Spectral norm minimization via semidefinite programming

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Given symmetric matrices $A_0, A_1, dots, A_n in mathbb R^m times m$, let $A(x) := A_0 + x_1 A_1 +cdots + x_n A_n$. How to formulate the following unconstrained spectral minimization problem as a semidefinite program?



$$min_x in mathbb R^n |A(x)|_2$$



Can anyone please help on this problem? Thanks!







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    Given symmetric matrices $A_0, A_1, dots, A_n in mathbb R^m times m$, let $A(x) := A_0 + x_1 A_1 +cdots + x_n A_n$. How to formulate the following unconstrained spectral minimization problem as a semidefinite program?



    $$min_x in mathbb R^n |A(x)|_2$$



    Can anyone please help on this problem? Thanks!







    share|cite|improve this question
























      up vote
      4
      down vote

      favorite
      1









      up vote
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      Given symmetric matrices $A_0, A_1, dots, A_n in mathbb R^m times m$, let $A(x) := A_0 + x_1 A_1 +cdots + x_n A_n$. How to formulate the following unconstrained spectral minimization problem as a semidefinite program?



      $$min_x in mathbb R^n |A(x)|_2$$



      Can anyone please help on this problem? Thanks!







      share|cite|improve this question














      Given symmetric matrices $A_0, A_1, dots, A_n in mathbb R^m times m$, let $A(x) := A_0 + x_1 A_1 +cdots + x_n A_n$. How to formulate the following unconstrained spectral minimization problem as a semidefinite program?



      $$min_x in mathbb R^n |A(x)|_2$$



      Can anyone please help on this problem? Thanks!









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 21 at 10:38









      Rodrigo de Azevedo

      12.6k41751




      12.6k41751










      asked Feb 9 '17 at 10:05









      Marcus118

      233




      233




















          1 Answer
          1






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          oldest

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          up vote
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          accepted










          Introducing variable $s > 0$ and rewriting the minimization problem in epigraph form,



          $$beginarrayll textminimize & s\ textsubject to & | mathrm A (mathrm x) |_2 leq sendarray$$



          Note that $| mathrm A (mathrm x) |_2 leq s$ is equivalent to $sigma_max left( mathrm A (mathrm x) right) leq s$, which is equivalent to



          $$lambda_max left( (mathrm A (mathrm x))^top mathrm A (mathrm x) right) leq s^2$$



          Hence,



          $$s^2 - lambda_max left( (mathrm A (mathrm x))^top mathrm A (mathrm x) right) = lambda_min left( s^2 mathrm I_m - (mathrm A (mathrm x))^top mathrm A (mathrm x) right) geq 0$$



          and, thus, we obtain



          $$s^2 mathrm I_m - (mathrm A (mathrm x))^top mathrm A (mathrm x) succeq mathrm O_m$$



          Dividing both sides by $s > 0$,



          $$s mathrm I_m - (mathrm A (mathrm x))^top left( s mathrm I_m right)^-1 mathrm A (mathrm x) succeq mathrm O_m$$



          Using the Schur complement test for positive semidefiniteness, the inequality above can be rewritten as the following linear matrix inequality (LMI)



          $$beginbmatrix s mathrm I_m & mathrm A (mathrm x)\ (mathrm A (mathrm x))^top & s mathrm I_mendbmatrix succeq mathrm O_2m$$



          Thus, we obtain the following semidefinite program (SDP) in $mathrm x in mathbb R^n$ and $s > 0$



          $$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m & mathrm A (mathrm x)\ (mathrm A (mathrm x))^top & s mathrm I_mendbmatrix succeq mathrm O_2mendarray$$



          Alternatively, since $mathrm A (mathrm x)$ is symmetric for all $mathrm x in mathbb R^n$, we can use the following SDP



          $$beginarrayll textminimize & s\ textsubject to & -s mathrm I_m preceq mathrm A (mathrm x) preceq s mathrm I_mendarray$$



          which can be rewritten as follows



          $$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m - mathrm A (mathrm x) & mathrm O_m\ mathrm O_m & s mathrm I_m + mathrm A (mathrm x)endbmatrix succeq mathrm O_2mendarray$$






          share|cite|improve this answer






















          • why is there suddenly a $lambda_min$
            – Sridhar Thiagarajan
            Aug 26 at 9:42






          • 1




            @SridharThiagarajan $$s^2 - max lambda_1, lambda_2, dots, lambda_m = min s^2 - lambda_1, s^2 - lambda_2, dots, s^2 - lambda_m $$ Would you agree?
            – Rodrigo de Azevedo
            Aug 27 at 14:33











          Your Answer




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          1 Answer
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          up vote
          6
          down vote



          accepted










          Introducing variable $s > 0$ and rewriting the minimization problem in epigraph form,



          $$beginarrayll textminimize & s\ textsubject to & | mathrm A (mathrm x) |_2 leq sendarray$$



          Note that $| mathrm A (mathrm x) |_2 leq s$ is equivalent to $sigma_max left( mathrm A (mathrm x) right) leq s$, which is equivalent to



          $$lambda_max left( (mathrm A (mathrm x))^top mathrm A (mathrm x) right) leq s^2$$



          Hence,



          $$s^2 - lambda_max left( (mathrm A (mathrm x))^top mathrm A (mathrm x) right) = lambda_min left( s^2 mathrm I_m - (mathrm A (mathrm x))^top mathrm A (mathrm x) right) geq 0$$



          and, thus, we obtain



          $$s^2 mathrm I_m - (mathrm A (mathrm x))^top mathrm A (mathrm x) succeq mathrm O_m$$



          Dividing both sides by $s > 0$,



          $$s mathrm I_m - (mathrm A (mathrm x))^top left( s mathrm I_m right)^-1 mathrm A (mathrm x) succeq mathrm O_m$$



          Using the Schur complement test for positive semidefiniteness, the inequality above can be rewritten as the following linear matrix inequality (LMI)



          $$beginbmatrix s mathrm I_m & mathrm A (mathrm x)\ (mathrm A (mathrm x))^top & s mathrm I_mendbmatrix succeq mathrm O_2m$$



          Thus, we obtain the following semidefinite program (SDP) in $mathrm x in mathbb R^n$ and $s > 0$



          $$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m & mathrm A (mathrm x)\ (mathrm A (mathrm x))^top & s mathrm I_mendbmatrix succeq mathrm O_2mendarray$$



          Alternatively, since $mathrm A (mathrm x)$ is symmetric for all $mathrm x in mathbb R^n$, we can use the following SDP



          $$beginarrayll textminimize & s\ textsubject to & -s mathrm I_m preceq mathrm A (mathrm x) preceq s mathrm I_mendarray$$



          which can be rewritten as follows



          $$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m - mathrm A (mathrm x) & mathrm O_m\ mathrm O_m & s mathrm I_m + mathrm A (mathrm x)endbmatrix succeq mathrm O_2mendarray$$






          share|cite|improve this answer






















          • why is there suddenly a $lambda_min$
            – Sridhar Thiagarajan
            Aug 26 at 9:42






          • 1




            @SridharThiagarajan $$s^2 - max lambda_1, lambda_2, dots, lambda_m = min s^2 - lambda_1, s^2 - lambda_2, dots, s^2 - lambda_m $$ Would you agree?
            – Rodrigo de Azevedo
            Aug 27 at 14:33















          up vote
          6
          down vote



          accepted










          Introducing variable $s > 0$ and rewriting the minimization problem in epigraph form,



          $$beginarrayll textminimize & s\ textsubject to & | mathrm A (mathrm x) |_2 leq sendarray$$



          Note that $| mathrm A (mathrm x) |_2 leq s$ is equivalent to $sigma_max left( mathrm A (mathrm x) right) leq s$, which is equivalent to



          $$lambda_max left( (mathrm A (mathrm x))^top mathrm A (mathrm x) right) leq s^2$$



          Hence,



          $$s^2 - lambda_max left( (mathrm A (mathrm x))^top mathrm A (mathrm x) right) = lambda_min left( s^2 mathrm I_m - (mathrm A (mathrm x))^top mathrm A (mathrm x) right) geq 0$$



          and, thus, we obtain



          $$s^2 mathrm I_m - (mathrm A (mathrm x))^top mathrm A (mathrm x) succeq mathrm O_m$$



          Dividing both sides by $s > 0$,



          $$s mathrm I_m - (mathrm A (mathrm x))^top left( s mathrm I_m right)^-1 mathrm A (mathrm x) succeq mathrm O_m$$



          Using the Schur complement test for positive semidefiniteness, the inequality above can be rewritten as the following linear matrix inequality (LMI)



          $$beginbmatrix s mathrm I_m & mathrm A (mathrm x)\ (mathrm A (mathrm x))^top & s mathrm I_mendbmatrix succeq mathrm O_2m$$



          Thus, we obtain the following semidefinite program (SDP) in $mathrm x in mathbb R^n$ and $s > 0$



          $$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m & mathrm A (mathrm x)\ (mathrm A (mathrm x))^top & s mathrm I_mendbmatrix succeq mathrm O_2mendarray$$



          Alternatively, since $mathrm A (mathrm x)$ is symmetric for all $mathrm x in mathbb R^n$, we can use the following SDP



          $$beginarrayll textminimize & s\ textsubject to & -s mathrm I_m preceq mathrm A (mathrm x) preceq s mathrm I_mendarray$$



          which can be rewritten as follows



          $$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m - mathrm A (mathrm x) & mathrm O_m\ mathrm O_m & s mathrm I_m + mathrm A (mathrm x)endbmatrix succeq mathrm O_2mendarray$$






          share|cite|improve this answer






















          • why is there suddenly a $lambda_min$
            – Sridhar Thiagarajan
            Aug 26 at 9:42






          • 1




            @SridharThiagarajan $$s^2 - max lambda_1, lambda_2, dots, lambda_m = min s^2 - lambda_1, s^2 - lambda_2, dots, s^2 - lambda_m $$ Would you agree?
            – Rodrigo de Azevedo
            Aug 27 at 14:33













          up vote
          6
          down vote



          accepted







          up vote
          6
          down vote



          accepted






          Introducing variable $s > 0$ and rewriting the minimization problem in epigraph form,



          $$beginarrayll textminimize & s\ textsubject to & | mathrm A (mathrm x) |_2 leq sendarray$$



          Note that $| mathrm A (mathrm x) |_2 leq s$ is equivalent to $sigma_max left( mathrm A (mathrm x) right) leq s$, which is equivalent to



          $$lambda_max left( (mathrm A (mathrm x))^top mathrm A (mathrm x) right) leq s^2$$



          Hence,



          $$s^2 - lambda_max left( (mathrm A (mathrm x))^top mathrm A (mathrm x) right) = lambda_min left( s^2 mathrm I_m - (mathrm A (mathrm x))^top mathrm A (mathrm x) right) geq 0$$



          and, thus, we obtain



          $$s^2 mathrm I_m - (mathrm A (mathrm x))^top mathrm A (mathrm x) succeq mathrm O_m$$



          Dividing both sides by $s > 0$,



          $$s mathrm I_m - (mathrm A (mathrm x))^top left( s mathrm I_m right)^-1 mathrm A (mathrm x) succeq mathrm O_m$$



          Using the Schur complement test for positive semidefiniteness, the inequality above can be rewritten as the following linear matrix inequality (LMI)



          $$beginbmatrix s mathrm I_m & mathrm A (mathrm x)\ (mathrm A (mathrm x))^top & s mathrm I_mendbmatrix succeq mathrm O_2m$$



          Thus, we obtain the following semidefinite program (SDP) in $mathrm x in mathbb R^n$ and $s > 0$



          $$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m & mathrm A (mathrm x)\ (mathrm A (mathrm x))^top & s mathrm I_mendbmatrix succeq mathrm O_2mendarray$$



          Alternatively, since $mathrm A (mathrm x)$ is symmetric for all $mathrm x in mathbb R^n$, we can use the following SDP



          $$beginarrayll textminimize & s\ textsubject to & -s mathrm I_m preceq mathrm A (mathrm x) preceq s mathrm I_mendarray$$



          which can be rewritten as follows



          $$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m - mathrm A (mathrm x) & mathrm O_m\ mathrm O_m & s mathrm I_m + mathrm A (mathrm x)endbmatrix succeq mathrm O_2mendarray$$






          share|cite|improve this answer














          Introducing variable $s > 0$ and rewriting the minimization problem in epigraph form,



          $$beginarrayll textminimize & s\ textsubject to & | mathrm A (mathrm x) |_2 leq sendarray$$



          Note that $| mathrm A (mathrm x) |_2 leq s$ is equivalent to $sigma_max left( mathrm A (mathrm x) right) leq s$, which is equivalent to



          $$lambda_max left( (mathrm A (mathrm x))^top mathrm A (mathrm x) right) leq s^2$$



          Hence,



          $$s^2 - lambda_max left( (mathrm A (mathrm x))^top mathrm A (mathrm x) right) = lambda_min left( s^2 mathrm I_m - (mathrm A (mathrm x))^top mathrm A (mathrm x) right) geq 0$$



          and, thus, we obtain



          $$s^2 mathrm I_m - (mathrm A (mathrm x))^top mathrm A (mathrm x) succeq mathrm O_m$$



          Dividing both sides by $s > 0$,



          $$s mathrm I_m - (mathrm A (mathrm x))^top left( s mathrm I_m right)^-1 mathrm A (mathrm x) succeq mathrm O_m$$



          Using the Schur complement test for positive semidefiniteness, the inequality above can be rewritten as the following linear matrix inequality (LMI)



          $$beginbmatrix s mathrm I_m & mathrm A (mathrm x)\ (mathrm A (mathrm x))^top & s mathrm I_mendbmatrix succeq mathrm O_2m$$



          Thus, we obtain the following semidefinite program (SDP) in $mathrm x in mathbb R^n$ and $s > 0$



          $$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m & mathrm A (mathrm x)\ (mathrm A (mathrm x))^top & s mathrm I_mendbmatrix succeq mathrm O_2mendarray$$



          Alternatively, since $mathrm A (mathrm x)$ is symmetric for all $mathrm x in mathbb R^n$, we can use the following SDP



          $$beginarrayll textminimize & s\ textsubject to & -s mathrm I_m preceq mathrm A (mathrm x) preceq s mathrm I_mendarray$$



          which can be rewritten as follows



          $$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m - mathrm A (mathrm x) & mathrm O_m\ mathrm O_m & s mathrm I_m + mathrm A (mathrm x)endbmatrix succeq mathrm O_2mendarray$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 4 at 17:19

























          answered Feb 9 '17 at 23:44









          Rodrigo de Azevedo

          12.6k41751




          12.6k41751











          • why is there suddenly a $lambda_min$
            – Sridhar Thiagarajan
            Aug 26 at 9:42






          • 1




            @SridharThiagarajan $$s^2 - max lambda_1, lambda_2, dots, lambda_m = min s^2 - lambda_1, s^2 - lambda_2, dots, s^2 - lambda_m $$ Would you agree?
            – Rodrigo de Azevedo
            Aug 27 at 14:33

















          • why is there suddenly a $lambda_min$
            – Sridhar Thiagarajan
            Aug 26 at 9:42






          • 1




            @SridharThiagarajan $$s^2 - max lambda_1, lambda_2, dots, lambda_m = min s^2 - lambda_1, s^2 - lambda_2, dots, s^2 - lambda_m $$ Would you agree?
            – Rodrigo de Azevedo
            Aug 27 at 14:33
















          why is there suddenly a $lambda_min$
          – Sridhar Thiagarajan
          Aug 26 at 9:42




          why is there suddenly a $lambda_min$
          – Sridhar Thiagarajan
          Aug 26 at 9:42




          1




          1




          @SridharThiagarajan $$s^2 - max lambda_1, lambda_2, dots, lambda_m = min s^2 - lambda_1, s^2 - lambda_2, dots, s^2 - lambda_m $$ Would you agree?
          – Rodrigo de Azevedo
          Aug 27 at 14:33





          @SridharThiagarajan $$s^2 - max lambda_1, lambda_2, dots, lambda_m = min s^2 - lambda_1, s^2 - lambda_2, dots, s^2 - lambda_m $$ Would you agree?
          – Rodrigo de Azevedo
          Aug 27 at 14:33













           

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