# Vector Space

• Jan 3rd 2009, 02:59 PM
Zero266
Vector 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 PM
ThePerfectHacker
Quote:

Originally Posted by Zero266
If u have a 4x3 matrix that is linearly independent, why can't you solve for Ax = b for EVERY b ?

Define $T:\mathbb{R}^3 \to \mathbb{R}^4$ by $T(\bold{x}) = A\bold{x}$.
It is straightforward to show that $T$ is a linear transformation.

Since the dimension of $\mathbb{R}^3$ over $\mathbb{R}$ is less than the dimension of $\mathbb{R}^4$ over $\mathbb{R}$ is means there is no surjective linear transformation. Thus, there exists $\bold{y}\in \mathbb{R}^4$ such that $A\bold{x} = \bold{y}$ is not solvable.
• Jan 3rd 2009, 04:07 PM
Zero266
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 PM
ThePerfectHacker
Quote:

Originally Posted by Zero266
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

Now if $A$ has linearly independent rows (or coloumns) then $A$ is invertible and so $\bold{x} = A^{-1}\bold{b}$.
Thus, it is solvable for 4x4 matrices.
• Jan 4th 2009, 07:59 PM
math2009
$b$ may be not in image of $Ax$
• Jan 7th 2009, 03:55 AM
HallsofIvy
Quote:

Originally Posted by Zero266
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

The point made before is that a n by m matrix, with independent columns, maps $R^n$ to an n dimensional subspace of $R^m$. If n< m, that is NOT all of $R^m$ 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 $R^n$.