Linear Algebra - Lecture 30 - Basis of a Subspace

In this video, I give the definition of "basis" for a subspace. Then, I work through the process for finding a basis for the null space and column space of any matrix.

[aubreydebruyn]

Thank you for amazing. Firstly, I enjoy that the lectures are numbered, so if a one was searching for concepts and finds your vids, it can be linked to the series. Secondly the reinforcements style, always writing out previous knowledge . Thirdly and most important, you teach exam technics.

[starcrosswongyl]

isnt it supposed to be Rn? in 4:52?

Answer: Sorry i didn't realise that the matrix is of the form n x m, thus Rm.

[weisanpang]

[ 1 4 0 2 0
0 0 1 -1 0
0 0 0 0 1
0 0 0 0 0]

[ 1 -2 0 -1 3
0 0 1 2 -2
0 0 0 0 0 ]

Hi professor James, above are matrices used in the 2 examples of finding the basis vectors of column space of a matrix, the first matrix is the original matrix, the second matrix is in reduced row echelon form.
My question is:
Even though it's easy to establish the basis using the pivot columns due to their unique "1" locations in the columns which allow them to be linearly combined to form any other vectors easily, however, from these 2 examples, it could be seen that the non-pivot columns could also be used as the basis. Example for the first matrix, if we choose column 2 and 4 as the basis, column 1 is (1/4 x column 2), and column 3 is (1/2 x column 2 - column 4), and column 2 and 4 (and 5) are linearly independent, so theoretically column 2, 4 and 5 can be used as the basis for this matrix, with the outcome that the basis vectors would have been very different from the pivot column solution.
Likewise, the second matrix could use column 2 and 4 as basis, which would see that column 1 is (-1/2 x column 2), and column 3 is (-1/4 x column 2 + 1/2 x column 3), and column 5 is (-1 x column2 + (-1 x column 4)), and column 2 and 4 are linearly independent.
That said, other than the reason (which I presumed) the pivot columns are easy to work with, why don't we choose the non-pivot columns as potential basis ? I reckon that the non-pivot columns may not be linearly independent all the time, but what are the other reasons ? From the view point of change of basis, choosing different sets of basis means choosing a totally different matrix transformation, so I am thinking it's not a trivial consideration. Thank you very much if you take the time to answer my doubt.

[jgehiuj]

Professor James, could you suggest me a book reference that you followed?