interpreting dual norm of quadratic norm with change of basis
I'm trying to compute
$$
|z|_* := sup_{|x|_Aleq1} |z^T x|
$$
for $|x|_A := x^T A x$ with $A succ 0$, symmetric. I see that we're looking at $z^T in mathbb{R}^n$ as a linear functional, so this is really the operator norm with $|cdot|$ over $mathbb{R}$ and $|cdot|_A$ over $mathbb{R}^n$.
I make a change of coordinates $y = A^{1/2}x$ and substitute to get
$$
|z|_* := sup_{|y|_2leq1} |z^T A^{-1/2}y|.
$$
Based on @max_zorn's comment, I use symmetry of $A$ so that
$$
|z|_* = sup_{y^T y=1}|A^{-1/2}z^T y| = sup_{|y|_2=1}|A^{-1/2}z^T y| leq |A^{-1/2}z|_2 |y|_2
$$
with equality at $y = A^{-1/2}z / |A^{-1/2}z|_2$ so $|z|_* = |A^{-1/2}z|_2$.
What is the function analysis explanation for why the dual norm should be computable as the euclidean norm of the transformed vector $A^{-1/2}z$? How can I think about duality more intuitively in this case?
functional-analysis norm convex-optimization duality-theorems
asked Jan 6 at 19:21
jjjjjj
1,100515
add a comment |
I'm trying to compute
$$
|z|_* := sup_{|x|_Aleq1} |z^T x|
$$
for $|x|_A := x^T A x$ with $A succ 0$, symmetric. I see that we're looking at $z^T in mathbb{R}^n$ as a linear functional, so this is really the operator norm with $|cdot|$ over $mathbb{R}$ and $|cdot|_A$ over $mathbb{R}^n$.
I make a change of coordinates $y = A^{1/2}x$ and substitute to get
$$
|z|_* := sup_{|y|_2leq1} |z^T A^{-1/2}y|.
$$
Based on @max_zorn's comment, I use symmetry of $A$ so that
$$
|z|_* = sup_{y^T y=1}|A^{-1/2}z^T y| = sup_{|y|_2=1}|A^{-1/2}z^T y| leq |A^{-1/2}z|_2 |y|_2
$$
with equality at $y = A^{-1/2}z / |A^{-1/2}z|_2$ so $|z|_* = |A^{-1/2}z|_2$.
What is the function analysis explanation for why the dual norm should be computable as the euclidean norm of the transformed vector $A^{-1/2}z$? How can I think about duality more intuitively in this case?
functional-analysis norm convex-optimization duality-theorems
asked Jan 6 at 19:21
jjjjjj
1,100515
$langle z,A^{-1/2}yrangle = langle A^{-1/2}z,yrangle$
– max_zorn
Jan 6 at 19:24
So you have $|z|_*=sup_{y^Ty=1} |A^{-1/2}z^T y| $. Then what?
– jjjjjj
Jan 6 at 20:12
You are done. This computes the Euclidean norm of $A^{-1/2}{z}$.
– max_zorn
Jan 6 at 20:19
Sorry, I'm still not seeing how. Can you clarify?
– jjjjjj
Jan 6 at 21:22
1
$max_{|y|leq 1} langle w,yrangle = |w|$. To see this, Observe that Cauchy-Schwarz gives LHS $leq$ RHS. To get equality, consider $y=w/|w|$.
– max_zorn
Jan 6 at 21:23
add a comment |
I'm trying to compute
$$
|z|_* := sup_{|x|_Aleq1} |z^T x|
$$
for $|x|_A := x^T A x$ with $A succ 0$, symmetric. I see that we're looking at $z^T in mathbb{R}^n$ as a linear functional, so this is really the operator norm with $|cdot|$ over $mathbb{R}$ and $|cdot|_A$ over $mathbb{R}^n$.
I make a change of coordinates $y = A^{1/2}x$ and substitute to get
$$
|z|_* := sup_{|y|_2leq1} |z^T A^{-1/2}y|.
$$
Based on @max_zorn's comment, I use symmetry of $A$ so that
$$
|z|_* = sup_{y^T y=1}|A^{-1/2}z^T y| = sup_{|y|_2=1}|A^{-1/2}z^T y| leq |A^{-1/2}z|_2 |y|_2
$$
with equality at $y = A^{-1/2}z / |A^{-1/2}z|_2$ so $|z|_* = |A^{-1/2}z|_2$.
What is the function analysis explanation for why the dual norm should be computable as the euclidean norm of the transformed vector $A^{-1/2}z$? How can I think about duality more intuitively in this case?
functional-analysis norm convex-optimization duality-theorems
asked Jan 6 at 19:21
jjjjjj
1,100515
I'm trying to compute
$$
|z|_* := sup_{|x|_Aleq1} |z^T x|
$$
for $|x|_A := x^T A x$ with $A succ 0$, symmetric. I see that we're looking at $z^T in mathbb{R}^n$ as a linear functional, so this is really the operator norm with $|cdot|$ over $mathbb{R}$ and $|cdot|_A$ over $mathbb{R}^n$.
I make a change of coordinates $y = A^{1/2}x$ and substitute to get
$$
|z|_* := sup_{|y|_2leq1} |z^T A^{-1/2}y|.
$$
Based on @max_zorn's comment, I use symmetry of $A$ so that
$$
|z|_* = sup_{y^T y=1}|A^{-1/2}z^T y| = sup_{|y|_2=1}|A^{-1/2}z^T y| leq |A^{-1/2}z|_2 |y|_2
$$
with equality at $y = A^{-1/2}z / |A^{-1/2}z|_2$ so $|z|_* = |A^{-1/2}z|_2$.
What is the function analysis explanation for why the dual norm should be computable as the euclidean norm of the transformed vector $A^{-1/2}z$? How can I think about duality more intuitively in this case?
functional-analysis norm convex-optimization duality-theorems
functional-analysis norm convex-optimization duality-theorems
asked Jan 6 at 19:21
jjjjjj
1,100515
asked Jan 6 at 19:21
jjjjjj
1,100515
edited Jan 7 at 18:11
asked Jan 6 at 19:21
jjjjjj
1,100515
asked Jan 6 at 19:21
jjjjjj
1,100515
asked Jan 6 at 19:21
jjjjjj
1,100515
1,100515
$langle z,A^{-1/2}yrangle = langle A^{-1/2}z,yrangle$
– max_zorn
Jan 6 at 19:24
So you have $|z|_*=sup_{y^Ty=1} |A^{-1/2}z^T y| $. Then what?
– jjjjjj
Jan 6 at 20:12
You are done. This computes the Euclidean norm of $A^{-1/2}{z}$.
– max_zorn
Jan 6 at 20:19
Sorry, I'm still not seeing how. Can you clarify?
– jjjjjj
Jan 6 at 21:22
1
$max_{|y|leq 1} langle w,yrangle = |w|$. To see this, Observe that Cauchy-Schwarz gives LHS $leq$ RHS. To get equality, consider $y=w/|w|$.
– max_zorn
Jan 6 at 21:23
add a comment |
$langle z,A^{-1/2}yrangle = langle A^{-1/2}z,yrangle$
– max_zorn
Jan 6 at 19:24
So you have $|z|_*=sup_{y^Ty=1} |A^{-1/2}z^T y| $. Then what?
– jjjjjj
Jan 6 at 20:12
You are done. This computes the Euclidean norm of $A^{-1/2}{z}$.
– max_zorn
Jan 6 at 20:19
Sorry, I'm still not seeing how. Can you clarify?
– jjjjjj
Jan 6 at 21:22
1
$max_{|y|leq 1} langle w,yrangle = |w|$. To see this, Observe that Cauchy-Schwarz gives LHS $leq$ RHS. To get equality, consider $y=w/|w|$.
– max_zorn
Jan 6 at 21:23
$langle z,A^{-1/2}yrangle = langle A^{-1/2}z,yrangle$
– max_zorn
Jan 6 at 19:24
$langle z,A^{-1/2}yrangle = langle A^{-1/2}z,yrangle$
– max_zorn
Jan 6 at 19:24
So you have $|z|_*=sup_{y^Ty=1} |A^{-1/2}z^T y| $. Then what?
– jjjjjj
Jan 6 at 20:12
So you have $|z|_*=sup_{y^Ty=1} |A^{-1/2}z^T y| $. Then what?
– jjjjjj
Jan 6 at 20:12
You are done. This computes the Euclidean norm of $A^{-1/2}{z}$.
– max_zorn
Jan 6 at 20:19
You are done. This computes the Euclidean norm of $A^{-1/2}{z}$.
– max_zorn
Jan 6 at 20:19
Sorry, I'm still not seeing how. Can you clarify?
– jjjjjj
Jan 6 at 21:22
Sorry, I'm still not seeing how. Can you clarify?
– jjjjjj
Jan 6 at 21:22
1
1
$max_{|y|leq 1} langle w,yrangle = |w|$. To see this, Observe that Cauchy-Schwarz gives LHS $leq$ RHS. To get equality, consider $y=w/|w|$.
– max_zorn
Jan 6 at 21:23
$max_{|y|leq 1} langle w,yrangle = |w|$. To see this, Observe that Cauchy-Schwarz gives LHS $leq$ RHS. To get equality, consider $y=w/|w|$.
– max_zorn
Jan 6 at 21:23
add a comment |
1 Answer
1
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votes
I suppose that you drop a power exponent of 1/2 in the term $||x||_A = (x^TAx)^{1/2}$
answered Nov 23 at 14:08
Caldera
11
I'm not sure this actually answers the question
– MRobinson
Nov 23 at 14:32
Well, I didn't put it clearly, I mean that the question do help me a lot, and my point is in fact the latter. Feel sorry to reedit my answer.
– Caldera
Nov 24 at 11:26
add a comment |
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1 Answer
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votes
I suppose that you drop a power exponent of 1/2 in the term $||x||_A = (x^TAx)^{1/2}$
answered Nov 23 at 14:08
Caldera
11
I'm not sure this actually answers the question
– MRobinson
Nov 23 at 14:32
Well, I didn't put it clearly, I mean that the question do help me a lot, and my point is in fact the latter. Feel sorry to reedit my answer.
– Caldera
Nov 24 at 11:26
add a comment |
I suppose that you drop a power exponent of 1/2 in the term $||x||_A = (x^TAx)^{1/2}$
answered Nov 23 at 14:08
Caldera
11
I'm not sure this actually answers the question
– MRobinson
Nov 23 at 14:32
Well, I didn't put it clearly, I mean that the question do help me a lot, and my point is in fact the latter. Feel sorry to reedit my answer.
– Caldera
Nov 24 at 11:26
add a comment |
I suppose that you drop a power exponent of 1/2 in the term $||x||_A = (x^TAx)^{1/2}$
answered Nov 23 at 14:08
Caldera
11
I suppose that you drop a power exponent of 1/2 in the term $||x||_A = (x^TAx)^{1/2}$
answered Nov 23 at 14:08
Caldera
11
edited Nov 24 at 4:35
answered Nov 23 at 14:08
Caldera
11
answered Nov 23 at 14:08
Caldera
11
answered Nov 23 at 14:08
Caldera
11
11
I'm not sure this actually answers the question
– MRobinson
Nov 23 at 14:32
Well, I didn't put it clearly, I mean that the question do help me a lot, and my point is in fact the latter. Feel sorry to reedit my answer.
– Caldera
Nov 24 at 11:26
add a comment |
I'm not sure this actually answers the question
– MRobinson
Nov 23 at 14:32
Well, I didn't put it clearly, I mean that the question do help me a lot, and my point is in fact the latter. Feel sorry to reedit my answer.
– Caldera
Nov 24 at 11:26
I'm not sure this actually answers the question
– MRobinson
Nov 23 at 14:32
I'm not sure this actually answers the question
– MRobinson
Nov 23 at 14:32
Well, I didn't put it clearly, I mean that the question do help me a lot, and my point is in fact the latter. Feel sorry to reedit my answer.
– Caldera
Nov 24 at 11:26
Well, I didn't put it clearly, I mean that the question do help me a lot, and my point is in fact the latter. Feel sorry to reedit my answer.
– Caldera
Nov 24 at 11:26
add a comment |
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$langle z,A^{-1/2}yrangle = langle A^{-1/2}z,yrangle$
– max_zorn
Jan 6 at 19:24
So you have $|z|_*=sup_{y^Ty=1} |A^{-1/2}z^T y| $. Then what?
– jjjjjj
Jan 6 at 20:12
You are done. This computes the Euclidean norm of $A^{-1/2}{z}$.
– max_zorn
Jan 6 at 20:19
Sorry, I'm still not seeing how. Can you clarify?
– jjjjjj
Jan 6 at 21:22
1
$max_{|y|leq 1} langle w,yrangle = |w|$. To see this, Observe that Cauchy-Schwarz gives LHS $leq$ RHS. To get equality, consider $y=w/|w|$.
– max_zorn
Jan 6 at 21:23