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Math Help - Proving matrix inverse properties

  1. #1
    Junior Member
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    Proving matrix inverse properties

    Assume that A inR^(nxn) and without using the invertible matrix theorem, prove the following:

    3.1.
    Spanning Sets. If A is an n x n matrix and A^(-1) exists, then the columns of A span R^n.
    3.2.
    Pivot Structure. If A is an n x n matrix and Ax = b has a solution for each b inR^n, then A is invertible.

    3.3.
    Linear Independence. If the matrix A is invertible, then the columns of A^(1)are linearly independent.

    I'm not sure how to do any of these. I don't know where to begin or anything. If someone could guide me through them that would be very helpful. Thank you
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by SpiffyEh View Post
    Assume that A inR^(nxn) and without using the invertible matrix theorem, prove the following:

    3.1.
    Spanning Sets. If A is an n x n matrix and A^(-1) exists, then the columns of A span R^n.
    3.2.
    Pivot Structure. If A is an n x n matrix and Ax = b has a solution for each b inR^n, then A is invertible.

    3.3.
    Linear Independence. If the matrix A is invertible, then the columns of A^(1)are linearly independent.

    I'm not sure how to do any of these. I don't know where to begin or anything. If someone could guide me through them that would be very helpful. Thank you
    3.1: I suggest that you try to prove the converse: that if the columns of A do not span \mathbb{R}^n, then A cannot be invertible: because, in that case, there would exist a vector y\in \mathbb{R}^n\backslash \mathrm{span}(A) for which there is no x\in \mathbb{R}^n with Ax = y, which contradicts invertibility of A.

    3.2 To say that Ax=b has a solution for each b in \mathbb{R}^n amounts to the same thing as saying that the span of A is the whole of \mathbb{R}^n. So this follows directly from 3.1.

    3.3. If the columns of A^{-1} are not linearly independent, then the columns of A^{-1} cannot span the whole of \mathbb{R}^n, hence by 3.1 A^{-1} is not invertible, thus contradicting the invertibility of A itself.
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