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I have a matrix that is its own inverse, $A=A^{-1}$. I want to calculate the inverse of $(A+A^{-1})$, for which I would like to use the following chain of equalities: $$(A+A^{-1})^{-1}=(2A)^{-1}=2A^{-1}=2A$$

It doesn't seem to work though, why is that? I can't find anything related among the arithmetic rules in our course literature.

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    $\begingroup$ You need to invert the $2$ as well. $(2^{-1}A)(A+A^{-1})=2^{-1}(AA+AA^{-1})=2^{-1}(2I)=I$. $\endgroup$
    – user583896
    Commented Aug 14, 2018 at 11:44
  • $\begingroup$ Ah, of course, thanks! $\endgroup$
    – Chisq
    Commented Aug 14, 2018 at 11:46

2 Answers 2

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$A+A^{-1} = 2A$ The inverse of $2A$ is $2^{-1} A^{-1}$ i.e the inverse of $A+A^{-1}$ is $2^{-1} A^{-1}$

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  • $\begingroup$ Also known as $\frac{1}{2}A$ $\endgroup$
    – Leonhard Euler
    Commented Aug 14, 2018 at 11:55
  • $\begingroup$ You mean $\frac{1}{2}A^{-1}$? $\endgroup$
    – A. Pongrácz
    Commented Aug 14, 2018 at 12:06
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We just need to find a matrix $B$ so as $B(A+A^{-1})=I$.After some trials we see that $1/2A$ is what we are looking for.

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