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by wjomlex on July 8th, 2008

wjomlex

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Let A be an m*n matrix. Show that, if and only if rank A = m, there exists an n*m matrix such that AB = Im (Identity matrix of size m)

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  • by Couchyam on July 23rd, 2009

    Couchyam

    An intuitive way of looking at it that might help for your outline of the proof is this: an mxn matrix with rank m means that matrix A has m linearly independent vectors within a space R^n. If AB = I, then A reverts coordinates shifted in mapping B back to their original locations. Thus, B must be an nxm matrix, so that it has an m dimensional input yielding an n dimensional output, which can be fed into A to yield the same m dimensional input (or a symmetrical one).
    If rank A didn't equal m, then there wouldn't be enough independent vectors to span R^m, and so any mapping AB would be short of one or more degrees of freedom, leaving the identity matrix with a zero row.

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