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Prove or disprove: There exist two linear operator $T$ and $U$ on $\mathbb{R^{2}}$ such that $TU = I$ but $UT \neq I$

I am not able to find any such linear transformation.

If any of U or T is invertible then the statement is false. But, I am not able to obtain any information for non-invertible matrices.

gaufler
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  • @Arthur That the transformation is one-one,onto and more precisely the inverse of the matrix representation of the transformation exists. – gaufler Sep 23 '19 at 14:42
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    If either of $U$ or $T$ is non-invertible, then $\det(UT)=\det(TU)=0\neq\det(I)$ by invertible matrix theorem. Therefore you need only look at cases where they are both invertible. – 79037662 Sep 23 '19 at 14:43
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    At some point you must make use of the fact that $\Bbb R^2$ is finite-dimensional (because for the infinite-dimensional space of polynomials, $T=$ formal derivative and $U=$formal integral are an example), in other words, you may want to argue with bases. – Hagen von Eitzen Sep 23 '19 at 14:43
  • @HagenvonEitzen Either bases, or surjective $\iff$ injective. – Arthur Sep 23 '19 at 14:44
  • https://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i

    The question is answered here.

     – 79037662 Sep 23 '19 at 14:55

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