- #1
Max Fleiss
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Greetings everyone!
I have a set of tasks I need to solve using using operator norms, inner product... and have some problems with the task in the attachment. I would really appreciate your help and advice.
This is what I have been thinking about so far:
I have to calculate a non trivial upper bound, so maybe it could be done by:
[tex] b=max( ||A||_1,||A||_2,||A||_\infty ) [/tex]
Since [tex] A [/tex] is a difference operator I estimated the following:
[tex] ||A||_1= 4 [/tex]
[tex] ||A||_\infty= 2 [/tex]
But how can I estimate [tex]||A||_2=?[/tex]
If I know that abs row sum is 2 (besides 0 there appears only one 1 and one -1 in the rows) and abs column sum is 4 (it is two times the size of row length dim(A)=2mn x mn). Can I estimate [tex]||A||_2[/tex] by:
[tex]||A||_2=\sqrt{rows^2+columns^2 }=\sqrt{(2 \cdot 2mn)^2+(4 \cdot mn)^2}[/tex][tex]=4 \sqrt{(mn)^2+(mn)^2}=4 \sqrt{2} \sqrt{m^2n^2}[/tex] since [tex]mn[/tex] are positive [tex]||A||_2=4 \sqrt{2} mn[/tex]
So I would say [tex]b=max(L_1,L_2,L_\infty)=L_2=4 \sqrt{2} mn[/tex]
Is my conclusion, approximation of a non trivial upper bound b right?
Thank you in advance for your help!
I have a set of tasks I need to solve using using operator norms, inner product... and have some problems with the task in the attachment. I would really appreciate your help and advice.
This is what I have been thinking about so far:
I have to calculate a non trivial upper bound, so maybe it could be done by:
[tex] b=max( ||A||_1,||A||_2,||A||_\infty ) [/tex]
Since [tex] A [/tex] is a difference operator I estimated the following:
[tex] ||A||_1= 4 [/tex]
[tex] ||A||_\infty= 2 [/tex]
But how can I estimate [tex]||A||_2=?[/tex]
If I know that abs row sum is 2 (besides 0 there appears only one 1 and one -1 in the rows) and abs column sum is 4 (it is two times the size of row length dim(A)=2mn x mn). Can I estimate [tex]||A||_2[/tex] by:
[tex]||A||_2=\sqrt{rows^2+columns^2 }=\sqrt{(2 \cdot 2mn)^2+(4 \cdot mn)^2}[/tex][tex]=4 \sqrt{(mn)^2+(mn)^2}=4 \sqrt{2} \sqrt{m^2n^2}[/tex] since [tex]mn[/tex] are positive [tex]||A||_2=4 \sqrt{2} mn[/tex]
So I would say [tex]b=max(L_1,L_2,L_\infty)=L_2=4 \sqrt{2} mn[/tex]
Is my conclusion, approximation of a non trivial upper bound b right?
Thank you in advance for your help!
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