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Proving that a right (or left) inverse of a square matrix is unique using only basic matrix operations

-- i.e. without any reference to higher-order matters like rank, vector spaces or whatever ( :)).

More precisely, armed with the knowledge only of:

  • rules of matrix equality check, addition, multiplication, distributive law and friends
  • Gauss-Jordan elimination and appropriate equation system solution cases for the reduced row-echelon form

Thanks in advance.

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    The idea I've got is that since elementary transforms of I will never give a zero row, any matrix A which is not row-equivalent to I, will produce an unsolvable set of equations for at least one column of B in A*B = I. Hence, for that matrix equation to hold, A must be row-equvalent to I and thus there's only one B sufficing the equation.2012-02-17
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    Or is it really that exotic?2012-02-17
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    "...and thus there's only one B sufficing the equation." How did you conclude that? Why do you need to show that A is row equivalent to I? Focus on the question: uniqueness.2012-02-17
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    Whatever row operations we apply to A to get it into reduced row-echelon form, if the same row operations are applied to I, the resulting matrix will have no zero rows. Then, if and only if A is row-equivalent to I, there will be a single set of solutions (i.e. B's column values) for every equation A_*B=I_, where A_ and I_ are transformed A and I and denotes j column. If there's a zero row i in A_, there will be at least one such j that I_ has a non-zero element in row i, and thus the system of equations has no solution.2012-02-17
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    @Arturo Thanks for keeping the comment.2012-02-17
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    But how do I deal with the left inverse? (Is there a simmetry I have overlooked?)2012-02-17
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    @mlvljr: If you can solve the problem for inverses on one side, taking transposes solves the problem for inverses on the other side: $AB=I$ if and only if $B^TA^T=I$, and if you know that right inverses are unique when they exist, then it follows that left inverses are unique when they exist, and vice-versa.2012-02-17
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    @Arturo That's what I was thinking to try, thanks! (Once again you help me :) )2012-02-17

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