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A friend of mine asked me today for this example:

If there is the inverse operator of the operator A, then $(A^{-1})^{-1}=A$?

But I do not have the ability to help, so I told him that his example will be posted on this site, and I am convinced that a number of colleagues will help us. I thank you for your answers preliminarily.

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  • $\begingroup$ What kind of operator are we talking about? What are your spaces? $\endgroup$
    – icurays1
    Commented Jul 27, 2014 at 16:47
  • $\begingroup$ $A$ is a linear operator $\endgroup$
    – Madrit Zhaku
    Commented Jul 27, 2014 at 16:57
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    $\begingroup$ It's defined as $A^{-1}A=AA^{-1}=I$ $\endgroup$
    – Shabbeh
    Commented Jul 27, 2014 at 16:59

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Be careful because some operators $A$ only have a right or left inverse and in this case it is not necessarily true that $(A^{-1})^{-1} = A.$ For example consider the Hilbert space $l_2$ of all square summable (complex) sequences. Define $A : l_2 \to l_2$ by $A(a_1, a_2, a_3, ...) = (0,a_1, a_2, ...)$ the right shift operator. Now it has a one sided inverse, namely the left shift operator, $B : l_2 \to l_2$ by $B(a_1, a_2, a_3, ...) = (a_2, a_3,...)$. Clearly, $BA(a_1, a_2, ...) = (a_1, a_2, ...)$, so $BA = I$. However, it is easy to see that $AB \neq I$.

If your operator $A$ comes from a group, then the statement is true.

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  • $\begingroup$ But $B$ is not the inverse of $A$, so the hypotheses of the problem are not satisfied. $\endgroup$
    – Jonas Meyer
    Commented Jul 27, 2014 at 17:25
  • $\begingroup$ thanks sir, please help me for the last part, it is easy to see that $AB≠I$. $\endgroup$
    – Madrit Zhaku
    Commented Jul 27, 2014 at 17:25
  • $\begingroup$ @ Jonas Meyer: i.e., The above solution is not correct $\endgroup$
    – Madrit Zhaku
    Commented Jul 27, 2014 at 17:26
  • $\begingroup$ $AB(a_1, a_2, a_3,...) = A(a_2, a_3, ..) = (0, a_2, a_3, ..)$ so you lose the first coordinate. $\endgroup$
    – Mustafa Said
    Commented Jul 27, 2014 at 17:26
  • $\begingroup$ can you please explain this to add to your solution to be more clear $\endgroup$
    – Madrit Zhaku
    Commented Jul 27, 2014 at 17:29

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