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Math Help - Vector Space

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    Vector Space

    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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    Quote Originally Posted by Zero266 View Post
    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.
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    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
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    Quote Originally Posted by Zero266 View Post
    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.
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    b may be not in image of Ax
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    Quote Originally Posted by Zero266 View Post
    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.
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