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Math Help - proof using invertible and spanning

  1. #1
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    proof using invertible and spanning

    Let A be an n x n matrix. Denote it's columns b v1, v2, ... vn. Keep in mind these are column vectors in Rn.

    Prove that A is invertible if and only if {v1, v2, ... vn} spans Rn.
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  2. #2
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    Re: proof using invertible and spanning

    I have no idea what you know and can use. But here is a quick proof using things I feel ought to be done before proving this theorem.



    A spans \mathbb{R}^n

    iff

    Ax=b is consistent for all b \in \mathbb{R}^n

    iff

    A has n pivot rows (and also n pivot columns since A is square)

    iff

    A is row equivalent to the n\times n identity matrix

    iff

    A is invertible.
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  3. #3
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    Re: proof using invertible and spanning

    First show that if you take A times \begin{bmatrix}1 \\ 0 \\ 0 \\ \cdot\cdot\cdot \\ 0\end{bmatrix} you get the first column of A, that if you take A times \begin{bmatrix} 0 \\ 1 \\ 0 \\ \cdot\cdot\cdot \\ 0 \end{bmatrix} you get the second column, etc.

    In other words, A maps the standard basis for Rn into the columns of A. Those vectors span the range of A.
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