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I have came across through a statement that "There exists few operators for which only left inverse exists" and it holds for only right also. But considering the fact that any operator belonging to a finite linear vector space can be represented in matrix form. When we represent the above mentioned special operators in matrix form, it implies that right inverse also exists. How to explain this?

Vidya Sagar V
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    Such operators are meant to be between infinite-dimensional spaces, otherwise, as you say, left invertibility implies right invertibility. Look here for examples. – Crostul Sep 06 '16 at 09:10
  • I think the question kind of answers itself. – Tunococ Sep 06 '16 at 09:15
  • @Crostul Do you mean to say that operators with with only one inverse do not exist in finite linear vector space? – Vidya Sagar V Sep 06 '16 at 12:28
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    @VidyaSagar This is exactly what I mean; it's the same thing you said: square matrices are left invertible if and only if they are right invertible (provided that they are finite matrices). – Crostul Sep 06 '16 at 12:31
  • If we consider operators from one vector space to another, then anything can happen, infinite dimensions or no – Ben Grossmann Sep 06 '16 at 12:51

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The matrix $\pmatrix{1\\0}$ has a left-inverse, but no right-inverse. On the other hand, the matrix $\pmatrix{0&1}$ has a right-inverse, but no left-inverse.

A square matrix has a left inverse iff it has a right inverse iff it is invertible.

Ben Grossmann
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For square matrices, there is no doubt, $AB=I$ implies $detA.detB \neq 0$ so both matrices are invertible. You could have rectangular matrices with a one sided inversion (and obvioulsy no inversion on the other side). In infinite dimension you can have pretty much whatever you want : derivative and integral on the polynomial space for example.

marmouset
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