I read the following statement in a exercise, which I don't think is correct:
$NM=I \rightarrow MN=I$, where $M,N$ are real square matrix of finite size.
However, I cannot think of a counter-example to this.
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@MathLover There is no such condition given so that N or M is invertible tho – Daniel Li Sep 12 '17 at 22:41
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1Left multiply by $N^{-1}$ and then right multiply by N. – Donald Splutterwit Sep 12 '17 at 22:42
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By definition, $N$ is invertible; now multiply by $N^{-1}$ on the LHS and $N$ on the RHS of both equations. – Steven Stadnicki Sep 12 '17 at 22:42
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Oh yeah det(M)*det(N)=1 – Daniel Li Sep 12 '17 at 22:42
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1@MathLover Clearly. However, this is about proving that there is such a thing as "the" inverse, and not separate left inverse and right inverse. – Arthur Sep 12 '17 at 22:42
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1The reason it is hard to prove is that it isn't true for "infinite" square matrices. – Thomas Andrews Sep 12 '17 at 22:48
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1Yes, via the invertible matrix theorem. For $M$ if $MN =I$ then $N$ is unique in doing so and $NM = I$. http://mathworld.wolfram.com/InvertibleMatrixTheorem.html – fleablood Sep 12 '17 at 22:59
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^Learning the proof of (relevant parts of) this theorem will answer the OP's question. – G Tony Jacobs Sep 12 '17 at 23:54