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!
optimization convex-optimization semidefinite-programming lmis spectral-norm
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up vote
4
down vote
favorite
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!
optimization convex-optimization semidefinite-programming lmis spectral-norm
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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!
optimization convex-optimization semidefinite-programming lmis spectral-norm
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!
optimization convex-optimization semidefinite-programming lmis spectral-norm
edited Aug 21 at 10:38
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Feb 9 '17 at 10:05
Marcus118
233
233
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1 Answer
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oldest
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up vote
6
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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
$$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m - mathrm A (mathrm x) & mathrm O_m\ mathrm O_m & s mathrm I_m + mathrm A (mathrm x)endbmatrix succeq mathrm O_2mendarray$$
why is there suddenly a $lambda_min$
â 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?
â Rodrigo de Azevedo
Aug 27 at 14:33
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m - mathrm A (mathrm x) & mathrm O_m\ mathrm O_m & s mathrm I_m + mathrm A (mathrm x)endbmatrix succeq mathrm O_2mendarray$$
why is there suddenly a $lambda_min$
â 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?
â Rodrigo de Azevedo
Aug 27 at 14:33
add a comment |Â
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
$$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m - mathrm A (mathrm x) & mathrm O_m\ mathrm O_m & s mathrm I_m + mathrm A (mathrm x)endbmatrix succeq mathrm O_2mendarray$$
why is there suddenly a $lambda_min$
â 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?
â Rodrigo de Azevedo
Aug 27 at 14:33
add a comment |Â
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
$$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m - mathrm A (mathrm x) & mathrm O_m\ mathrm O_m & s mathrm I_m + mathrm A (mathrm x)endbmatrix succeq mathrm O_2mendarray$$
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
$$beginarrayll textminimize & s\ textsubject to & beginbmatrix s mathrm I_m - mathrm A (mathrm x) & mathrm O_m\ mathrm O_m & s mathrm I_m + mathrm A (mathrm x)endbmatrix succeq mathrm O_2mendarray$$
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 $lambda_min$
â 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?
â Rodrigo de Azevedo
Aug 27 at 14:33
add a comment |Â
why is there suddenly a $lambda_min$
â 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?
â Rodrigo de Azevedo
Aug 27 at 14:33
why is there suddenly a $lambda_min$
â Sridhar Thiagarajan
Aug 26 at 9:42
why is there suddenly a $lambda_min$
â 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?
â 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?
â Rodrigo de Azevedo
Aug 27 at 14:33
add a comment |Â
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