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
Define $\displaystyle T:\mathbb{R}^3 \to \mathbb{R}^4$ by $\displaystyle T(\bold{x}) = A\bold{x}$.

It is straightforward to show that $\displaystyle T$ is a linear transformation.

Since the dimension of $\displaystyle \mathbb{R}^3$ over $\displaystyle \mathbb{R}$ is less than the dimension of $\displaystyle \mathbb{R}^4$ over $\displaystyle \mathbb{R}$ is means there is no surjective linear transformation. Thus, there exists $\displaystyle \bold{y}\in \mathbb{R}^4$ such that $\displaystyle A\bold{x} = \bold{y}$ is not solvable. - 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
$\displaystyle b$ may be not in image of $\displaystyle Ax$

- Jan 7th 2009, 03:55 AMHallsofIvy
The point made before is that a n by m matrix, with independent columns, maps $\displaystyle R^n$ to an n dimensional subspace of $\displaystyle R^m$. If n< m, that is NOT all of $\displaystyle R^m$ 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 $\displaystyle R^n$.