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Show that a matrix has a right inverse

  1. Apr 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##A## be an ##m \times n## matrix with rank ##m##. Prove that there exists an ##n \times m## matrix ##B## such that ##AB= I_m##

    2. Relevant equations


    3. The attempt at a solution
    So here is how far I get. I am given that ##A## has rank ##m##. Since ##L_A(x) = Ax## is a map ##\mathbb{R}^n \rightarrow \mathbb{R}^m##, this means that ##L_A## is a surjective map.
    I know that in set theory, surjective maps must have right-inverses, so I get the sense that I am on the right track. But I am not sure how to continue the proof.
     
  2. jcsd
  3. Apr 3, 2017 #2
    You should be more precise in your expression. The fact that you have a map ##\mathbb R^n\to\mathbb R^m ## doesn't imply it's surjective.

    What do we know about ##m, n ##? Right iverses certainly exist [in fact, there's a whole family of such right inverses] when there are more columns than rows and the matrix ##A## has a full row rank.

    The Rank-Nullity theorem provides a solution.
     
  4. Apr 3, 2017 #3
    The Rank-Nullity theorem seems to tell me that ##\text{nullity}(A) = n-m>0##. But I don't see how this helps me
     
  5. Apr 3, 2017 #4

    StoneTemplePython

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    Gold Member

    my general approach is to make use of orthogonality for these things -- it is a much cleaner cut than general linear independence. Since we are in reals, consider using gram-schmidt and get:

    ##\mathbf A = \mathbf{T V}^T##

    (this is an atypical factorization but one I used in another of your threads recently), or more typically:

    ##\mathbf A^T = \mathbf{Q R}##, hence ##\mathbf A = \Big(\mathbf R^T \mathbf Q^T \Big)##.

    Where ##\mathbf Q## is orthogonal, but not full rank (it is n x m) and ##\mathbf R## is square m x m matrix.

    What happens if you multiply

    ##\Big( \mathbf R^T \mathbf Q^T\Big) \Big(\mathbf Q\Big) = \mathbf R^T \big(\mathbf Q^T \mathbf Q\big) = \mathbf R^T##. So we need to be able to invert ##\mathbf R^T##... how do I know that ##\mathbf R^T## is invertible?

    You could also use Singular Value Decomposition to get this.
     
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