Prove that $sum_{i,j=1}^Nx_ialpha_{ij}x_j>0$ for a non-symmetric matrix












1












$begingroup$


Let $A=(alpha)_{i,j=1}^N$ a real square matrix. The matrix $A$ is not necessary symmetric. I want to prove that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j>0,;;;forall (x_1,ldots,x_N)inmathbb{R}^Nbackslashleftlbrace 0rightrbrace.$$



If matrix $A$ was symmetric, one way to do this, is to prove that all eigenvalues are positive. How can I deal with this case?






linear-algebra matrices







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
    $endgroup$
    – anomaly
    Dec 28 '18 at 21:22
















1












$begingroup$


Let $A=(alpha)_{i,j=1}^N$ a real square matrix. The matrix $A$ is not necessary symmetric. I want to prove that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j>0,;;;forall (x_1,ldots,x_N)inmathbb{R}^Nbackslashleftlbrace 0rightrbrace.$$



If matrix $A$ was symmetric, one way to do this, is to prove that all eigenvalues are positive. How can I deal with this case?






linear-algebra matrices







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
    $endgroup$
    – anomaly
    Dec 28 '18 at 21:22














1












1








1





$begingroup$


Let $A=(alpha)_{i,j=1}^N$ a real square matrix. The matrix $A$ is not necessary symmetric. I want to prove that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j>0,;;;forall (x_1,ldots,x_N)inmathbb{R}^Nbackslashleftlbrace 0rightrbrace.$$



If matrix $A$ was symmetric, one way to do this, is to prove that all eigenvalues are positive. How can I deal with this case?






linear-algebra matrices







share|cite|improve this question









$endgroup$




Let $A=(alpha)_{i,j=1}^N$ a real square matrix. The matrix $A$ is not necessary symmetric. I want to prove that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j>0,;;;forall (x_1,ldots,x_N)inmathbb{R}^Nbackslashleftlbrace 0rightrbrace.$$



If matrix $A$ was symmetric, one way to do this, is to prove that all eigenvalues are positive. How can I deal with this case?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 28 '18 at 19:31









math_lovermath_lover

9710




9710








  • 1




    $begingroup$
    Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
    $endgroup$
    – anomaly
    Dec 28 '18 at 21:22














  • 1




    $begingroup$
    Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
    $endgroup$
    – anomaly
    Dec 28 '18 at 21:22








1




1




$begingroup$
Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
$endgroup$
– anomaly
Dec 28 '18 at 21:22




$begingroup$
Replace $alpha$ with $frac{1}{2}(alpha + alpha^t)$.
$endgroup$
– anomaly
Dec 28 '18 at 21:22










2 Answers
2






active

oldest

votes


















1












$begingroup$

Any real square matrix $A$ can be written (uniquely) as the sum of a symmetric matrix and a skew-symmetric matrix



$$A = A_{text{sym}} + A_{text{sq}}$$



where
$$A_{text{sym}} = frac{1}{2} ( A + A^t) \
A_{text{sq}} = frac{1}{2} ( A - A^t) $$



Say $A_{text{sym}}= (tilde alpha_{ij})$. It is easy to see that
$$sum_{i,j=1}^Nx_ialpha_{ij}x_j=sum_{i,j=1}^Nx_itildealpha_{ij}x_j $$ ( that is, $A$ and $A_{text{sym}}$ produce the same quadratic form). Now you are reduced to checking whether the real symmetric matrix $A_{text{sym}}$ is positive definite, which is standard.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    This is false. Consider $A=-I_N$ and $x_1=x_2=cdots=x_N=1$. Then
    $$sum_{i,j=1}^Nx_ia_{i,j}x_j=-N.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
      $endgroup$
      – anomaly
      Dec 28 '18 at 21:23












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Any real square matrix $A$ can be written (uniquely) as the sum of a symmetric matrix and a skew-symmetric matrix



    $$A = A_{text{sym}} + A_{text{sq}}$$



    where
    $$A_{text{sym}} = frac{1}{2} ( A + A^t) \
    A_{text{sq}} = frac{1}{2} ( A - A^t) $$



    Say $A_{text{sym}}= (tilde alpha_{ij})$. It is easy to see that
    $$sum_{i,j=1}^Nx_ialpha_{ij}x_j=sum_{i,j=1}^Nx_itildealpha_{ij}x_j $$ ( that is, $A$ and $A_{text{sym}}$ produce the same quadratic form). Now you are reduced to checking whether the real symmetric matrix $A_{text{sym}}$ is positive definite, which is standard.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Any real square matrix $A$ can be written (uniquely) as the sum of a symmetric matrix and a skew-symmetric matrix



      $$A = A_{text{sym}} + A_{text{sq}}$$



      where
      $$A_{text{sym}} = frac{1}{2} ( A + A^t) \
      A_{text{sq}} = frac{1}{2} ( A - A^t) $$



      Say $A_{text{sym}}= (tilde alpha_{ij})$. It is easy to see that
      $$sum_{i,j=1}^Nx_ialpha_{ij}x_j=sum_{i,j=1}^Nx_itildealpha_{ij}x_j $$ ( that is, $A$ and $A_{text{sym}}$ produce the same quadratic form). Now you are reduced to checking whether the real symmetric matrix $A_{text{sym}}$ is positive definite, which is standard.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Any real square matrix $A$ can be written (uniquely) as the sum of a symmetric matrix and a skew-symmetric matrix



        $$A = A_{text{sym}} + A_{text{sq}}$$



        where
        $$A_{text{sym}} = frac{1}{2} ( A + A^t) \
        A_{text{sq}} = frac{1}{2} ( A - A^t) $$



        Say $A_{text{sym}}= (tilde alpha_{ij})$. It is easy to see that
        $$sum_{i,j=1}^Nx_ialpha_{ij}x_j=sum_{i,j=1}^Nx_itildealpha_{ij}x_j $$ ( that is, $A$ and $A_{text{sym}}$ produce the same quadratic form). Now you are reduced to checking whether the real symmetric matrix $A_{text{sym}}$ is positive definite, which is standard.






        share|cite|improve this answer









        $endgroup$



        Any real square matrix $A$ can be written (uniquely) as the sum of a symmetric matrix and a skew-symmetric matrix



        $$A = A_{text{sym}} + A_{text{sq}}$$



        where
        $$A_{text{sym}} = frac{1}{2} ( A + A^t) \
        A_{text{sq}} = frac{1}{2} ( A - A^t) $$



        Say $A_{text{sym}}= (tilde alpha_{ij})$. It is easy to see that
        $$sum_{i,j=1}^Nx_ialpha_{ij}x_j=sum_{i,j=1}^Nx_itildealpha_{ij}x_j $$ ( that is, $A$ and $A_{text{sym}}$ produce the same quadratic form). Now you are reduced to checking whether the real symmetric matrix $A_{text{sym}}$ is positive definite, which is standard.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 21:14









        Orest BucicovschiOrest Bucicovschi

        28.6k31748




        28.6k31748























            0












            $begingroup$

            This is false. Consider $A=-I_N$ and $x_1=x_2=cdots=x_N=1$. Then
            $$sum_{i,j=1}^Nx_ia_{i,j}x_j=-N.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
              $endgroup$
              – anomaly
              Dec 28 '18 at 21:23
















            0












            $begingroup$

            This is false. Consider $A=-I_N$ and $x_1=x_2=cdots=x_N=1$. Then
            $$sum_{i,j=1}^Nx_ia_{i,j}x_j=-N.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
              $endgroup$
              – anomaly
              Dec 28 '18 at 21:23














            0












            0








            0





            $begingroup$

            This is false. Consider $A=-I_N$ and $x_1=x_2=cdots=x_N=1$. Then
            $$sum_{i,j=1}^Nx_ia_{i,j}x_j=-N.$$






            share|cite|improve this answer









            $endgroup$



            This is false. Consider $A=-I_N$ and $x_1=x_2=cdots=x_N=1$. Then
            $$sum_{i,j=1}^Nx_ia_{i,j}x_j=-N.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 28 '18 at 19:35









            Ben WBen W

            2,734918




            2,734918












            • $begingroup$
              It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
              $endgroup$
              – anomaly
              Dec 28 '18 at 21:23


















            • $begingroup$
              It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
              $endgroup$
              – anomaly
              Dec 28 '18 at 21:23
















            $begingroup$
            It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
            $endgroup$
            – anomaly
            Dec 28 '18 at 21:23




            $begingroup$
            It isn't clear, but I think the OP has a particular $alpha$ in mind or is asking about techniques for proving this sort of result in general. (Of course, symmetric does not imply positive-semidefinite.)
            $endgroup$
            – anomaly
            Dec 28 '18 at 21:23


















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