Mixing
Solve optimization problem using KKT conditions
$begingroup$
I'm trying to understand the solution to Boyd and Vandenberghe Problem 5.30:
Boyd and Vandenberghe Problem 5.30
The Lagrangian is $$L(X,nu)=text{tr}X-logdet X+nu'left(Xs-yright),$$ so the stationary KKT condition is $$nabla L(X,nu)=I-X^{-1}+snu'=0,$$ but the solution implies $$nabla L(X,nu)=I-X^{-1}+dfrac{nu s'+snu'}{2}.$$ Did I compute the gradient incorrectly?
convex-optimization vector-analysis lagrange-multiplier duality-theorems karush-kuhn-tucker
asked Nov 28 '18 at 23:02
L WheelsL Wheels
212
$endgroup$
add a comment |
$begingroup$
I'm trying to understand the solution to Boyd and Vandenberghe Problem 5.30:
Boyd and Vandenberghe Problem 5.30
The Lagrangian is $$L(X,nu)=text{tr}X-logdet X+nu'left(Xs-yright),$$ so the stationary KKT condition is $$nabla L(X,nu)=I-X^{-1}+snu'=0,$$ but the solution implies $$nabla L(X,nu)=I-X^{-1}+dfrac{nu s'+snu'}{2}.$$ Did I compute the gradient incorrectly?
convex-optimization vector-analysis lagrange-multiplier duality-theorems karush-kuhn-tucker
asked Nov 28 '18 at 23:02
L WheelsL Wheels
212
$endgroup$
1
$begingroup$
$X$ is symmetric, so the gradient of a quadratic form involving $X$ should also be symmetric.
$endgroup$
– Brian Borchers
Nov 28 '18 at 23:53
add a comment |
$begingroup$
I'm trying to understand the solution to Boyd and Vandenberghe Problem 5.30:
Boyd and Vandenberghe Problem 5.30
The Lagrangian is $$L(X,nu)=text{tr}X-logdet X+nu'left(Xs-yright),$$ so the stationary KKT condition is $$nabla L(X,nu)=I-X^{-1}+snu'=0,$$ but the solution implies $$nabla L(X,nu)=I-X^{-1}+dfrac{nu s'+snu'}{2}.$$ Did I compute the gradient incorrectly?
convex-optimization vector-analysis lagrange-multiplier duality-theorems karush-kuhn-tucker
asked Nov 28 '18 at 23:02
L WheelsL Wheels
212
$endgroup$
I'm trying to understand the solution to Boyd and Vandenberghe Problem 5.30:
Boyd and Vandenberghe Problem 5.30
The Lagrangian is $$L(X,nu)=text{tr}X-logdet X+nu'left(Xs-yright),$$ so the stationary KKT condition is $$nabla L(X,nu)=I-X^{-1}+snu'=0,$$ but the solution implies $$nabla L(X,nu)=I-X^{-1}+dfrac{nu s'+snu'}{2}.$$ Did I compute the gradient incorrectly?
convex-optimization vector-analysis lagrange-multiplier duality-theorems karush-kuhn-tucker
convex-optimization vector-analysis lagrange-multiplier duality-theorems karush-kuhn-tucker
asked Nov 28 '18 at 23:02
L WheelsL Wheels
212
asked Nov 28 '18 at 23:02
L WheelsL Wheels
212
asked Nov 28 '18 at 23:02
L WheelsL Wheels
212
asked Nov 28 '18 at 23:02
L WheelsL Wheels
212
asked Nov 28 '18 at 23:02
L WheelsL Wheels
212
212
1
$begingroup$
$X$ is symmetric, so the gradient of a quadratic form involving $X$ should also be symmetric.
$endgroup$
– Brian Borchers
Nov 28 '18 at 23:53
add a comment |
1
$begingroup$
$X$ is symmetric, so the gradient of a quadratic form involving $X$ should also be symmetric.
$endgroup$
– Brian Borchers
Nov 28 '18 at 23:53
1
1
$begingroup$
$X$ is symmetric, so the gradient of a quadratic form involving $X$ should also be symmetric.
$endgroup$
– Brian Borchers
Nov 28 '18 at 23:53
$begingroup$
$X$ is symmetric, so the gradient of a quadratic form involving $X$ should also be symmetric.
$endgroup$
– Brian Borchers
Nov 28 '18 at 23:53
add a comment |
1 Answer
1
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$begingroup$
I think you have to use this property:
$$ nu^TXs = sum X_{ij} nu_i s_j = tr(nu s^T X) = tr(X^T (nu s^T)^T) $$
X belongs to symmetric matrices $(X in S^n)$ so: $X^T = X$ then:
$$ tr(Xsnu^T) = tr(snu^TX) $$
we have obtained that $tr(nu s^TX)=tr(snu^TX)$, consequently:
$$ tr(nu s^TX) = [tr(nu s^TX) + tr(snu^TX)]/2 = tr(frac{nu s^T + snu^T}{2}cdot X) $$
and $nabla_X$ of that is the term: $(nu s^T + snu^T)/2$
answered Jan 14 at 18:13
Abradolf LinclerAbradolf Lincler
11
$endgroup$
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/… in order to improve your formatting.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 18:18
add a comment |
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$begingroup$
I think you have to use this property:
$$ nu^TXs = sum X_{ij} nu_i s_j = tr(nu s^T X) = tr(X^T (nu s^T)^T) $$
X belongs to symmetric matrices $(X in S^n)$ so: $X^T = X$ then:
$$ tr(Xsnu^T) = tr(snu^TX) $$
we have obtained that $tr(nu s^TX)=tr(snu^TX)$, consequently:
$$ tr(nu s^TX) = [tr(nu s^TX) + tr(snu^TX)]/2 = tr(frac{nu s^T + snu^T}{2}cdot X) $$
and $nabla_X$ of that is the term: $(nu s^T + snu^T)/2$
answered Jan 14 at 18:13
Abradolf LinclerAbradolf Lincler
11
$endgroup$
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/… in order to improve your formatting.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 18:18
add a comment |
$begingroup$
I think you have to use this property:
$$ nu^TXs = sum X_{ij} nu_i s_j = tr(nu s^T X) = tr(X^T (nu s^T)^T) $$
X belongs to symmetric matrices $(X in S^n)$ so: $X^T = X$ then:
$$ tr(Xsnu^T) = tr(snu^TX) $$
we have obtained that $tr(nu s^TX)=tr(snu^TX)$, consequently:
$$ tr(nu s^TX) = [tr(nu s^TX) + tr(snu^TX)]/2 = tr(frac{nu s^T + snu^T}{2}cdot X) $$
and $nabla_X$ of that is the term: $(nu s^T + snu^T)/2$
answered Jan 14 at 18:13
Abradolf LinclerAbradolf Lincler
11
$endgroup$
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/… in order to improve your formatting.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 18:18
add a comment |
$begingroup$
I think you have to use this property:
$$ nu^TXs = sum X_{ij} nu_i s_j = tr(nu s^T X) = tr(X^T (nu s^T)^T) $$
X belongs to symmetric matrices $(X in S^n)$ so: $X^T = X$ then:
$$ tr(Xsnu^T) = tr(snu^TX) $$
we have obtained that $tr(nu s^TX)=tr(snu^TX)$, consequently:
$$ tr(nu s^TX) = [tr(nu s^TX) + tr(snu^TX)]/2 = tr(frac{nu s^T + snu^T}{2}cdot X) $$
and $nabla_X$ of that is the term: $(nu s^T + snu^T)/2$
answered Jan 14 at 18:13
Abradolf LinclerAbradolf Lincler
11
$endgroup$
I think you have to use this property:
$$ nu^TXs = sum X_{ij} nu_i s_j = tr(nu s^T X) = tr(X^T (nu s^T)^T) $$
X belongs to symmetric matrices $(X in S^n)$ so: $X^T = X$ then:
$$ tr(Xsnu^T) = tr(snu^TX) $$
we have obtained that $tr(nu s^TX)=tr(snu^TX)$, consequently:
$$ tr(nu s^TX) = [tr(nu s^TX) + tr(snu^TX)]/2 = tr(frac{nu s^T + snu^T}{2}cdot X) $$
and $nabla_X$ of that is the term: $(nu s^T + snu^T)/2$
answered Jan 14 at 18:13
Abradolf LinclerAbradolf Lincler
11
edited Jan 14 at 18:28
answered Jan 14 at 18:13
Abradolf LinclerAbradolf Lincler
11
answered Jan 14 at 18:13
Abradolf LinclerAbradolf Lincler
11
answered Jan 14 at 18:13
Abradolf LinclerAbradolf Lincler
11
11
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/… in order to improve your formatting.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 18:18
add a comment |
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/… in order to improve your formatting.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 18:18
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/… in order to improve your formatting.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 18:18
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/… in order to improve your formatting.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 18:18
add a comment |
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$begingroup$
$X$ is symmetric, so the gradient of a quadratic form involving $X$ should also be symmetric.
$endgroup$
– Brian Borchers
Nov 28 '18 at 23:53