Maximize $det X$, subject to $X_{ii}leq P_i$, where $X>0$












2












$begingroup$


Given $P_1,P_2,cdots,P_N$.



begin{array}{ll} text{maximize} & det X\ text{subject to} & mathrm X_{ii}leq P_i \forall i=1,2,cdots,nend{array}



$Xinmathbb{R}^{ntimes n}$, $X>0$ (i.e. positive definite).



For $n=2$, maximum is achieved when $X_{ii}=P_i$ and $X$ is diagonal matrix.



Then for $n=3$, Let $X=begin{bmatrix}X_1 & X_2\X_2^T &x_3end{bmatrix}$, where $X_1inmathbb{R}^{2times2}$.



$det(X)=det(X_1)times det(x_3-X_2^TX_1^{-1}X_2)leq det(X_1)times x_3leq P_1P_2P_3$.



Then following same logic my conjecture is $max{ det X}=P_1cdots P_n$, when $X_{ii}=P_i$.



Is it correct? If not, can you please point my mistake or give me counter example? Thanks






optimization determinant positive-definite







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$endgroup$

















    2












    $begingroup$


    Given $P_1,P_2,cdots,P_N$.



    begin{array}{ll} text{maximize} & det X\ text{subject to} & mathrm X_{ii}leq P_i \forall i=1,2,cdots,nend{array}



    $Xinmathbb{R}^{ntimes n}$, $X>0$ (i.e. positive definite).



    For $n=2$, maximum is achieved when $X_{ii}=P_i$ and $X$ is diagonal matrix.



    Then for $n=3$, Let $X=begin{bmatrix}X_1 & X_2\X_2^T &x_3end{bmatrix}$, where $X_1inmathbb{R}^{2times2}$.



    $det(X)=det(X_1)times det(x_3-X_2^TX_1^{-1}X_2)leq det(X_1)times x_3leq P_1P_2P_3$.



    Then following same logic my conjecture is $max{ det X}=P_1cdots P_n$, when $X_{ii}=P_i$.



    Is it correct? If not, can you please point my mistake or give me counter example? Thanks






    optimization determinant positive-definite







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Given $P_1,P_2,cdots,P_N$.



      begin{array}{ll} text{maximize} & det X\ text{subject to} & mathrm X_{ii}leq P_i \forall i=1,2,cdots,nend{array}



      $Xinmathbb{R}^{ntimes n}$, $X>0$ (i.e. positive definite).



      For $n=2$, maximum is achieved when $X_{ii}=P_i$ and $X$ is diagonal matrix.



      Then for $n=3$, Let $X=begin{bmatrix}X_1 & X_2\X_2^T &x_3end{bmatrix}$, where $X_1inmathbb{R}^{2times2}$.



      $det(X)=det(X_1)times det(x_3-X_2^TX_1^{-1}X_2)leq det(X_1)times x_3leq P_1P_2P_3$.



      Then following same logic my conjecture is $max{ det X}=P_1cdots P_n$, when $X_{ii}=P_i$.



      Is it correct? If not, can you please point my mistake or give me counter example? Thanks






      optimization determinant positive-definite







      share|cite|improve this question











      $endgroup$




      Given $P_1,P_2,cdots,P_N$.



      begin{array}{ll} text{maximize} & det X\ text{subject to} & mathrm X_{ii}leq P_i \forall i=1,2,cdots,nend{array}



      $Xinmathbb{R}^{ntimes n}$, $X>0$ (i.e. positive definite).



      For $n=2$, maximum is achieved when $X_{ii}=P_i$ and $X$ is diagonal matrix.



      Then for $n=3$, Let $X=begin{bmatrix}X_1 & X_2\X_2^T &x_3end{bmatrix}$, where $X_1inmathbb{R}^{2times2}$.



      $det(X)=det(X_1)times det(x_3-X_2^TX_1^{-1}X_2)leq det(X_1)times x_3leq P_1P_2P_3$.



      Then following same logic my conjecture is $max{ det X}=P_1cdots P_n$, when $X_{ii}=P_i$.



      Is it correct? If not, can you please point my mistake or give me counter example? Thanks






      optimization determinant positive-definite




      optimization determinant positive-definite






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      edited Dec 15 '18 at 3:48









      Tianlalu

      3,08621038




      3,08621038










      asked Dec 15 '18 at 3:17









      LeeLee

      330111




      330111






















          1 Answer
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          $begingroup$

          Your problem is equivalent to maximizing $log det X$ such that $X_{ii} leq P_i$ and $X$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is
          $$L(X,lambda) = logdet X - sum_i lambda_i (X_{ii} - P_i)$$
          The derivative of $logdet X$ is $(X^{-1})^T$, so the KKT conditions are:
          $$(X^{-1})_{ii} - lambda_i = 0$$
          $$(X^{-1})_{ij} = 0 quad (i neq j)$$
          $$lambda_i (X_{ii} - P_i)=0$$
          $$X_{ii} leq P_i$$
          $$lambda geq 0.$$
          The point you found satisfies these conditions and is therefore optimal.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            what if rank$ X< n$? I think in this case these points are not optimal
            $endgroup$
            – Lee
            Dec 18 '18 at 9:24








          • 1




            $begingroup$
            @Lee a positive definite matrix has full rank.
            $endgroup$
            – LinAlg
            Dec 18 '18 at 15:13











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          2












          $begingroup$

          Your problem is equivalent to maximizing $log det X$ such that $X_{ii} leq P_i$ and $X$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is
          $$L(X,lambda) = logdet X - sum_i lambda_i (X_{ii} - P_i)$$
          The derivative of $logdet X$ is $(X^{-1})^T$, so the KKT conditions are:
          $$(X^{-1})_{ii} - lambda_i = 0$$
          $$(X^{-1})_{ij} = 0 quad (i neq j)$$
          $$lambda_i (X_{ii} - P_i)=0$$
          $$X_{ii} leq P_i$$
          $$lambda geq 0.$$
          The point you found satisfies these conditions and is therefore optimal.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            what if rank$ X< n$? I think in this case these points are not optimal
            $endgroup$
            – Lee
            Dec 18 '18 at 9:24








          • 1




            $begingroup$
            @Lee a positive definite matrix has full rank.
            $endgroup$
            – LinAlg
            Dec 18 '18 at 15:13
















          2












          $begingroup$

          Your problem is equivalent to maximizing $log det X$ such that $X_{ii} leq P_i$ and $X$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is
          $$L(X,lambda) = logdet X - sum_i lambda_i (X_{ii} - P_i)$$
          The derivative of $logdet X$ is $(X^{-1})^T$, so the KKT conditions are:
          $$(X^{-1})_{ii} - lambda_i = 0$$
          $$(X^{-1})_{ij} = 0 quad (i neq j)$$
          $$lambda_i (X_{ii} - P_i)=0$$
          $$X_{ii} leq P_i$$
          $$lambda geq 0.$$
          The point you found satisfies these conditions and is therefore optimal.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            what if rank$ X< n$? I think in this case these points are not optimal
            $endgroup$
            – Lee
            Dec 18 '18 at 9:24








          • 1




            $begingroup$
            @Lee a positive definite matrix has full rank.
            $endgroup$
            – LinAlg
            Dec 18 '18 at 15:13














          2












          2








          2





          $begingroup$

          Your problem is equivalent to maximizing $log det X$ such that $X_{ii} leq P_i$ and $X$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is
          $$L(X,lambda) = logdet X - sum_i lambda_i (X_{ii} - P_i)$$
          The derivative of $logdet X$ is $(X^{-1})^T$, so the KKT conditions are:
          $$(X^{-1})_{ii} - lambda_i = 0$$
          $$(X^{-1})_{ij} = 0 quad (i neq j)$$
          $$lambda_i (X_{ii} - P_i)=0$$
          $$X_{ii} leq P_i$$
          $$lambda geq 0.$$
          The point you found satisfies these conditions and is therefore optimal.






          share|cite|improve this answer









          $endgroup$



          Your problem is equivalent to maximizing $log det X$ such that $X_{ii} leq P_i$ and $X$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is
          $$L(X,lambda) = logdet X - sum_i lambda_i (X_{ii} - P_i)$$
          The derivative of $logdet X$ is $(X^{-1})^T$, so the KKT conditions are:
          $$(X^{-1})_{ii} - lambda_i = 0$$
          $$(X^{-1})_{ij} = 0 quad (i neq j)$$
          $$lambda_i (X_{ii} - P_i)=0$$
          $$X_{ii} leq P_i$$
          $$lambda geq 0.$$
          The point you found satisfies these conditions and is therefore optimal.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 16:34









          LinAlgLinAlg

          9,5491521




          9,5491521












          • $begingroup$
            what if rank$ X< n$? I think in this case these points are not optimal
            $endgroup$
            – Lee
            Dec 18 '18 at 9:24








          • 1




            $begingroup$
            @Lee a positive definite matrix has full rank.
            $endgroup$
            – LinAlg
            Dec 18 '18 at 15:13


















          • $begingroup$
            what if rank$ X< n$? I think in this case these points are not optimal
            $endgroup$
            – Lee
            Dec 18 '18 at 9:24








          • 1




            $begingroup$
            @Lee a positive definite matrix has full rank.
            $endgroup$
            – LinAlg
            Dec 18 '18 at 15:13
















          $begingroup$
          what if rank$ X< n$? I think in this case these points are not optimal
          $endgroup$
          – Lee
          Dec 18 '18 at 9:24






          $begingroup$
          what if rank$ X< n$? I think in this case these points are not optimal
          $endgroup$
          – Lee
          Dec 18 '18 at 9:24






          1




          1




          $begingroup$
          @Lee a positive definite matrix has full rank.
          $endgroup$
          – LinAlg
          Dec 18 '18 at 15:13




          $begingroup$
          @Lee a positive definite matrix has full rank.
          $endgroup$
          – LinAlg
          Dec 18 '18 at 15:13


















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