The inverse of $(A+A^-1)$ when $A=A^-1$

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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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  • 3




    You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
    – user583896
    Aug 14 at 11:44











  • Ah, of course, thanks!
    – Chisq
    Aug 14 at 11:46














up vote
2
down vote

favorite












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.







share|cite|improve this question
















  • 3




    You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
    – user583896
    Aug 14 at 11:44











  • Ah, of course, thanks!
    – Chisq
    Aug 14 at 11:46












up vote
2
down vote

favorite









up vote
2
down vote

favorite











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.







share|cite|improve this question












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.









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 14 at 11:43









Chisq

1467




1467







  • 3




    You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
    – user583896
    Aug 14 at 11:44











  • Ah, of course, thanks!
    – Chisq
    Aug 14 at 11:46












  • 3




    You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
    – user583896
    Aug 14 at 11:44











  • Ah, of course, thanks!
    – Chisq
    Aug 14 at 11:46







3




3




You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
– user583896
Aug 14 at 11:44





You need to invert the $2$ as well. $(2^-1A)(A+A^-1)=2^-1(AA+AA^-1)=2^-1(2I)=I$.
– user583896
Aug 14 at 11:44













Ah, of course, thanks!
– Chisq
Aug 14 at 11:46




Ah, of course, thanks!
– Chisq
Aug 14 at 11:46










2 Answers
2






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up vote
2
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accepted










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






share|cite|improve this answer






















  • Also known as $frac12A$
    – stuart stevenson
    Aug 14 at 11:55










  • You mean $frac12A^-1$?
    – A. Pongrácz
    Aug 14 at 12:06

















up vote
0
down vote













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.






share|cite|improve this answer




















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










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






    share|cite|improve this answer






















    • Also known as $frac12A$
      – stuart stevenson
      Aug 14 at 11:55










    • You mean $frac12A^-1$?
      – A. Pongrácz
      Aug 14 at 12:06














    up vote
    2
    down vote



    accepted










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






    share|cite|improve this answer






















    • Also known as $frac12A$
      – stuart stevenson
      Aug 14 at 11:55










    • You mean $frac12A^-1$?
      – A. Pongrácz
      Aug 14 at 12:06












    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






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






    share|cite|improve this answer














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







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 14 at 12:02









    A. Pongrácz

    3,682624




    3,682624










    answered Aug 14 at 11:48









    Sadil Khan

    3297




    3297











    • Also known as $frac12A$
      – stuart stevenson
      Aug 14 at 11:55










    • You mean $frac12A^-1$?
      – A. Pongrácz
      Aug 14 at 12:06
















    • Also known as $frac12A$
      – stuart stevenson
      Aug 14 at 11:55










    • You mean $frac12A^-1$?
      – A. Pongrácz
      Aug 14 at 12:06















    Also known as $frac12A$
    – stuart stevenson
    Aug 14 at 11:55




    Also known as $frac12A$
    – stuart stevenson
    Aug 14 at 11:55












    You mean $frac12A^-1$?
    – A. Pongrácz
    Aug 14 at 12:06




    You mean $frac12A^-1$?
    – A. Pongrácz
    Aug 14 at 12:06










    up vote
    0
    down vote













    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.






    share|cite|improve this answer
























      up vote
      0
      down vote













      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.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 14 at 11:51









        dmtri

        571316




        571316






















             

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