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Whether the set of a symmetric copositive matrices $x^T A x geq 0$ for $forall x in mathbbR^n$ $x geq 0$ is convex?



The Next CEO of Stack OverflowNotation for the set of symmetric matrices and symmetric positive definite matricesIs this set of matrices convex?How to prove that the set of real $ n times n $ PSD matrices of rank $leq r < n $ and unit trace, is not contractible?Convex hull of rotation matrices is closed and contains the originIs the the set of symmetric positive-definite matrices in the space of all symmetric matrices strictly convex?Given $P$ a real symmetric positive semidefinite matrix, what matrix $A$ ensure that $A^TPA$ is PSD?Use the definition of convexity to show that $A$ is a convex set.Show that a set of positive semidefinite (PSD) matrices is a convex setProjection of a Symmetric Matrix onto the space of Positive Semi-Definite MatricesHow can I prove that $mathbbS_+^n$ is a closed and convex set?










0












$begingroup$



Whether the set of a $n times n$ symmetric copositive matrices $x^T A x geq 0$ $forall x in mathbbR^n$ and $x geq 0$ is convex?





Attempt:



If I show that the set of the positive semidefinite (psd) matrices is convex, then the intersection of the set of psd matrices and $x geq 0$ is also convex. Is that correct?










share|cite|improve this question











$endgroup$











  • $begingroup$
    What is "copositive" ?
    $endgroup$
    – kimchi lover
    yesterday










  • $begingroup$
    @kimchilover en.wikipedia.org/wiki/Copositive_matrix
    $endgroup$
    – Yanko
    yesterday















0












$begingroup$



Whether the set of a $n times n$ symmetric copositive matrices $x^T A x geq 0$ $forall x in mathbbR^n$ and $x geq 0$ is convex?





Attempt:



If I show that the set of the positive semidefinite (psd) matrices is convex, then the intersection of the set of psd matrices and $x geq 0$ is also convex. Is that correct?










share|cite|improve this question











$endgroup$











  • $begingroup$
    What is "copositive" ?
    $endgroup$
    – kimchi lover
    yesterday










  • $begingroup$
    @kimchilover en.wikipedia.org/wiki/Copositive_matrix
    $endgroup$
    – Yanko
    yesterday













0












0








0





$begingroup$



Whether the set of a $n times n$ symmetric copositive matrices $x^T A x geq 0$ $forall x in mathbbR^n$ and $x geq 0$ is convex?





Attempt:



If I show that the set of the positive semidefinite (psd) matrices is convex, then the intersection of the set of psd matrices and $x geq 0$ is also convex. Is that correct?










share|cite|improve this question











$endgroup$





Whether the set of a $n times n$ symmetric copositive matrices $x^T A x geq 0$ $forall x in mathbbR^n$ and $x geq 0$ is convex?





Attempt:



If I show that the set of the positive semidefinite (psd) matrices is convex, then the intersection of the set of psd matrices and $x geq 0$ is also convex. Is that correct?







matrices convex-analysis copositivity






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share|cite|improve this question













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share|cite|improve this question








edited yesterday









Rodrigo de Azevedo

13.2k41960




13.2k41960










asked yesterday









learninglearning

617




617











  • $begingroup$
    What is "copositive" ?
    $endgroup$
    – kimchi lover
    yesterday










  • $begingroup$
    @kimchilover en.wikipedia.org/wiki/Copositive_matrix
    $endgroup$
    – Yanko
    yesterday
















  • $begingroup$
    What is "copositive" ?
    $endgroup$
    – kimchi lover
    yesterday










  • $begingroup$
    @kimchilover en.wikipedia.org/wiki/Copositive_matrix
    $endgroup$
    – Yanko
    yesterday















$begingroup$
What is "copositive" ?
$endgroup$
– kimchi lover
yesterday




$begingroup$
What is "copositive" ?
$endgroup$
– kimchi lover
yesterday












$begingroup$
@kimchilover en.wikipedia.org/wiki/Copositive_matrix
$endgroup$
– Yanko
yesterday




$begingroup$
@kimchilover en.wikipedia.org/wiki/Copositive_matrix
$endgroup$
– Yanko
yesterday










2 Answers
2






active

oldest

votes


















2












$begingroup$

No need to do that! Note that if $A,B$ are copositive, then for any $0<lambda<1$ and $xge 0$:$$x^TBigg(lambda A+(1-lambda)BBigg)x=lambda x^TAx+(1-lambda )x^TBxge 0$$which implies that also $lambda A+(1-lambda)B$ is copositive and hence the set of copositive matrices is convex.






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    You can follow these steps.



    Let $mathcalF$ denote the family of all the set of a $ntimes n$ symmetric copositive matrices.



    Now prove the following two assertions: The first one is,




    For any $lambda>0$, $AinmathcalF$ implies $lambda AinmathcalA$.




    Which follows from the fact that $x^T (lambda A) x = lambda (x^T Ax)$.



    The second is,




    For any $A,BinmathcalF$, $A+BinmathcalF$.




    Which you can do by calculation $x^T (A+B)x$.



    Once done. You can combine those two to show that




    For any $A,BinmathcalF$ and $lambdain[0,1]$ we have $lambda A + (1-lambda) B inmathcalF$




    which means, by definition, that $mathcalF$ is convex.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      No need to do that! Note that if $A,B$ are copositive, then for any $0<lambda<1$ and $xge 0$:$$x^TBigg(lambda A+(1-lambda)BBigg)x=lambda x^TAx+(1-lambda )x^TBxge 0$$which implies that also $lambda A+(1-lambda)B$ is copositive and hence the set of copositive matrices is convex.






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        No need to do that! Note that if $A,B$ are copositive, then for any $0<lambda<1$ and $xge 0$:$$x^TBigg(lambda A+(1-lambda)BBigg)x=lambda x^TAx+(1-lambda )x^TBxge 0$$which implies that also $lambda A+(1-lambda)B$ is copositive and hence the set of copositive matrices is convex.






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          No need to do that! Note that if $A,B$ are copositive, then for any $0<lambda<1$ and $xge 0$:$$x^TBigg(lambda A+(1-lambda)BBigg)x=lambda x^TAx+(1-lambda )x^TBxge 0$$which implies that also $lambda A+(1-lambda)B$ is copositive and hence the set of copositive matrices is convex.






          share|cite|improve this answer









          $endgroup$



          No need to do that! Note that if $A,B$ are copositive, then for any $0<lambda<1$ and $xge 0$:$$x^TBigg(lambda A+(1-lambda)BBigg)x=lambda x^TAx+(1-lambda )x^TBxge 0$$which implies that also $lambda A+(1-lambda)B$ is copositive and hence the set of copositive matrices is convex.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Mostafa AyazMostafa Ayaz

          18.1k31040




          18.1k31040





















              1












              $begingroup$

              You can follow these steps.



              Let $mathcalF$ denote the family of all the set of a $ntimes n$ symmetric copositive matrices.



              Now prove the following two assertions: The first one is,




              For any $lambda>0$, $AinmathcalF$ implies $lambda AinmathcalA$.




              Which follows from the fact that $x^T (lambda A) x = lambda (x^T Ax)$.



              The second is,




              For any $A,BinmathcalF$, $A+BinmathcalF$.




              Which you can do by calculation $x^T (A+B)x$.



              Once done. You can combine those two to show that




              For any $A,BinmathcalF$ and $lambdain[0,1]$ we have $lambda A + (1-lambda) B inmathcalF$




              which means, by definition, that $mathcalF$ is convex.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                You can follow these steps.



                Let $mathcalF$ denote the family of all the set of a $ntimes n$ symmetric copositive matrices.



                Now prove the following two assertions: The first one is,




                For any $lambda>0$, $AinmathcalF$ implies $lambda AinmathcalA$.




                Which follows from the fact that $x^T (lambda A) x = lambda (x^T Ax)$.



                The second is,




                For any $A,BinmathcalF$, $A+BinmathcalF$.




                Which you can do by calculation $x^T (A+B)x$.



                Once done. You can combine those two to show that




                For any $A,BinmathcalF$ and $lambdain[0,1]$ we have $lambda A + (1-lambda) B inmathcalF$




                which means, by definition, that $mathcalF$ is convex.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  You can follow these steps.



                  Let $mathcalF$ denote the family of all the set of a $ntimes n$ symmetric copositive matrices.



                  Now prove the following two assertions: The first one is,




                  For any $lambda>0$, $AinmathcalF$ implies $lambda AinmathcalA$.




                  Which follows from the fact that $x^T (lambda A) x = lambda (x^T Ax)$.



                  The second is,




                  For any $A,BinmathcalF$, $A+BinmathcalF$.




                  Which you can do by calculation $x^T (A+B)x$.



                  Once done. You can combine those two to show that




                  For any $A,BinmathcalF$ and $lambdain[0,1]$ we have $lambda A + (1-lambda) B inmathcalF$




                  which means, by definition, that $mathcalF$ is convex.






                  share|cite|improve this answer









                  $endgroup$



                  You can follow these steps.



                  Let $mathcalF$ denote the family of all the set of a $ntimes n$ symmetric copositive matrices.



                  Now prove the following two assertions: The first one is,




                  For any $lambda>0$, $AinmathcalF$ implies $lambda AinmathcalA$.




                  Which follows from the fact that $x^T (lambda A) x = lambda (x^T Ax)$.



                  The second is,




                  For any $A,BinmathcalF$, $A+BinmathcalF$.




                  Which you can do by calculation $x^T (A+B)x$.



                  Once done. You can combine those two to show that




                  For any $A,BinmathcalF$ and $lambdain[0,1]$ we have $lambda A + (1-lambda) B inmathcalF$




                  which means, by definition, that $mathcalF$ is convex.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  YankoYanko

                  8,1282830




                  8,1282830



























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