p norm of a matrix [closed]
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i have the definition $||A||_p =max_{x{_neq 0}}frac{||Ax||_p}{||x||_p}$ and have to show that $max_{x{_neq 0}}frac{||Ax||_p}{||x||_p}=max_{||x||_p=1}||Ax||_p$. I would be very grateful if someone could give me some hints how to solve this problem.
norm
asked Jan 21 at 23:35
tim123tim123
194
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closed as off-topic by RRL, Lee David Chung Lin, Gibbs, Kemono Chen, metamorphy Jan 22 at 11:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lee David Chung Lin, Gibbs, Kemono Chen, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
i have the definition $||A||_p =max_{x{_neq 0}}frac{||Ax||_p}{||x||_p}$ and have to show that $max_{x{_neq 0}}frac{||Ax||_p}{||x||_p}=max_{||x||_p=1}||Ax||_p$. I would be very grateful if someone could give me some hints how to solve this problem.
norm
asked Jan 21 at 23:35
tim123tim123
194
$endgroup$
closed as off-topic by RRL, Lee David Chung Lin, Gibbs, Kemono Chen, metamorphy Jan 22 at 11:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lee David Chung Lin, Gibbs, Kemono Chen, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
i have the definition $||A||_p =max_{x{_neq 0}}frac{||Ax||_p}{||x||_p}$ and have to show that $max_{x{_neq 0}}frac{||Ax||_p}{||x||_p}=max_{||x||_p=1}||Ax||_p$. I would be very grateful if someone could give me some hints how to solve this problem.
norm
asked Jan 21 at 23:35
tim123tim123
194
$endgroup$
i have the definition $||A||_p =max_{x{_neq 0}}frac{||Ax||_p}{||x||_p}$ and have to show that $max_{x{_neq 0}}frac{||Ax||_p}{||x||_p}=max_{||x||_p=1}||Ax||_p$. I would be very grateful if someone could give me some hints how to solve this problem.
norm
norm
asked Jan 21 at 23:35
tim123tim123
194
asked Jan 21 at 23:35
tim123tim123
194
asked Jan 21 at 23:35
tim123tim123
194
asked Jan 21 at 23:35
tim123tim123
194
asked Jan 21 at 23:35
tim123tim123
194
194
closed as off-topic by RRL, Lee David Chung Lin, Gibbs, Kemono Chen, metamorphy Jan 22 at 11:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lee David Chung Lin, Gibbs, Kemono Chen, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Lee David Chung Lin, Gibbs, Kemono Chen, metamorphy Jan 22 at 11:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lee David Chung Lin, Gibbs, Kemono Chen, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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RHS $leq $ LHS is obvious. Given $x neq 0$ define $y =frac 1 {|x|_p} x$. Verify that $|y|_p=1$. This gives RHS $geq {|Ay|}$. If you write this in terms of $x$ you will get RHS $geq |Ax|_p$. Take max over $x$ to get RHS $geq $ LHS.
answered Jan 21 at 23:43
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
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whats actually RHS and LHS?
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– tim123
Jan 21 at 23:49
$begingroup$
Right Hand Side and Left Hand Side
$endgroup$
– Stan Tendijck
Jan 21 at 23:50
$begingroup$
@tim123 RHS and LHS are the right and left sides of the equality you are asked to prove.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
RHS $leq $ LHS is obvious. Given $x neq 0$ define $y =frac 1 {|x|_p} x$. Verify that $|y|_p=1$. This gives RHS $geq {|Ay|}$. If you write this in terms of $x$ you will get RHS $geq |Ax|_p$. Take max over $x$ to get RHS $geq $ LHS.
answered Jan 21 at 23:43
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
$begingroup$
whats actually RHS and LHS?
$endgroup$
– tim123
Jan 21 at 23:49
$begingroup$
Right Hand Side and Left Hand Side
$endgroup$
– Stan Tendijck
Jan 21 at 23:50
$begingroup$
@tim123 RHS and LHS are the right and left sides of the equality you are asked to prove.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:50
add a comment |
$begingroup$
RHS $leq $ LHS is obvious. Given $x neq 0$ define $y =frac 1 {|x|_p} x$. Verify that $|y|_p=1$. This gives RHS $geq {|Ay|}$. If you write this in terms of $x$ you will get RHS $geq |Ax|_p$. Take max over $x$ to get RHS $geq $ LHS.
answered Jan 21 at 23:43
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
$begingroup$
whats actually RHS and LHS?
$endgroup$
– tim123
Jan 21 at 23:49
$begingroup$
Right Hand Side and Left Hand Side
$endgroup$
– Stan Tendijck
Jan 21 at 23:50
$begingroup$
@tim123 RHS and LHS are the right and left sides of the equality you are asked to prove.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:50
add a comment |
$begingroup$
RHS $leq $ LHS is obvious. Given $x neq 0$ define $y =frac 1 {|x|_p} x$. Verify that $|y|_p=1$. This gives RHS $geq {|Ay|}$. If you write this in terms of $x$ you will get RHS $geq |Ax|_p$. Take max over $x$ to get RHS $geq $ LHS.
answered Jan 21 at 23:43
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
RHS $leq $ LHS is obvious. Given $x neq 0$ define $y =frac 1 {|x|_p} x$. Verify that $|y|_p=1$. This gives RHS $geq {|Ay|}$. If you write this in terms of $x$ you will get RHS $geq |Ax|_p$. Take max over $x$ to get RHS $geq $ LHS.
answered Jan 21 at 23:43
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 21 at 23:43
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 21 at 23:43
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 21 at 23:43
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
62.7k42262
$begingroup$
whats actually RHS and LHS?
$endgroup$
– tim123
Jan 21 at 23:49
$begingroup$
Right Hand Side and Left Hand Side
$endgroup$
– Stan Tendijck
Jan 21 at 23:50
$begingroup$
@tim123 RHS and LHS are the right and left sides of the equality you are asked to prove.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:50
add a comment |
$begingroup$
whats actually RHS and LHS?
$endgroup$
– tim123
Jan 21 at 23:49
$begingroup$
Right Hand Side and Left Hand Side
$endgroup$
– Stan Tendijck
Jan 21 at 23:50
$begingroup$
@tim123 RHS and LHS are the right and left sides of the equality you are asked to prove.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:50
$begingroup$
whats actually RHS and LHS?
$endgroup$
– tim123
Jan 21 at 23:49
$begingroup$
whats actually RHS and LHS?
$endgroup$
– tim123
Jan 21 at 23:49
$begingroup$
Right Hand Side and Left Hand Side
$endgroup$
– Stan Tendijck
Jan 21 at 23:50
$begingroup$
Right Hand Side and Left Hand Side
$endgroup$
– Stan Tendijck
Jan 21 at 23:50
$begingroup$
@tim123 RHS and LHS are the right and left sides of the equality you are asked to prove.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:50
$begingroup$
@tim123 RHS and LHS are the right and left sides of the equality you are asked to prove.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:50
add a comment |
