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Show that every point is an extreme point of the given convex set
Clash Royale CLAN TAG#URR8PPP
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1
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Show that for any $a,b in mathbbR$ every point
$p=beginpmatrix s\ t endpmatrix$ with property
$fracs^2a^2+fract^2b^2=1$ is an extreme point of the convex
set $E(a,b)=leftbeginpmatrix x_1\ x_2 endpmatrix in
mathbbR^2 , Biggvert , fracx_1^2a^2+fracx_2^2b^2 leq
1right$
This is a task from an old exam but I like to know how it's solved and I'm very unsure about my solution:
So the extreme point of a strictly convex function on a convex set is always found at one of the extreme points of the set.
Luckily we already know that $E(a,b)$ is a convex set. Now what needs to be shown is that $fracs^2a^2+fract^2b^2=1$ is strictly convex. It is strictly convex if its Hessian matrix is positive definite. So we need to calculate the Hessian matrix and show it's positive definite:
Here is the part I'm not sure about, the Hessian matrix should be $beginpmatrix
frac2a^2+fract^2b^2-1 & frac2sa^2+frac2tb^2-1\
frac2sa^2+frac2tb^2-1 & fracs^2a^2+frac2b^2-1
endpmatrix$ and it's positive definite if all its eigenvalues are greater than zero (because it's a symmetric matrix it's enough to check only this).
But here I have huge problem because it's too complicated to determine the eigenvalues and it also doesn't really seem to be greater than zero anyway because $a,b in mathbbR$...? : /
Maybe there is another better way of doing it?
linear-algebra matrices geometry convex-analysis convex-geometry
asked 1 hour ago
roblind
501410
add a comment |Â
up vote
1
down vote
favorite
Show that for any $a,b in mathbbR$ every point
$p=beginpmatrix s\ t endpmatrix$ with property
$fracs^2a^2+fract^2b^2=1$ is an extreme point of the convex
set $E(a,b)=leftbeginpmatrix x_1\ x_2 endpmatrix in
mathbbR^2 , Biggvert , fracx_1^2a^2+fracx_2^2b^2 leq
1right$
This is a task from an old exam but I like to know how it's solved and I'm very unsure about my solution:
So the extreme point of a strictly convex function on a convex set is always found at one of the extreme points of the set.
Luckily we already know that $E(a,b)$ is a convex set. Now what needs to be shown is that $fracs^2a^2+fract^2b^2=1$ is strictly convex. It is strictly convex if its Hessian matrix is positive definite. So we need to calculate the Hessian matrix and show it's positive definite:
Here is the part I'm not sure about, the Hessian matrix should be $beginpmatrix
frac2a^2+fract^2b^2-1 & frac2sa^2+frac2tb^2-1\
frac2sa^2+frac2tb^2-1 & fracs^2a^2+frac2b^2-1
endpmatrix$ and it's positive definite if all its eigenvalues are greater than zero (because it's a symmetric matrix it's enough to check only this).
But here I have huge problem because it's too complicated to determine the eigenvalues and it also doesn't really seem to be greater than zero anyway because $a,b in mathbbR$...? : /
Maybe there is another better way of doing it?
linear-algebra matrices geometry convex-analysis convex-geometry
asked 1 hour ago
roblind
501410
2
The question does not mention convex functions. In your attempt at a solution you seem to be trying to show that the function $f$ given by $f(s,t)=s^2/a^2+t^2/b^2-1$ is a strictly convex function (which is not what you are required to do). In doing so, you have also calculated the second derivatives incorrectly. (The Hessian of $f$ is actually $beginpmatrix2/a^2 & 0 \ 0 & 2/b^2endpmatrix$. It is easily seen that this is positive definite by direct application of the definition or because $2/a^2>0$ and the determinant is positive).
– smcc
1 hour ago
2
This follows easily from strict convexity of $x to x^2$.
– Kavi Rama Murthy
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that for any $a,b in mathbbR$ every point
$p=beginpmatrix s\ t endpmatrix$ with property
$fracs^2a^2+fract^2b^2=1$ is an extreme point of the convex
set $E(a,b)=leftbeginpmatrix x_1\ x_2 endpmatrix in
mathbbR^2 , Biggvert , fracx_1^2a^2+fracx_2^2b^2 leq
1right$
This is a task from an old exam but I like to know how it's solved and I'm very unsure about my solution:
So the extreme point of a strictly convex function on a convex set is always found at one of the extreme points of the set.
Luckily we already know that $E(a,b)$ is a convex set. Now what needs to be shown is that $fracs^2a^2+fract^2b^2=1$ is strictly convex. It is strictly convex if its Hessian matrix is positive definite. So we need to calculate the Hessian matrix and show it's positive definite:
Here is the part I'm not sure about, the Hessian matrix should be $beginpmatrix
frac2a^2+fract^2b^2-1 & frac2sa^2+frac2tb^2-1\
frac2sa^2+frac2tb^2-1 & fracs^2a^2+frac2b^2-1
endpmatrix$ and it's positive definite if all its eigenvalues are greater than zero (because it's a symmetric matrix it's enough to check only this).
But here I have huge problem because it's too complicated to determine the eigenvalues and it also doesn't really seem to be greater than zero anyway because $a,b in mathbbR$...? : /
Maybe there is another better way of doing it?
linear-algebra matrices geometry convex-analysis convex-geometry
asked 1 hour ago
roblind
501410
Show that for any $a,b in mathbbR$ every point
$p=beginpmatrix s\ t endpmatrix$ with property
$fracs^2a^2+fract^2b^2=1$ is an extreme point of the convex
set $E(a,b)=leftbeginpmatrix x_1\ x_2 endpmatrix in
mathbbR^2 , Biggvert , fracx_1^2a^2+fracx_2^2b^2 leq
1right$
This is a task from an old exam but I like to know how it's solved and I'm very unsure about my solution:
So the extreme point of a strictly convex function on a convex set is always found at one of the extreme points of the set.
Luckily we already know that $E(a,b)$ is a convex set. Now what needs to be shown is that $fracs^2a^2+fract^2b^2=1$ is strictly convex. It is strictly convex if its Hessian matrix is positive definite. So we need to calculate the Hessian matrix and show it's positive definite:
Here is the part I'm not sure about, the Hessian matrix should be $beginpmatrix
frac2a^2+fract^2b^2-1 & frac2sa^2+frac2tb^2-1\
frac2sa^2+frac2tb^2-1 & fracs^2a^2+frac2b^2-1
endpmatrix$ and it's positive definite if all its eigenvalues are greater than zero (because it's a symmetric matrix it's enough to check only this).
But here I have huge problem because it's too complicated to determine the eigenvalues and it also doesn't really seem to be greater than zero anyway because $a,b in mathbbR$...? : /
Maybe there is another better way of doing it?
linear-algebra matrices geometry convex-analysis convex-geometry
linear-algebra matrices geometry convex-analysis convex-geometry
asked 1 hour ago
roblind
501410
asked 1 hour ago
roblind
501410
asked 1 hour ago
roblind
501410
asked 1 hour ago
roblind
501410
asked 1 hour ago
roblind
501410
501410
2
The question does not mention convex functions. In your attempt at a solution you seem to be trying to show that the function $f$ given by $f(s,t)=s^2/a^2+t^2/b^2-1$ is a strictly convex function (which is not what you are required to do). In doing so, you have also calculated the second derivatives incorrectly. (The Hessian of $f$ is actually $beginpmatrix2/a^2 & 0 \ 0 & 2/b^2endpmatrix$. It is easily seen that this is positive definite by direct application of the definition or because $2/a^2>0$ and the determinant is positive).
– smcc
1 hour ago
2
This follows easily from strict convexity of $x to x^2$.
– Kavi Rama Murthy
1 hour ago
add a comment |Â
2
The question does not mention convex functions. In your attempt at a solution you seem to be trying to show that the function $f$ given by $f(s,t)=s^2/a^2+t^2/b^2-1$ is a strictly convex function (which is not what you are required to do). In doing so, you have also calculated the second derivatives incorrectly. (The Hessian of $f$ is actually $beginpmatrix2/a^2 & 0 \ 0 & 2/b^2endpmatrix$. It is easily seen that this is positive definite by direct application of the definition or because $2/a^2>0$ and the determinant is positive).
– smcc
1 hour ago
2
This follows easily from strict convexity of $x to x^2$.
– Kavi Rama Murthy
1 hour ago
2
2
The question does not mention convex functions. In your attempt at a solution you seem to be trying to show that the function $f$ given by $f(s,t)=s^2/a^2+t^2/b^2-1$ is a strictly convex function (which is not what you are required to do). In doing so, you have also calculated the second derivatives incorrectly. (The Hessian of $f$ is actually $beginpmatrix2/a^2 & 0 \ 0 & 2/b^2endpmatrix$. It is easily seen that this is positive definite by direct application of the definition or because $2/a^2>0$ and the determinant is positive).
– smcc
1 hour ago
The question does not mention convex functions. In your attempt at a solution you seem to be trying to show that the function $f$ given by $f(s,t)=s^2/a^2+t^2/b^2-1$ is a strictly convex function (which is not what you are required to do). In doing so, you have also calculated the second derivatives incorrectly. (The Hessian of $f$ is actually $beginpmatrix2/a^2 & 0 \ 0 & 2/b^2endpmatrix$. It is easily seen that this is positive definite by direct application of the definition or because $2/a^2>0$ and the determinant is positive).
– smcc
1 hour ago
2
2
This follows easily from strict convexity of $x to x^2$.
– Kavi Rama Murthy
1 hour ago
This follows easily from strict convexity of $x to x^2$.
– Kavi Rama Murthy
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
For $(x,y)in E$,
$$ 0le frac(x-s)^2a^2+frac(y-t)^2b^2=underbracefracx^2a^2+fracy^2b^2_le 1+underbracefracs^2a^2+fract^2b^2_=1-2fracxsa^2-2fracytb^2$$
with equality only for $(x,y)=(s,t)$.
Hence
$$ frac sa^2cdot x+frac tb^2cdot yle 1$$with equality only for $(x,y)=(s,t)$.
We conclude that $E$ is contained in one of the closed half planes given by the line given with the equation $frac sa^2cdot x+frac tb^2cdot y=1$ and that $(s,t)$ is the only point of $E$ on that line.
answered 1 hour ago
Hagen von Eitzen
272k21264492
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
For $(x,y)in E$,
$$ 0le frac(x-s)^2a^2+frac(y-t)^2b^2=underbracefracx^2a^2+fracy^2b^2_le 1+underbracefracs^2a^2+fract^2b^2_=1-2fracxsa^2-2fracytb^2$$
with equality only for $(x,y)=(s,t)$.
Hence
$$ frac sa^2cdot x+frac tb^2cdot yle 1$$with equality only for $(x,y)=(s,t)$.
We conclude that $E$ is contained in one of the closed half planes given by the line given with the equation $frac sa^2cdot x+frac tb^2cdot y=1$ and that $(s,t)$ is the only point of $E$ on that line.
answered 1 hour ago
Hagen von Eitzen
272k21264492
add a comment |Â
up vote
4
down vote
accepted
For $(x,y)in E$,
$$ 0le frac(x-s)^2a^2+frac(y-t)^2b^2=underbracefracx^2a^2+fracy^2b^2_le 1+underbracefracs^2a^2+fract^2b^2_=1-2fracxsa^2-2fracytb^2$$
with equality only for $(x,y)=(s,t)$.
Hence
$$ frac sa^2cdot x+frac tb^2cdot yle 1$$with equality only for $(x,y)=(s,t)$.
We conclude that $E$ is contained in one of the closed half planes given by the line given with the equation $frac sa^2cdot x+frac tb^2cdot y=1$ and that $(s,t)$ is the only point of $E$ on that line.
answered 1 hour ago
Hagen von Eitzen
272k21264492
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
For $(x,y)in E$,
$$ 0le frac(x-s)^2a^2+frac(y-t)^2b^2=underbracefracx^2a^2+fracy^2b^2_le 1+underbracefracs^2a^2+fract^2b^2_=1-2fracxsa^2-2fracytb^2$$
with equality only for $(x,y)=(s,t)$.
Hence
$$ frac sa^2cdot x+frac tb^2cdot yle 1$$with equality only for $(x,y)=(s,t)$.
We conclude that $E$ is contained in one of the closed half planes given by the line given with the equation $frac sa^2cdot x+frac tb^2cdot y=1$ and that $(s,t)$ is the only point of $E$ on that line.
answered 1 hour ago
Hagen von Eitzen
272k21264492
For $(x,y)in E$,
$$ 0le frac(x-s)^2a^2+frac(y-t)^2b^2=underbracefracx^2a^2+fracy^2b^2_le 1+underbracefracs^2a^2+fract^2b^2_=1-2fracxsa^2-2fracytb^2$$
with equality only for $(x,y)=(s,t)$.
Hence
$$ frac sa^2cdot x+frac tb^2cdot yle 1$$with equality only for $(x,y)=(s,t)$.
We conclude that $E$ is contained in one of the closed half planes given by the line given with the equation $frac sa^2cdot x+frac tb^2cdot y=1$ and that $(s,t)$ is the only point of $E$ on that line.
answered 1 hour ago
Hagen von Eitzen
272k21264492
answered 1 hour ago
Hagen von Eitzen
272k21264492
answered 1 hour ago
Hagen von Eitzen
272k21264492
answered 1 hour ago
Hagen von Eitzen
272k21264492
272k21264492
add a comment |Â
add a comment |Â
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2
The question does not mention convex functions. In your attempt at a solution you seem to be trying to show that the function $f$ given by $f(s,t)=s^2/a^2+t^2/b^2-1$ is a strictly convex function (which is not what you are required to do). In doing so, you have also calculated the second derivatives incorrectly. (The Hessian of $f$ is actually $beginpmatrix2/a^2 & 0 \ 0 & 2/b^2endpmatrix$. It is easily seen that this is positive definite by direct application of the definition or because $2/a^2>0$ and the determinant is positive).
– smcc
1 hour ago
2
This follows easily from strict convexity of $x to x^2$.
– Kavi Rama Murthy
1 hour ago