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solve $underset{d in mathbb{R}^n}{min}$ $g^Td$ subject to $d^THd = 1$ where H $in mathbb{R}^{n times n}$ is...
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I would like to solve the optimization problem $$underset{d in mathbb{R}^n}{min} g^Td$$ subject to $$d^THd = 1$$ where H $in mathbb{R}^{n times n}$ is positive definite and symmetric $textbf{without}$ using Lagrange multipliers.
Does someone know of a way to obtain the solution without relying on Lagrange Multipliers? It would be sufficient to show that $d^* := -frac{H^{-1}g}{||H^{-1}g||_H}$ where $|| cdot ||_H = sqrt{d^THd}$ is the optimal solution (i.e. $g^Td^* leq g^Td forall d in mathbb{R}^n$)
optimization
asked Jan 28 at 15:12
geo17geo17
1038
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add a comment |
$begingroup$
I would like to solve the optimization problem $$underset{d in mathbb{R}^n}{min} g^Td$$ subject to $$d^THd = 1$$ where H $in mathbb{R}^{n times n}$ is positive definite and symmetric $textbf{without}$ using Lagrange multipliers.
Does someone know of a way to obtain the solution without relying on Lagrange Multipliers? It would be sufficient to show that $d^* := -frac{H^{-1}g}{||H^{-1}g||_H}$ where $|| cdot ||_H = sqrt{d^THd}$ is the optimal solution (i.e. $g^Td^* leq g^Td forall d in mathbb{R}^n$)
optimization
asked Jan 28 at 15:12
geo17geo17
1038
$endgroup$
add a comment |
$begingroup$
I would like to solve the optimization problem $$underset{d in mathbb{R}^n}{min} g^Td$$ subject to $$d^THd = 1$$ where H $in mathbb{R}^{n times n}$ is positive definite and symmetric $textbf{without}$ using Lagrange multipliers.
Does someone know of a way to obtain the solution without relying on Lagrange Multipliers? It would be sufficient to show that $d^* := -frac{H^{-1}g}{||H^{-1}g||_H}$ where $|| cdot ||_H = sqrt{d^THd}$ is the optimal solution (i.e. $g^Td^* leq g^Td forall d in mathbb{R}^n$)
optimization
asked Jan 28 at 15:12
geo17geo17
1038
$endgroup$
I would like to solve the optimization problem $$underset{d in mathbb{R}^n}{min} g^Td$$ subject to $$d^THd = 1$$ where H $in mathbb{R}^{n times n}$ is positive definite and symmetric $textbf{without}$ using Lagrange multipliers.
Does someone know of a way to obtain the solution without relying on Lagrange Multipliers? It would be sufficient to show that $d^* := -frac{H^{-1}g}{||H^{-1}g||_H}$ where $|| cdot ||_H = sqrt{d^THd}$ is the optimal solution (i.e. $g^Td^* leq g^Td forall d in mathbb{R}^n$)
optimization
optimization
asked Jan 28 at 15:12
geo17geo17
1038
asked Jan 28 at 15:12
geo17geo17
1038
asked Jan 28 at 15:12
geo17geo17
1038
asked Jan 28 at 15:12
geo17geo17
1038
asked Jan 28 at 15:12
geo17geo17
1038
1038
add a comment |
add a comment |
1 Answer
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Let us introduce a new inner product $[cdot,cdot]$ on $mathbb{R}^n$ by setting $[u,v]=(u,Hv)equiv u^THv$. This is possible, because $H$ is symmetric and positive definite. We have $||u||^2_H=[u,u]$. Now the constraints say that $d$ lies on the unit sphere (with respect to the new inner product), and we have $g^Td=[d,H^{-1}g]$. Where does $[d,H^{-1}g]$ have minimum on the unit sphere? Obviously, the minimum is attained at the unit vector that has the direction opposite to $H^{-1}g$, i.e., at the vector $$d^*=-H^{-1}g/||H^{-1}g||_H.$$
Indeed, we have $[d^*,H^{-1}g]=-||H^{-1}g||_H$, and for any $d$ on the unit sphere one has $|[d,H^{-1}g]|le ||d||_H||H^{-1}g||_H=||H^{-1}g||_H$ and hence $[d,H^{-1}g]ge-||H^{-1}g||_H$.
answered Jan 28 at 15:20
VladimirVladimir
5,413618
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1 Answer
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1 Answer
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oldest
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active
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active
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$begingroup$
Let us introduce a new inner product $[cdot,cdot]$ on $mathbb{R}^n$ by setting $[u,v]=(u,Hv)equiv u^THv$. This is possible, because $H$ is symmetric and positive definite. We have $||u||^2_H=[u,u]$. Now the constraints say that $d$ lies on the unit sphere (with respect to the new inner product), and we have $g^Td=[d,H^{-1}g]$. Where does $[d,H^{-1}g]$ have minimum on the unit sphere? Obviously, the minimum is attained at the unit vector that has the direction opposite to $H^{-1}g$, i.e., at the vector $$d^*=-H^{-1}g/||H^{-1}g||_H.$$
Indeed, we have $[d^*,H^{-1}g]=-||H^{-1}g||_H$, and for any $d$ on the unit sphere one has $|[d,H^{-1}g]|le ||d||_H||H^{-1}g||_H=||H^{-1}g||_H$ and hence $[d,H^{-1}g]ge-||H^{-1}g||_H$.
answered Jan 28 at 15:20
VladimirVladimir
5,413618
$endgroup$
add a comment |
$begingroup$
Let us introduce a new inner product $[cdot,cdot]$ on $mathbb{R}^n$ by setting $[u,v]=(u,Hv)equiv u^THv$. This is possible, because $H$ is symmetric and positive definite. We have $||u||^2_H=[u,u]$. Now the constraints say that $d$ lies on the unit sphere (with respect to the new inner product), and we have $g^Td=[d,H^{-1}g]$. Where does $[d,H^{-1}g]$ have minimum on the unit sphere? Obviously, the minimum is attained at the unit vector that has the direction opposite to $H^{-1}g$, i.e., at the vector $$d^*=-H^{-1}g/||H^{-1}g||_H.$$
Indeed, we have $[d^*,H^{-1}g]=-||H^{-1}g||_H$, and for any $d$ on the unit sphere one has $|[d,H^{-1}g]|le ||d||_H||H^{-1}g||_H=||H^{-1}g||_H$ and hence $[d,H^{-1}g]ge-||H^{-1}g||_H$.
answered Jan 28 at 15:20
VladimirVladimir
5,413618
$endgroup$
add a comment |
$begingroup$
Let us introduce a new inner product $[cdot,cdot]$ on $mathbb{R}^n$ by setting $[u,v]=(u,Hv)equiv u^THv$. This is possible, because $H$ is symmetric and positive definite. We have $||u||^2_H=[u,u]$. Now the constraints say that $d$ lies on the unit sphere (with respect to the new inner product), and we have $g^Td=[d,H^{-1}g]$. Where does $[d,H^{-1}g]$ have minimum on the unit sphere? Obviously, the minimum is attained at the unit vector that has the direction opposite to $H^{-1}g$, i.e., at the vector $$d^*=-H^{-1}g/||H^{-1}g||_H.$$
Indeed, we have $[d^*,H^{-1}g]=-||H^{-1}g||_H$, and for any $d$ on the unit sphere one has $|[d,H^{-1}g]|le ||d||_H||H^{-1}g||_H=||H^{-1}g||_H$ and hence $[d,H^{-1}g]ge-||H^{-1}g||_H$.
answered Jan 28 at 15:20
VladimirVladimir
5,413618
$endgroup$
Let us introduce a new inner product $[cdot,cdot]$ on $mathbb{R}^n$ by setting $[u,v]=(u,Hv)equiv u^THv$. This is possible, because $H$ is symmetric and positive definite. We have $||u||^2_H=[u,u]$. Now the constraints say that $d$ lies on the unit sphere (with respect to the new inner product), and we have $g^Td=[d,H^{-1}g]$. Where does $[d,H^{-1}g]$ have minimum on the unit sphere? Obviously, the minimum is attained at the unit vector that has the direction opposite to $H^{-1}g$, i.e., at the vector $$d^*=-H^{-1}g/||H^{-1}g||_H.$$
Indeed, we have $[d^*,H^{-1}g]=-||H^{-1}g||_H$, and for any $d$ on the unit sphere one has $|[d,H^{-1}g]|le ||d||_H||H^{-1}g||_H=||H^{-1}g||_H$ and hence $[d,H^{-1}g]ge-||H^{-1}g||_H$.
answered Jan 28 at 15:20
VladimirVladimir
5,413618
edited Jan 28 at 16:48
answered Jan 28 at 15:20
VladimirVladimir
5,413618
answered Jan 28 at 15:20
VladimirVladimir
5,413618
answered Jan 28 at 15:20
VladimirVladimir
5,413618
5,413618
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