How to deal with this 2-norm? [closed]












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$A$ is an $m×n$ matrix, verify the inequality $|A|_inftyleqsqrt n|A|_2$ and give an example of a nonzero matrix if the equality is achieved.










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closed as off-topic by Saad, Leucippus, Eevee Trainer, A Blumenthal, mrtaurho Dec 18 '18 at 5:39


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." – Saad, Leucippus, Eevee Trainer, A Blumenthal, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.
















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    Where are you stuck? Maybe start with reviewing the definitions of these norms.
    $endgroup$
    – Dave
    Dec 18 '18 at 1:51










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    The 2-norm of a matrix is its maximal singular value? How to represent the singular value of A?
    $endgroup$
    – Kristy
    Dec 18 '18 at 1:54
















0












$begingroup$


$A$ is an $m×n$ matrix, verify the inequality $|A|_inftyleqsqrt n|A|_2$ and give an example of a nonzero matrix if the equality is achieved.










share|cite|improve this question









$endgroup$



closed as off-topic by Saad, Leucippus, Eevee Trainer, A Blumenthal, mrtaurho Dec 18 '18 at 5:39


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." – Saad, Leucippus, Eevee Trainer, A Blumenthal, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Where are you stuck? Maybe start with reviewing the definitions of these norms.
    $endgroup$
    – Dave
    Dec 18 '18 at 1:51










  • $begingroup$
    The 2-norm of a matrix is its maximal singular value? How to represent the singular value of A?
    $endgroup$
    – Kristy
    Dec 18 '18 at 1:54














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$begingroup$


$A$ is an $m×n$ matrix, verify the inequality $|A|_inftyleqsqrt n|A|_2$ and give an example of a nonzero matrix if the equality is achieved.










share|cite|improve this question









$endgroup$




$A$ is an $m×n$ matrix, verify the inequality $|A|_inftyleqsqrt n|A|_2$ and give an example of a nonzero matrix if the equality is achieved.







linear-algebra






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asked Dec 18 '18 at 0:41









KristyKristy

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closed as off-topic by Saad, Leucippus, Eevee Trainer, A Blumenthal, mrtaurho Dec 18 '18 at 5:39


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." – Saad, Leucippus, Eevee Trainer, A Blumenthal, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Saad, Leucippus, Eevee Trainer, A Blumenthal, mrtaurho Dec 18 '18 at 5:39


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." – Saad, Leucippus, Eevee Trainer, A Blumenthal, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Where are you stuck? Maybe start with reviewing the definitions of these norms.
    $endgroup$
    – Dave
    Dec 18 '18 at 1:51










  • $begingroup$
    The 2-norm of a matrix is its maximal singular value? How to represent the singular value of A?
    $endgroup$
    – Kristy
    Dec 18 '18 at 1:54


















  • $begingroup$
    Where are you stuck? Maybe start with reviewing the definitions of these norms.
    $endgroup$
    – Dave
    Dec 18 '18 at 1:51










  • $begingroup$
    The 2-norm of a matrix is its maximal singular value? How to represent the singular value of A?
    $endgroup$
    – Kristy
    Dec 18 '18 at 1:54
















$begingroup$
Where are you stuck? Maybe start with reviewing the definitions of these norms.
$endgroup$
– Dave
Dec 18 '18 at 1:51




$begingroup$
Where are you stuck? Maybe start with reviewing the definitions of these norms.
$endgroup$
– Dave
Dec 18 '18 at 1:51












$begingroup$
The 2-norm of a matrix is its maximal singular value? How to represent the singular value of A?
$endgroup$
– Kristy
Dec 18 '18 at 1:54




$begingroup$
The 2-norm of a matrix is its maximal singular value? How to represent the singular value of A?
$endgroup$
– Kristy
Dec 18 '18 at 1:54










1 Answer
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For any $0neq xinmathbb R^n$, we have $$Vert xVert_2=sqrt{x_1^2+cdots+x_n^2}leq sqrt{nmax_{1leq jleq n}x_j^2}=sqrt{n}Vert xVert_infty$$ and $$Vert AxVert_infty=max_{1leq jleq n}|(Ax)_j|=sqrt{max_{1leq jleq n}(Ax)_j^2}leqsqrt{(Ax)_1^2+cdots+(Ax)_n^2}=Vert AxVert_2$$ which gives $$frac{Vert AxVert_infty}{Vert xVert_infty}leqsqrt{n}frac{Vert AxVert_2}{Vert xVert_2}$$






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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    For any $0neq xinmathbb R^n$, we have $$Vert xVert_2=sqrt{x_1^2+cdots+x_n^2}leq sqrt{nmax_{1leq jleq n}x_j^2}=sqrt{n}Vert xVert_infty$$ and $$Vert AxVert_infty=max_{1leq jleq n}|(Ax)_j|=sqrt{max_{1leq jleq n}(Ax)_j^2}leqsqrt{(Ax)_1^2+cdots+(Ax)_n^2}=Vert AxVert_2$$ which gives $$frac{Vert AxVert_infty}{Vert xVert_infty}leqsqrt{n}frac{Vert AxVert_2}{Vert xVert_2}$$






    share|cite|improve this answer









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      0












      $begingroup$

      For any $0neq xinmathbb R^n$, we have $$Vert xVert_2=sqrt{x_1^2+cdots+x_n^2}leq sqrt{nmax_{1leq jleq n}x_j^2}=sqrt{n}Vert xVert_infty$$ and $$Vert AxVert_infty=max_{1leq jleq n}|(Ax)_j|=sqrt{max_{1leq jleq n}(Ax)_j^2}leqsqrt{(Ax)_1^2+cdots+(Ax)_n^2}=Vert AxVert_2$$ which gives $$frac{Vert AxVert_infty}{Vert xVert_infty}leqsqrt{n}frac{Vert AxVert_2}{Vert xVert_2}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        For any $0neq xinmathbb R^n$, we have $$Vert xVert_2=sqrt{x_1^2+cdots+x_n^2}leq sqrt{nmax_{1leq jleq n}x_j^2}=sqrt{n}Vert xVert_infty$$ and $$Vert AxVert_infty=max_{1leq jleq n}|(Ax)_j|=sqrt{max_{1leq jleq n}(Ax)_j^2}leqsqrt{(Ax)_1^2+cdots+(Ax)_n^2}=Vert AxVert_2$$ which gives $$frac{Vert AxVert_infty}{Vert xVert_infty}leqsqrt{n}frac{Vert AxVert_2}{Vert xVert_2}$$






        share|cite|improve this answer









        $endgroup$



        For any $0neq xinmathbb R^n$, we have $$Vert xVert_2=sqrt{x_1^2+cdots+x_n^2}leq sqrt{nmax_{1leq jleq n}x_j^2}=sqrt{n}Vert xVert_infty$$ and $$Vert AxVert_infty=max_{1leq jleq n}|(Ax)_j|=sqrt{max_{1leq jleq n}(Ax)_j^2}leqsqrt{(Ax)_1^2+cdots+(Ax)_n^2}=Vert AxVert_2$$ which gives $$frac{Vert AxVert_infty}{Vert xVert_infty}leqsqrt{n}frac{Vert AxVert_2}{Vert xVert_2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 2:10









        DaveDave

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        9,05311033















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