Question about an inequation described by matrices.












1












$begingroup$


$A=(a_{ij})_{1 le i, j le n}$ is a matrix that$sum_limits{i=1}^{n} a_{ij}=1$ for every j and $sum_limits{j=1}^n a_{ij} = 1$ for every i and $a_{ij} ge 0$.And
$$begin{equation}
begin{pmatrix}
y_1 \
vdots \
y_n \
end{pmatrix}
=mathbf{A}
begin{pmatrix}
x_1 \
vdots \
x_n
end{pmatrix}
end{equation}$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$



It may somehow matter to convex function.






analysis inequality convex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    can you write out the inequality fully? That is, replace each $y_i$ with its expression in terms of the $x_i$'s
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:08










  • $begingroup$
    @mathworker21 i have written it out fully but it doesn't help. Or you means to edit the question to make it more clear.
    $endgroup$
    – X.T Chen
    Dec 25 '18 at 9:12










  • $begingroup$
    just write it out fully in the question. i don't have paper on me
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:13












  • $begingroup$
    @mathworker21 $ prod_limits{i=1}^{n} (sum_limits{j=1}^{n} a_{ij}x_j) = prod_limits{j=1}^{n} x_j $
    $endgroup$
    – X.T Chen
    Dec 25 '18 at 9:39












  • $begingroup$
    ok, so the LHS has the terms $(prod_{j=1}^n a_{f(j)j})prod_{j=1}^n x_j$, for any $f: {1,dots,n} to {1,dots,n}$. Now just add all of these terms up. You should get $prod_{j=1}^n x_j$, right?
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:49
















1












$begingroup$


$A=(a_{ij})_{1 le i, j le n}$ is a matrix that$sum_limits{i=1}^{n} a_{ij}=1$ for every j and $sum_limits{j=1}^n a_{ij} = 1$ for every i and $a_{ij} ge 0$.And
$$begin{equation}
begin{pmatrix}
y_1 \
vdots \
y_n \
end{pmatrix}
=mathbf{A}
begin{pmatrix}
x_1 \
vdots \
x_n
end{pmatrix}
end{equation}$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$



It may somehow matter to convex function.






analysis inequality convex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    can you write out the inequality fully? That is, replace each $y_i$ with its expression in terms of the $x_i$'s
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:08










  • $begingroup$
    @mathworker21 i have written it out fully but it doesn't help. Or you means to edit the question to make it more clear.
    $endgroup$
    – X.T Chen
    Dec 25 '18 at 9:12










  • $begingroup$
    just write it out fully in the question. i don't have paper on me
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:13












  • $begingroup$
    @mathworker21 $ prod_limits{i=1}^{n} (sum_limits{j=1}^{n} a_{ij}x_j) = prod_limits{j=1}^{n} x_j $
    $endgroup$
    – X.T Chen
    Dec 25 '18 at 9:39












  • $begingroup$
    ok, so the LHS has the terms $(prod_{j=1}^n a_{f(j)j})prod_{j=1}^n x_j$, for any $f: {1,dots,n} to {1,dots,n}$. Now just add all of these terms up. You should get $prod_{j=1}^n x_j$, right?
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:49














1












1








1





$begingroup$


$A=(a_{ij})_{1 le i, j le n}$ is a matrix that$sum_limits{i=1}^{n} a_{ij}=1$ for every j and $sum_limits{j=1}^n a_{ij} = 1$ for every i and $a_{ij} ge 0$.And
$$begin{equation}
begin{pmatrix}
y_1 \
vdots \
y_n \
end{pmatrix}
=mathbf{A}
begin{pmatrix}
x_1 \
vdots \
x_n
end{pmatrix}
end{equation}$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$



It may somehow matter to convex function.






analysis inequality convex-analysis







share|cite|improve this question









$endgroup$




$A=(a_{ij})_{1 le i, j le n}$ is a matrix that$sum_limits{i=1}^{n} a_{ij}=1$ for every j and $sum_limits{j=1}^n a_{ij} = 1$ for every i and $a_{ij} ge 0$.And
$$begin{equation}
begin{pmatrix}
y_1 \
vdots \
y_n \
end{pmatrix}
=mathbf{A}
begin{pmatrix}
x_1 \
vdots \
x_n
end{pmatrix}
end{equation}$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$



It may somehow matter to convex function.






analysis inequality convex-analysis




analysis inequality convex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 25 '18 at 8:57









X.T ChenX.T Chen

988




988












  • $begingroup$
    can you write out the inequality fully? That is, replace each $y_i$ with its expression in terms of the $x_i$'s
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:08










  • $begingroup$
    @mathworker21 i have written it out fully but it doesn't help. Or you means to edit the question to make it more clear.
    $endgroup$
    – X.T Chen
    Dec 25 '18 at 9:12










  • $begingroup$
    just write it out fully in the question. i don't have paper on me
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:13












  • $begingroup$
    @mathworker21 $ prod_limits{i=1}^{n} (sum_limits{j=1}^{n} a_{ij}x_j) = prod_limits{j=1}^{n} x_j $
    $endgroup$
    – X.T Chen
    Dec 25 '18 at 9:39












  • $begingroup$
    ok, so the LHS has the terms $(prod_{j=1}^n a_{f(j)j})prod_{j=1}^n x_j$, for any $f: {1,dots,n} to {1,dots,n}$. Now just add all of these terms up. You should get $prod_{j=1}^n x_j$, right?
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:49


















  • $begingroup$
    can you write out the inequality fully? That is, replace each $y_i$ with its expression in terms of the $x_i$'s
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:08










  • $begingroup$
    @mathworker21 i have written it out fully but it doesn't help. Or you means to edit the question to make it more clear.
    $endgroup$
    – X.T Chen
    Dec 25 '18 at 9:12










  • $begingroup$
    just write it out fully in the question. i don't have paper on me
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:13












  • $begingroup$
    @mathworker21 $ prod_limits{i=1}^{n} (sum_limits{j=1}^{n} a_{ij}x_j) = prod_limits{j=1}^{n} x_j $
    $endgroup$
    – X.T Chen
    Dec 25 '18 at 9:39












  • $begingroup$
    ok, so the LHS has the terms $(prod_{j=1}^n a_{f(j)j})prod_{j=1}^n x_j$, for any $f: {1,dots,n} to {1,dots,n}$. Now just add all of these terms up. You should get $prod_{j=1}^n x_j$, right?
    $endgroup$
    – mathworker21
    Dec 25 '18 at 9:49
















$begingroup$
can you write out the inequality fully? That is, replace each $y_i$ with its expression in terms of the $x_i$'s
$endgroup$
– mathworker21
Dec 25 '18 at 9:08




$begingroup$
can you write out the inequality fully? That is, replace each $y_i$ with its expression in terms of the $x_i$'s
$endgroup$
– mathworker21
Dec 25 '18 at 9:08












$begingroup$
@mathworker21 i have written it out fully but it doesn't help. Or you means to edit the question to make it more clear.
$endgroup$
– X.T Chen
Dec 25 '18 at 9:12




$begingroup$
@mathworker21 i have written it out fully but it doesn't help. Or you means to edit the question to make it more clear.
$endgroup$
– X.T Chen
Dec 25 '18 at 9:12












$begingroup$
just write it out fully in the question. i don't have paper on me
$endgroup$
– mathworker21
Dec 25 '18 at 9:13






$begingroup$
just write it out fully in the question. i don't have paper on me
$endgroup$
– mathworker21
Dec 25 '18 at 9:13














$begingroup$
@mathworker21 $ prod_limits{i=1}^{n} (sum_limits{j=1}^{n} a_{ij}x_j) = prod_limits{j=1}^{n} x_j $
$endgroup$
– X.T Chen
Dec 25 '18 at 9:39






$begingroup$
@mathworker21 $ prod_limits{i=1}^{n} (sum_limits{j=1}^{n} a_{ij}x_j) = prod_limits{j=1}^{n} x_j $
$endgroup$
– X.T Chen
Dec 25 '18 at 9:39














$begingroup$
ok, so the LHS has the terms $(prod_{j=1}^n a_{f(j)j})prod_{j=1}^n x_j$, for any $f: {1,dots,n} to {1,dots,n}$. Now just add all of these terms up. You should get $prod_{j=1}^n x_j$, right?
$endgroup$
– mathworker21
Dec 25 '18 at 9:49




$begingroup$
ok, so the LHS has the terms $(prod_{j=1}^n a_{f(j)j})prod_{j=1}^n x_j$, for any $f: {1,dots,n} to {1,dots,n}$. Now just add all of these terms up. You should get $prod_{j=1}^n x_j$, right?
$endgroup$
– mathworker21
Dec 25 '18 at 9:49










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint:

If any $y_i$ is zero, from the condition on $a_{ij}$, at least one $x_k$ must be zero, so we may consider only positive $x_i,y_j$. Also the doubly stochastic matrix represents the majorization $x succ y$, so by Karamata’s Inequality, with the concave $log$ function, $sum log y_i geqslant sum log x_i$.






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    Clue: the right hand side of the given equation is a linear combination of the columns of $A$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I've already known that. Could you be more clear?
      $endgroup$
      – X.T Chen
      Dec 25 '18 at 9:43











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint:

    If any $y_i$ is zero, from the condition on $a_{ij}$, at least one $x_k$ must be zero, so we may consider only positive $x_i,y_j$. Also the doubly stochastic matrix represents the majorization $x succ y$, so by Karamata’s Inequality, with the concave $log$ function, $sum log y_i geqslant sum log x_i$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint:

      If any $y_i$ is zero, from the condition on $a_{ij}$, at least one $x_k$ must be zero, so we may consider only positive $x_i,y_j$. Also the doubly stochastic matrix represents the majorization $x succ y$, so by Karamata’s Inequality, with the concave $log$ function, $sum log y_i geqslant sum log x_i$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint:

        If any $y_i$ is zero, from the condition on $a_{ij}$, at least one $x_k$ must be zero, so we may consider only positive $x_i,y_j$. Also the doubly stochastic matrix represents the majorization $x succ y$, so by Karamata’s Inequality, with the concave $log$ function, $sum log y_i geqslant sum log x_i$.






        share|cite|improve this answer









        $endgroup$



        Hint:

        If any $y_i$ is zero, from the condition on $a_{ij}$, at least one $x_k$ must be zero, so we may consider only positive $x_i,y_j$. Also the doubly stochastic matrix represents the majorization $x succ y$, so by Karamata’s Inequality, with the concave $log$ function, $sum log y_i geqslant sum log x_i$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 15:04









        MacavityMacavity

        35.6k52554




        35.6k52554























            -1












            $begingroup$

            Clue: the right hand side of the given equation is a linear combination of the columns of $A$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I've already known that. Could you be more clear?
              $endgroup$
              – X.T Chen
              Dec 25 '18 at 9:43
















            -1












            $begingroup$

            Clue: the right hand side of the given equation is a linear combination of the columns of $A$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I've already known that. Could you be more clear?
              $endgroup$
              – X.T Chen
              Dec 25 '18 at 9:43














            -1












            -1








            -1





            $begingroup$

            Clue: the right hand side of the given equation is a linear combination of the columns of $A$.






            share|cite|improve this answer









            $endgroup$



            Clue: the right hand side of the given equation is a linear combination of the columns of $A$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 25 '18 at 9:31









            John MitchellJohn Mitchell

            377210




            377210












            • $begingroup$
              I've already known that. Could you be more clear?
              $endgroup$
              – X.T Chen
              Dec 25 '18 at 9:43


















            • $begingroup$
              I've already known that. Could you be more clear?
              $endgroup$
              – X.T Chen
              Dec 25 '18 at 9:43
















            $begingroup$
            I've already known that. Could you be more clear?
            $endgroup$
            – X.T Chen
            Dec 25 '18 at 9:43




            $begingroup$
            I've already known that. Could you be more clear?
            $endgroup$
            – X.T Chen
            Dec 25 '18 at 9:43


















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