Formulation of spectral norm minimization as a semidefinite program
$begingroup$
Given a matrix $F in mathbb{C}^{m times n}$ such that $m > n$ and other (non-symmetric) square matrix $A$ of size $n times n$, how can one formulate
$$ arg min_b left|A- {F}^{*} operatorname{diag} left( b right) , {F} right|_{2}$$
where $b in mathbb{C}^m$ is some vector and $*$ denotes the conjugate transpose, as a semidefinite program?
I started as follows. Writing the above problem in epigraph form by introducing a variable $x$,
begin{array}{ll} text{minimize} & x\ text{subject to} & left|A- {F}^{*} operatorname{diag} left( b right) , {F} right|_{2} leq xend{array}
which is equivalent to
begin{array}{ll} text{minimize} & x\ text{subject to} & sigma_{max}(A- {F}^{*} operatorname{diag} left( b right) , {F} ) leq xend{array}
which is equivalent to
begin{array}{ll} text{minimize} & x\ text{subject to} & lambda_{max}big((A- {F}^{*} operatorname{diag} left( b right) , {F} )^*(A- {F}^{*} operatorname{diag} left( b right) , {F} ) big) leq x^2end{array}
Can anybody tell me how I can proceed with this?
convex-optimization semidefinite-programming spectral-norm
edited Jan 19 at 8:42
Rodrigo de Azevedo
13.1k41960
asked Jan 16 at 15:00
abina shrabina shr
698
$endgroup$
|
show 2 more comments
$begingroup$
Given a matrix $F in mathbb{C}^{m times n}$ such that $m > n$ and other (non-symmetric) square matrix $A$ of size $n times n$, how can one formulate
$$ arg min_b left|A- {F}^{*} operatorname{diag} left( b right) , {F} right|_{2}$$
where $b in mathbb{C}^m$ is some vector and $*$ denotes the conjugate transpose, as a semidefinite program?
I started as follows. Writing the above problem in epigraph form by introducing a variable $x$,
begin{array}{ll} text{minimize} & x\ text{subject to} & left|A- {F}^{*} operatorname{diag} left( b right) , {F} right|_{2} leq xend{array}
which is equivalent to
begin{array}{ll} text{minimize} & x\ text{subject to} & sigma_{max}(A- {F}^{*} operatorname{diag} left( b right) , {F} ) leq xend{array}
which is equivalent to
begin{array}{ll} text{minimize} & x\ text{subject to} & lambda_{max}big((A- {F}^{*} operatorname{diag} left( b right) , {F} )^*(A- {F}^{*} operatorname{diag} left( b right) , {F} ) big) leq x^2end{array}
Can anybody tell me how I can proceed with this?
convex-optimization semidefinite-programming spectral-norm
edited Jan 19 at 8:42
Rodrigo de Azevedo
13.1k41960
asked Jan 16 at 15:00
abina shrabina shr
698
$endgroup$
$begingroup$
A key concept is that $F^{T}mbox{diag}(b)F=sum_{i=1}^{n} b_{i} (F_{i}^{T}F_{i})$.
$endgroup$
– Brian Borchers
Jan 16 at 16:24
$begingroup$
@BrianBorchers could you please elaborate a bit?
$endgroup$
– abina shr
Jan 17 at 12:44
$begingroup$
Take a look at this.
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:47
$begingroup$
@RodrigodeAzevedo So I have reached this step: $ begin{array}{ll} text{minimize} & x\ text{subject to} & begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0_{2n}end{array}$. How do I proceed after this?
$endgroup$
– abina shr
Jan 21 at 10:54
$begingroup$
@abinashr Don't you have an SDP in $x$ and $b$ already? Next step would be to solve it.
$endgroup$
– Rodrigo de Azevedo
Jan 21 at 11:10
|
show 2 more comments
$begingroup$
Given a matrix $F in mathbb{C}^{m times n}$ such that $m > n$ and other (non-symmetric) square matrix $A$ of size $n times n$, how can one formulate
$$ arg min_b left|A- {F}^{*} operatorname{diag} left( b right) , {F} right|_{2}$$
where $b in mathbb{C}^m$ is some vector and $*$ denotes the conjugate transpose, as a semidefinite program?
I started as follows. Writing the above problem in epigraph form by introducing a variable $x$,
begin{array}{ll} text{minimize} & x\ text{subject to} & left|A- {F}^{*} operatorname{diag} left( b right) , {F} right|_{2} leq xend{array}
which is equivalent to
begin{array}{ll} text{minimize} & x\ text{subject to} & sigma_{max}(A- {F}^{*} operatorname{diag} left( b right) , {F} ) leq xend{array}
which is equivalent to
begin{array}{ll} text{minimize} & x\ text{subject to} & lambda_{max}big((A- {F}^{*} operatorname{diag} left( b right) , {F} )^*(A- {F}^{*} operatorname{diag} left( b right) , {F} ) big) leq x^2end{array}
Can anybody tell me how I can proceed with this?
convex-optimization semidefinite-programming spectral-norm
edited Jan 19 at 8:42
Rodrigo de Azevedo
13.1k41960
asked Jan 16 at 15:00
abina shrabina shr
698
$endgroup$
Given a matrix $F in mathbb{C}^{m times n}$ such that $m > n$ and other (non-symmetric) square matrix $A$ of size $n times n$, how can one formulate
$$ arg min_b left|A- {F}^{*} operatorname{diag} left( b right) , {F} right|_{2}$$
where $b in mathbb{C}^m$ is some vector and $*$ denotes the conjugate transpose, as a semidefinite program?
I started as follows. Writing the above problem in epigraph form by introducing a variable $x$,
begin{array}{ll} text{minimize} & x\ text{subject to} & left|A- {F}^{*} operatorname{diag} left( b right) , {F} right|_{2} leq xend{array}
which is equivalent to
begin{array}{ll} text{minimize} & x\ text{subject to} & sigma_{max}(A- {F}^{*} operatorname{diag} left( b right) , {F} ) leq xend{array}
which is equivalent to
begin{array}{ll} text{minimize} & x\ text{subject to} & lambda_{max}big((A- {F}^{*} operatorname{diag} left( b right) , {F} )^*(A- {F}^{*} operatorname{diag} left( b right) , {F} ) big) leq x^2end{array}
Can anybody tell me how I can proceed with this?
convex-optimization semidefinite-programming spectral-norm
convex-optimization semidefinite-programming spectral-norm
edited Jan 19 at 8:42
Rodrigo de Azevedo
13.1k41960
asked Jan 16 at 15:00
abina shrabina shr
698
edited Jan 19 at 8:42
Rodrigo de Azevedo
13.1k41960
asked Jan 16 at 15:00
abina shrabina shr
698
edited Jan 19 at 8:42
Rodrigo de Azevedo
13.1k41960
edited Jan 19 at 8:42
Rodrigo de Azevedo
13.1k41960
edited Jan 19 at 8:42
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Jan 16 at 15:00
abina shrabina shr
698
asked Jan 16 at 15:00
abina shrabina shr
698
asked Jan 16 at 15:00
abina shrabina shr
698
698
$begingroup$
A key concept is that $F^{T}mbox{diag}(b)F=sum_{i=1}^{n} b_{i} (F_{i}^{T}F_{i})$.
$endgroup$
– Brian Borchers
Jan 16 at 16:24
$begingroup$
@BrianBorchers could you please elaborate a bit?
$endgroup$
– abina shr
Jan 17 at 12:44
$begingroup$
Take a look at this.
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:47
$begingroup$
@RodrigodeAzevedo So I have reached this step: $ begin{array}{ll} text{minimize} & x\ text{subject to} & begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0_{2n}end{array}$. How do I proceed after this?
$endgroup$
– abina shr
Jan 21 at 10:54
$begingroup$
@abinashr Don't you have an SDP in $x$ and $b$ already? Next step would be to solve it.
$endgroup$
– Rodrigo de Azevedo
Jan 21 at 11:10
|
show 2 more comments
$begingroup$
A key concept is that $F^{T}mbox{diag}(b)F=sum_{i=1}^{n} b_{i} (F_{i}^{T}F_{i})$.
$endgroup$
– Brian Borchers
Jan 16 at 16:24
$begingroup$
@BrianBorchers could you please elaborate a bit?
$endgroup$
– abina shr
Jan 17 at 12:44
$begingroup$
Take a look at this.
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:47
$begingroup$
@RodrigodeAzevedo So I have reached this step: $ begin{array}{ll} text{minimize} & x\ text{subject to} & begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0_{2n}end{array}$. How do I proceed after this?
$endgroup$
– abina shr
Jan 21 at 10:54
$begingroup$
@abinashr Don't you have an SDP in $x$ and $b$ already? Next step would be to solve it.
$endgroup$
– Rodrigo de Azevedo
Jan 21 at 11:10
$begingroup$
A key concept is that $F^{T}mbox{diag}(b)F=sum_{i=1}^{n} b_{i} (F_{i}^{T}F_{i})$.
$endgroup$
– Brian Borchers
Jan 16 at 16:24
$begingroup$
A key concept is that $F^{T}mbox{diag}(b)F=sum_{i=1}^{n} b_{i} (F_{i}^{T}F_{i})$.
$endgroup$
– Brian Borchers
Jan 16 at 16:24
$begingroup$
@BrianBorchers could you please elaborate a bit?
$endgroup$
– abina shr
Jan 17 at 12:44
$begingroup$
@BrianBorchers could you please elaborate a bit?
$endgroup$
– abina shr
Jan 17 at 12:44
$begingroup$
Take a look at this.
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:47
$begingroup$
Take a look at this.
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:47
$begingroup$
@RodrigodeAzevedo So I have reached this step: $ begin{array}{ll} text{minimize} & x\ text{subject to} & begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0_{2n}end{array}$. How do I proceed after this?
$endgroup$
– abina shr
Jan 21 at 10:54
$begingroup$
@RodrigodeAzevedo So I have reached this step: $ begin{array}{ll} text{minimize} & x\ text{subject to} & begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0_{2n}end{array}$. How do I proceed after this?
$endgroup$
– abina shr
Jan 21 at 10:54
$begingroup$
@abinashr Don't you have an SDP in $x$ and $b$ already? Next step would be to solve it.
$endgroup$
– Rodrigo de Azevedo
Jan 21 at 11:10
$begingroup$
@abinashr Don't you have an SDP in $x$ and $b$ already? Next step would be to solve it.
$endgroup$
– Rodrigo de Azevedo
Jan 21 at 11:10
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
There are two ways to approach this. One is to recognize that
$$sigma_max(X)leq y quadLongleftrightarrowquad begin{bmatrix} yI & X \ X^T & yI end{bmatrix}
succeq 0$$
So the constraint becomes
$$begin{bmatrix} x I & A - F^Tmathop{textrm{diag}}(b) F \ A - F^Tmathop{textrm{diag}}(b) F & x I end{bmatrix} succeq 0$$
Another way is to recognize that, for a symmetric matrix,
$$sigma_max(X) = max{-lambda_min(X),lambda_max(X)}$$
And with that, we could do
$$ -x I preceq A - F^Tmathop{textrm{diag}}(b)F preceq x I$$
The latter will be preferred because a pair of LMIs is more performant than one twice the size.
answered Jan 16 at 16:53
Michael GrantMichael Grant
15.2k12045
$endgroup$
$begingroup$
For complex case, does it simply become $ begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0$?
$endgroup$
– abina shr
Jan 17 at 16:39
$begingroup$
Shouldn't it be $begin{bmatrix} yI & X \ X^T & yI end{bmatrix} succeq 0$?
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:45
$begingroup$
yes, thanks Rodrigo!
$endgroup$
– Michael Grant
Jan 19 at 13:37
$begingroup$
@abinashr in the complex case, yes that's fine. But note that if $b$ is not real, you lose Hermitian symmetry, so you can't use the second more compact form.
$endgroup$
– Michael Grant
Jan 20 at 0:43
$begingroup$
@MichaelGrant could you please explain why can't I write it in the more compact form if $b$ is complex?
$endgroup$
– abina shr
Jan 21 at 10:45
|
show 3 more comments
Your Answer
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
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votes
$begingroup$
There are two ways to approach this. One is to recognize that
$$sigma_max(X)leq y quadLongleftrightarrowquad begin{bmatrix} yI & X \ X^T & yI end{bmatrix}
succeq 0$$
So the constraint becomes
$$begin{bmatrix} x I & A - F^Tmathop{textrm{diag}}(b) F \ A - F^Tmathop{textrm{diag}}(b) F & x I end{bmatrix} succeq 0$$
Another way is to recognize that, for a symmetric matrix,
$$sigma_max(X) = max{-lambda_min(X),lambda_max(X)}$$
And with that, we could do
$$ -x I preceq A - F^Tmathop{textrm{diag}}(b)F preceq x I$$
The latter will be preferred because a pair of LMIs is more performant than one twice the size.
answered Jan 16 at 16:53
Michael GrantMichael Grant
15.2k12045
$endgroup$
$begingroup$
For complex case, does it simply become $ begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0$?
$endgroup$
– abina shr
Jan 17 at 16:39
$begingroup$
Shouldn't it be $begin{bmatrix} yI & X \ X^T & yI end{bmatrix} succeq 0$?
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:45
$begingroup$
yes, thanks Rodrigo!
$endgroup$
– Michael Grant
Jan 19 at 13:37
$begingroup$
@abinashr in the complex case, yes that's fine. But note that if $b$ is not real, you lose Hermitian symmetry, so you can't use the second more compact form.
$endgroup$
– Michael Grant
Jan 20 at 0:43
$begingroup$
@MichaelGrant could you please explain why can't I write it in the more compact form if $b$ is complex?
$endgroup$
– abina shr
Jan 21 at 10:45
|
show 3 more comments
$begingroup$
There are two ways to approach this. One is to recognize that
$$sigma_max(X)leq y quadLongleftrightarrowquad begin{bmatrix} yI & X \ X^T & yI end{bmatrix}
succeq 0$$
So the constraint becomes
$$begin{bmatrix} x I & A - F^Tmathop{textrm{diag}}(b) F \ A - F^Tmathop{textrm{diag}}(b) F & x I end{bmatrix} succeq 0$$
Another way is to recognize that, for a symmetric matrix,
$$sigma_max(X) = max{-lambda_min(X),lambda_max(X)}$$
And with that, we could do
$$ -x I preceq A - F^Tmathop{textrm{diag}}(b)F preceq x I$$
The latter will be preferred because a pair of LMIs is more performant than one twice the size.
answered Jan 16 at 16:53
Michael GrantMichael Grant
15.2k12045
$endgroup$
$begingroup$
For complex case, does it simply become $ begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0$?
$endgroup$
– abina shr
Jan 17 at 16:39
$begingroup$
Shouldn't it be $begin{bmatrix} yI & X \ X^T & yI end{bmatrix} succeq 0$?
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:45
$begingroup$
yes, thanks Rodrigo!
$endgroup$
– Michael Grant
Jan 19 at 13:37
$begingroup$
@abinashr in the complex case, yes that's fine. But note that if $b$ is not real, you lose Hermitian symmetry, so you can't use the second more compact form.
$endgroup$
– Michael Grant
Jan 20 at 0:43
$begingroup$
@MichaelGrant could you please explain why can't I write it in the more compact form if $b$ is complex?
$endgroup$
– abina shr
Jan 21 at 10:45
|
show 3 more comments
$begingroup$
There are two ways to approach this. One is to recognize that
$$sigma_max(X)leq y quadLongleftrightarrowquad begin{bmatrix} yI & X \ X^T & yI end{bmatrix}
succeq 0$$
So the constraint becomes
$$begin{bmatrix} x I & A - F^Tmathop{textrm{diag}}(b) F \ A - F^Tmathop{textrm{diag}}(b) F & x I end{bmatrix} succeq 0$$
Another way is to recognize that, for a symmetric matrix,
$$sigma_max(X) = max{-lambda_min(X),lambda_max(X)}$$
And with that, we could do
$$ -x I preceq A - F^Tmathop{textrm{diag}}(b)F preceq x I$$
The latter will be preferred because a pair of LMIs is more performant than one twice the size.
answered Jan 16 at 16:53
Michael GrantMichael Grant
15.2k12045
$endgroup$
There are two ways to approach this. One is to recognize that
$$sigma_max(X)leq y quadLongleftrightarrowquad begin{bmatrix} yI & X \ X^T & yI end{bmatrix}
succeq 0$$
So the constraint becomes
$$begin{bmatrix} x I & A - F^Tmathop{textrm{diag}}(b) F \ A - F^Tmathop{textrm{diag}}(b) F & x I end{bmatrix} succeq 0$$
Another way is to recognize that, for a symmetric matrix,
$$sigma_max(X) = max{-lambda_min(X),lambda_max(X)}$$
And with that, we could do
$$ -x I preceq A - F^Tmathop{textrm{diag}}(b)F preceq x I$$
The latter will be preferred because a pair of LMIs is more performant than one twice the size.
answered Jan 16 at 16:53
Michael GrantMichael Grant
15.2k12045
edited Jan 20 at 0:40
answered Jan 16 at 16:53
Michael GrantMichael Grant
15.2k12045
answered Jan 16 at 16:53
Michael GrantMichael Grant
15.2k12045
answered Jan 16 at 16:53
Michael GrantMichael Grant
15.2k12045
15.2k12045
$begingroup$
For complex case, does it simply become $ begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0$?
$endgroup$
– abina shr
Jan 17 at 16:39
$begingroup$
Shouldn't it be $begin{bmatrix} yI & X \ X^T & yI end{bmatrix} succeq 0$?
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:45
$begingroup$
yes, thanks Rodrigo!
$endgroup$
– Michael Grant
Jan 19 at 13:37
$begingroup$
@abinashr in the complex case, yes that's fine. But note that if $b$ is not real, you lose Hermitian symmetry, so you can't use the second more compact form.
$endgroup$
– Michael Grant
Jan 20 at 0:43
$begingroup$
@MichaelGrant could you please explain why can't I write it in the more compact form if $b$ is complex?
$endgroup$
– abina shr
Jan 21 at 10:45
|
show 3 more comments
$begingroup$
For complex case, does it simply become $ begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0$?
$endgroup$
– abina shr
Jan 17 at 16:39
$begingroup$
Shouldn't it be $begin{bmatrix} yI & X \ X^T & yI end{bmatrix} succeq 0$?
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:45
$begingroup$
yes, thanks Rodrigo!
$endgroup$
– Michael Grant
Jan 19 at 13:37
$begingroup$
@abinashr in the complex case, yes that's fine. But note that if $b$ is not real, you lose Hermitian symmetry, so you can't use the second more compact form.
$endgroup$
– Michael Grant
Jan 20 at 0:43
$begingroup$
@MichaelGrant could you please explain why can't I write it in the more compact form if $b$ is complex?
$endgroup$
– abina shr
Jan 21 at 10:45
$begingroup$
For complex case, does it simply become $ begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0$?
$endgroup$
– abina shr
Jan 17 at 16:39
$begingroup$
For complex case, does it simply become $ begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0$?
$endgroup$
– abina shr
Jan 17 at 16:39
$begingroup$
Shouldn't it be $begin{bmatrix} yI & X \ X^T & yI end{bmatrix} succeq 0$?
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:45
$begingroup$
Shouldn't it be $begin{bmatrix} yI & X \ X^T & yI end{bmatrix} succeq 0$?
$endgroup$
– Rodrigo de Azevedo
Jan 19 at 8:45
$begingroup$
yes, thanks Rodrigo!
$endgroup$
– Michael Grant
Jan 19 at 13:37
$begingroup$
yes, thanks Rodrigo!
$endgroup$
– Michael Grant
Jan 19 at 13:37
$begingroup$
@abinashr in the complex case, yes that's fine. But note that if $b$ is not real, you lose Hermitian symmetry, so you can't use the second more compact form.
$endgroup$
– Michael Grant
Jan 20 at 0:43
$begingroup$
@abinashr in the complex case, yes that's fine. But note that if $b$ is not real, you lose Hermitian symmetry, so you can't use the second more compact form.
$endgroup$
– Michael Grant
Jan 20 at 0:43
$begingroup$
@MichaelGrant could you please explain why can't I write it in the more compact form if $b$ is complex?
$endgroup$
– abina shr
Jan 21 at 10:45
$begingroup$
@MichaelGrant could you please explain why can't I write it in the more compact form if $b$ is complex?
$endgroup$
– abina shr
Jan 21 at 10:45
|
show 3 more comments
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$begingroup$
A key concept is that $F^{T}mbox{diag}(b)F=sum_{i=1}^{n} b_{i} (F_{i}^{T}F_{i})$.
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– Brian Borchers
Jan 16 at 16:24
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@BrianBorchers could you please elaborate a bit?
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– abina shr
Jan 17 at 12:44
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Take a look at this.
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– Rodrigo de Azevedo
Jan 19 at 8:47
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@RodrigodeAzevedo So I have reached this step: $ begin{array}{ll} text{minimize} & x\ text{subject to} & begin{bmatrix} x I & A - F^*mathop{textrm{diag}}(b) F \ A^* - F^*mathop{textrm{diag}}(conj(b)) F & x I end{bmatrix} succeq 0_{2n}end{array}$. How do I proceed after this?
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– abina shr
Jan 21 at 10:54
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@abinashr Don't you have an SDP in $x$ and $b$ already? Next step would be to solve it.
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– Rodrigo de Azevedo
Jan 21 at 11:10