A special subset of $GL(n;Bbb R)$












0












$begingroup$


Let $M(n,Bbb R)$ denote set of all $ntimes n$ real matrices. For $Ain M(n,Bbb R)$ we denote $A^t$ as the transpose of $A$. Denote $GL(n,Bbb R)$ as set of all invertible real $ntimes n$ matrices. A matix $Sin M(n,Bbb R)$ is said to be skew-symmetric if $S^t=-S$.



Define $$G:={Ain GL(n,Bbb R) : A^tSA=S,text{ for all skew-symmetric matrix $S$}}.$$ It easy to show that if $A_1,A_2in G$ then $A_1A_2$ is also in $G$.



My question: if $Ain G$, then is it true that $A^{-1}in G$ where $A^{-1}$ denotes the multiplicative inverse of $A$ in $GL(n,Bbb R)$?






linear-algebra matrices vector-spaces symmetric-matrices







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Just out of curiosity, can you make an explicit example of an element of $G$, say for $n=2$? Are you sure that $G$ is not ${I}$?
    $endgroup$
    – Giuseppe Negro
    Jan 14 at 14:07










  • $begingroup$
    @GiuseppeNegro I've checked the form for $n=2$ It seems that for the purpose of $G$ are good any matrices with determinant $=1$. Additionally, inverse has also determinant $=1$.
    $endgroup$
    – Widawensen
    Jan 14 at 16:41












  • $begingroup$
    @Widawensen: It seems to me that this is true. Interesting. Thanks
    $endgroup$
    – Giuseppe Negro
    Jan 14 at 18:29
















0












$begingroup$


Let $M(n,Bbb R)$ denote set of all $ntimes n$ real matrices. For $Ain M(n,Bbb R)$ we denote $A^t$ as the transpose of $A$. Denote $GL(n,Bbb R)$ as set of all invertible real $ntimes n$ matrices. A matix $Sin M(n,Bbb R)$ is said to be skew-symmetric if $S^t=-S$.



Define $$G:={Ain GL(n,Bbb R) : A^tSA=S,text{ for all skew-symmetric matrix $S$}}.$$ It easy to show that if $A_1,A_2in G$ then $A_1A_2$ is also in $G$.



My question: if $Ain G$, then is it true that $A^{-1}in G$ where $A^{-1}$ denotes the multiplicative inverse of $A$ in $GL(n,Bbb R)$?






linear-algebra matrices vector-spaces symmetric-matrices







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Just out of curiosity, can you make an explicit example of an element of $G$, say for $n=2$? Are you sure that $G$ is not ${I}$?
    $endgroup$
    – Giuseppe Negro
    Jan 14 at 14:07










  • $begingroup$
    @GiuseppeNegro I've checked the form for $n=2$ It seems that for the purpose of $G$ are good any matrices with determinant $=1$. Additionally, inverse has also determinant $=1$.
    $endgroup$
    – Widawensen
    Jan 14 at 16:41












  • $begingroup$
    @Widawensen: It seems to me that this is true. Interesting. Thanks
    $endgroup$
    – Giuseppe Negro
    Jan 14 at 18:29














0












0








0





$begingroup$


Let $M(n,Bbb R)$ denote set of all $ntimes n$ real matrices. For $Ain M(n,Bbb R)$ we denote $A^t$ as the transpose of $A$. Denote $GL(n,Bbb R)$ as set of all invertible real $ntimes n$ matrices. A matix $Sin M(n,Bbb R)$ is said to be skew-symmetric if $S^t=-S$.



Define $$G:={Ain GL(n,Bbb R) : A^tSA=S,text{ for all skew-symmetric matrix $S$}}.$$ It easy to show that if $A_1,A_2in G$ then $A_1A_2$ is also in $G$.



My question: if $Ain G$, then is it true that $A^{-1}in G$ where $A^{-1}$ denotes the multiplicative inverse of $A$ in $GL(n,Bbb R)$?






linear-algebra matrices vector-spaces symmetric-matrices







share|cite|improve this question











$endgroup$




Let $M(n,Bbb R)$ denote set of all $ntimes n$ real matrices. For $Ain M(n,Bbb R)$ we denote $A^t$ as the transpose of $A$. Denote $GL(n,Bbb R)$ as set of all invertible real $ntimes n$ matrices. A matix $Sin M(n,Bbb R)$ is said to be skew-symmetric if $S^t=-S$.



Define $$G:={Ain GL(n,Bbb R) : A^tSA=S,text{ for all skew-symmetric matrix $S$}}.$$ It easy to show that if $A_1,A_2in G$ then $A_1A_2$ is also in $G$.



My question: if $Ain G$, then is it true that $A^{-1}in G$ where $A^{-1}$ denotes the multiplicative inverse of $A$ in $GL(n,Bbb R)$?






linear-algebra matrices vector-spaces symmetric-matrices




linear-algebra matrices vector-spaces symmetric-matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 14:02









amWhy

1




1










asked Jan 14 at 13:30









UserSUserS

1,5391112




1,5391112








  • 2




    $begingroup$
    Just out of curiosity, can you make an explicit example of an element of $G$, say for $n=2$? Are you sure that $G$ is not ${I}$?
    $endgroup$
    – Giuseppe Negro
    Jan 14 at 14:07










  • $begingroup$
    @GiuseppeNegro I've checked the form for $n=2$ It seems that for the purpose of $G$ are good any matrices with determinant $=1$. Additionally, inverse has also determinant $=1$.
    $endgroup$
    – Widawensen
    Jan 14 at 16:41












  • $begingroup$
    @Widawensen: It seems to me that this is true. Interesting. Thanks
    $endgroup$
    – Giuseppe Negro
    Jan 14 at 18:29














  • 2




    $begingroup$
    Just out of curiosity, can you make an explicit example of an element of $G$, say for $n=2$? Are you sure that $G$ is not ${I}$?
    $endgroup$
    – Giuseppe Negro
    Jan 14 at 14:07










  • $begingroup$
    @GiuseppeNegro I've checked the form for $n=2$ It seems that for the purpose of $G$ are good any matrices with determinant $=1$. Additionally, inverse has also determinant $=1$.
    $endgroup$
    – Widawensen
    Jan 14 at 16:41












  • $begingroup$
    @Widawensen: It seems to me that this is true. Interesting. Thanks
    $endgroup$
    – Giuseppe Negro
    Jan 14 at 18:29








2




2




$begingroup$
Just out of curiosity, can you make an explicit example of an element of $G$, say for $n=2$? Are you sure that $G$ is not ${I}$?
$endgroup$
– Giuseppe Negro
Jan 14 at 14:07




$begingroup$
Just out of curiosity, can you make an explicit example of an element of $G$, say for $n=2$? Are you sure that $G$ is not ${I}$?
$endgroup$
– Giuseppe Negro
Jan 14 at 14:07












$begingroup$
@GiuseppeNegro I've checked the form for $n=2$ It seems that for the purpose of $G$ are good any matrices with determinant $=1$. Additionally, inverse has also determinant $=1$.
$endgroup$
– Widawensen
Jan 14 at 16:41






$begingroup$
@GiuseppeNegro I've checked the form for $n=2$ It seems that for the purpose of $G$ are good any matrices with determinant $=1$. Additionally, inverse has also determinant $=1$.
$endgroup$
– Widawensen
Jan 14 at 16:41














$begingroup$
@Widawensen: It seems to me that this is true. Interesting. Thanks
$endgroup$
– Giuseppe Negro
Jan 14 at 18:29




$begingroup$
@Widawensen: It seems to me that this is true. Interesting. Thanks
$endgroup$
– Giuseppe Negro
Jan 14 at 18:29










2 Answers
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$begingroup$

If $A in G$ and $ A^tSA=S$, then $A^tS=SA^{-1}$, hence $S=(A^t)^{-1}SA^{-1}$.



Since $(A^t)^{-1} = (A^{-1})^t$, we get $A^{-1} in G$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Yes. It stems from the fact that $(A^t)^{-1} = (A^{-1})^t$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
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      $begingroup$

      If $A in G$ and $ A^tSA=S$, then $A^tS=SA^{-1}$, hence $S=(A^t)^{-1}SA^{-1}$.



      Since $(A^t)^{-1} = (A^{-1})^t$, we get $A^{-1} in G$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        If $A in G$ and $ A^tSA=S$, then $A^tS=SA^{-1}$, hence $S=(A^t)^{-1}SA^{-1}$.



        Since $(A^t)^{-1} = (A^{-1})^t$, we get $A^{-1} in G$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          If $A in G$ and $ A^tSA=S$, then $A^tS=SA^{-1}$, hence $S=(A^t)^{-1}SA^{-1}$.



          Since $(A^t)^{-1} = (A^{-1})^t$, we get $A^{-1} in G$.






          share|cite|improve this answer









          $endgroup$



          If $A in G$ and $ A^tSA=S$, then $A^tS=SA^{-1}$, hence $S=(A^t)^{-1}SA^{-1}$.



          Since $(A^t)^{-1} = (A^{-1})^t$, we get $A^{-1} in G$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 14 at 13:48









          FredFred

          45.5k1848




          45.5k1848























              3












              $begingroup$

              Yes. It stems from the fact that $(A^t)^{-1} = (A^{-1})^t$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Yes. It stems from the fact that $(A^t)^{-1} = (A^{-1})^t$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Yes. It stems from the fact that $(A^t)^{-1} = (A^{-1})^t$.






                  share|cite|improve this answer









                  $endgroup$



                  Yes. It stems from the fact that $(A^t)^{-1} = (A^{-1})^t$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 14 at 13:41









                  MariuslpMariuslp

                  717212




                  717212






























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