If u have a 4x3 matrix that is linearly independent, why can't you solve for Ax = b for EVERY b ?

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- Jan 3rd 2009, 02:59 PMZero266Vector Space
If u have a 4x3 matrix that is linearly independent, why can't you solve for Ax = b for EVERY b ?

- Jan 3rd 2009, 03:08 PMThePerfectHacker
- Jan 3rd 2009, 04:07 PMZero266
Okay but then why does a 4X4 matrix that is linearly independent with Ax= b can be solved for every b? i don't see how you use the fact that the matrix in question was 4x3

- Jan 3rd 2009, 06:32 PMThePerfectHacker
- Jan 4th 2009, 07:59 PMmath2009
may be not in image of

- Jan 7th 2009, 03:55 AMHallsofIvy
The point made before is that a n by m matrix, with independent columns, maps to an n dimensional subspace of . If n< m, that is NOT all of and there will be vectors b not in that subspace so Ax= b would have no solution.

But if n= m, that is, if A is an n by n matrix with independent columns, then its image is**all**of .