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$\begingroup$ @Bye_World: I suggest that you answer in the answer box and not in the comments. $\endgroup$– Martin ArgeramiNov 24, 2015 at 11:31
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$\begingroup$ @Bye_World Why did you delete your comment? I thought it was a good answer :( $\endgroup$– Chris GongNov 24, 2015 at 15:10
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Yes, it is perfectly possible. When you perform row reduction, you are set to make the first columns the pivot columns. But the column space does not depend on the order of the columns. Nothing prevents you from doing "row reduction" by working on the last column first.
answered Nov 24, 2015 at 11:30
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$\begingroup$ So is it fair to say that we usually use the pivot columns to find a basis for the column space because we know they're linearly independent, yet non pivot columns can also be generating a set for the column space? $\endgroup$ Nov 24, 2015 at 12:57
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