Prove that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$.












2














Given that $E$ is a finite dimensional space. Let $dim E=ngeq 1$ and ${e_i}^{n}_{i=1}$ be a basis for $E.$ Then, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that
begin{align}x=sum_{i=1}^{n}alpha_i e_i.end{align}
The problem is: I want to prove that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$.



I, therefore, post the proof in the answer section after it has been approved.






linear-algebra functional-analysis norm normed-spaces







share|cite|improve this question
























  • Your problem makes no sense. What are the $alpha_i$'s?
    – José Carlos Santos
    Dec 26 at 11:58










  • @ José Carlos Santos: They are unique scalars such ${e_i}^{n}_{i=1}$ is a basis for $E$.
    – Mike
    Dec 26 at 12:00












  • That answer makes no sense either.
    – José Carlos Santos
    Dec 26 at 12:01












  • Okay, let me edit my post then. Just some time!
    – Mike
    Dec 26 at 12:02










  • @José Carlos Santos: Kindly check, I made some edits. It should be fine now!
    – Mike
    Dec 26 at 12:07
















2














Given that $E$ is a finite dimensional space. Let $dim E=ngeq 1$ and ${e_i}^{n}_{i=1}$ be a basis for $E.$ Then, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that
begin{align}x=sum_{i=1}^{n}alpha_i e_i.end{align}
The problem is: I want to prove that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$.



I, therefore, post the proof in the answer section after it has been approved.






linear-algebra functional-analysis norm normed-spaces







share|cite|improve this question
























  • Your problem makes no sense. What are the $alpha_i$'s?
    – José Carlos Santos
    Dec 26 at 11:58










  • @ José Carlos Santos: They are unique scalars such ${e_i}^{n}_{i=1}$ is a basis for $E$.
    – Mike
    Dec 26 at 12:00












  • That answer makes no sense either.
    – José Carlos Santos
    Dec 26 at 12:01












  • Okay, let me edit my post then. Just some time!
    – Mike
    Dec 26 at 12:02










  • @José Carlos Santos: Kindly check, I made some edits. It should be fine now!
    – Mike
    Dec 26 at 12:07














2












2








2


1





Given that $E$ is a finite dimensional space. Let $dim E=ngeq 1$ and ${e_i}^{n}_{i=1}$ be a basis for $E.$ Then, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that
begin{align}x=sum_{i=1}^{n}alpha_i e_i.end{align}
The problem is: I want to prove that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$.



I, therefore, post the proof in the answer section after it has been approved.






linear-algebra functional-analysis norm normed-spaces







share|cite|improve this question















Given that $E$ is a finite dimensional space. Let $dim E=ngeq 1$ and ${e_i}^{n}_{i=1}$ be a basis for $E.$ Then, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that
begin{align}x=sum_{i=1}^{n}alpha_i e_i.end{align}
The problem is: I want to prove that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$.



I, therefore, post the proof in the answer section after it has been approved.






linear-algebra functional-analysis norm normed-spaces




linear-algebra functional-analysis norm normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 at 14:18

























asked Dec 26 at 11:52









Mike

1,278220




1,278220












  • Your problem makes no sense. What are the $alpha_i$'s?
    – José Carlos Santos
    Dec 26 at 11:58










  • @ José Carlos Santos: They are unique scalars such ${e_i}^{n}_{i=1}$ is a basis for $E$.
    – Mike
    Dec 26 at 12:00












  • That answer makes no sense either.
    – José Carlos Santos
    Dec 26 at 12:01












  • Okay, let me edit my post then. Just some time!
    – Mike
    Dec 26 at 12:02










  • @José Carlos Santos: Kindly check, I made some edits. It should be fine now!
    – Mike
    Dec 26 at 12:07


















  • Your problem makes no sense. What are the $alpha_i$'s?
    – José Carlos Santos
    Dec 26 at 11:58










  • @ José Carlos Santos: They are unique scalars such ${e_i}^{n}_{i=1}$ is a basis for $E$.
    – Mike
    Dec 26 at 12:00












  • That answer makes no sense either.
    – José Carlos Santos
    Dec 26 at 12:01












  • Okay, let me edit my post then. Just some time!
    – Mike
    Dec 26 at 12:02










  • @José Carlos Santos: Kindly check, I made some edits. It should be fine now!
    – Mike
    Dec 26 at 12:07
















Your problem makes no sense. What are the $alpha_i$'s?
– José Carlos Santos
Dec 26 at 11:58




Your problem makes no sense. What are the $alpha_i$'s?
– José Carlos Santos
Dec 26 at 11:58












@ José Carlos Santos: They are unique scalars such ${e_i}^{n}_{i=1}$ is a basis for $E$.
– Mike
Dec 26 at 12:00






@ José Carlos Santos: They are unique scalars such ${e_i}^{n}_{i=1}$ is a basis for $E$.
– Mike
Dec 26 at 12:00














That answer makes no sense either.
– José Carlos Santos
Dec 26 at 12:01






That answer makes no sense either.
– José Carlos Santos
Dec 26 at 12:01














Okay, let me edit my post then. Just some time!
– Mike
Dec 26 at 12:02




Okay, let me edit my post then. Just some time!
– Mike
Dec 26 at 12:02












@José Carlos Santos: Kindly check, I made some edits. It should be fine now!
– Mike
Dec 26 at 12:07




@José Carlos Santos: Kindly check, I made some edits. It should be fine now!
– Mike
Dec 26 at 12:07










1 Answer
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$a.qquad$ Let $xin E,$ then begin{align}|x|_0=0&iff maxlimits_{1leq ileq n}|alpha_i|=0 iff |alpha_i|=0,,forall ,{1leq ileq n}\&iff alpha_i=0,,forall ,{1leq ileq n}\&iff x=0,;forall;xin Eend{align}
$b.qquad$ Let $lambda in K$, then $lambda x=sum_{i=1}^{n}lambda, alpha_i e_i$ and



begin{align}|lambda x|_0&= maxlimits_{1leq ileq n}|lambda,alpha_i|=|lambda,|maxlimits_{1leq ileq n}|alpha_i|=|lambda,|| x|_0,;forall;xin Eend{align}



$c.qquad$ Let $x,yin E$, then $x=sum_{i=1}^{n}alpha_i e_i$ and $y=sum_{i=1}^{n}beta_i e_i,$ for some $alpha_i,beta_i,;iin{1,2,cdots,n}.$ Thus,



begin{align}| x+y|_0&= maxlimits_{1leq ileq n}|alpha_i+beta_i|leq maxlimits_{1leq ileq n}|alpha_i|+maxlimits_{1leq ileq n}|beta_i|=| x|_0+|y|_0,;forall;x,yin Eend{align}






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    $a.qquad$ Let $xin E,$ then begin{align}|x|_0=0&iff maxlimits_{1leq ileq n}|alpha_i|=0 iff |alpha_i|=0,,forall ,{1leq ileq n}\&iff alpha_i=0,,forall ,{1leq ileq n}\&iff x=0,;forall;xin Eend{align}
    $b.qquad$ Let $lambda in K$, then $lambda x=sum_{i=1}^{n}lambda, alpha_i e_i$ and



    begin{align}|lambda x|_0&= maxlimits_{1leq ileq n}|lambda,alpha_i|=|lambda,|maxlimits_{1leq ileq n}|alpha_i|=|lambda,|| x|_0,;forall;xin Eend{align}



    $c.qquad$ Let $x,yin E$, then $x=sum_{i=1}^{n}alpha_i e_i$ and $y=sum_{i=1}^{n}beta_i e_i,$ for some $alpha_i,beta_i,;iin{1,2,cdots,n}.$ Thus,



    begin{align}| x+y|_0&= maxlimits_{1leq ileq n}|alpha_i+beta_i|leq maxlimits_{1leq ileq n}|alpha_i|+maxlimits_{1leq ileq n}|beta_i|=| x|_0+|y|_0,;forall;x,yin Eend{align}






    share|cite|improve this answer


























      0














      $a.qquad$ Let $xin E,$ then begin{align}|x|_0=0&iff maxlimits_{1leq ileq n}|alpha_i|=0 iff |alpha_i|=0,,forall ,{1leq ileq n}\&iff alpha_i=0,,forall ,{1leq ileq n}\&iff x=0,;forall;xin Eend{align}
      $b.qquad$ Let $lambda in K$, then $lambda x=sum_{i=1}^{n}lambda, alpha_i e_i$ and



      begin{align}|lambda x|_0&= maxlimits_{1leq ileq n}|lambda,alpha_i|=|lambda,|maxlimits_{1leq ileq n}|alpha_i|=|lambda,|| x|_0,;forall;xin Eend{align}



      $c.qquad$ Let $x,yin E$, then $x=sum_{i=1}^{n}alpha_i e_i$ and $y=sum_{i=1}^{n}beta_i e_i,$ for some $alpha_i,beta_i,;iin{1,2,cdots,n}.$ Thus,



      begin{align}| x+y|_0&= maxlimits_{1leq ileq n}|alpha_i+beta_i|leq maxlimits_{1leq ileq n}|alpha_i|+maxlimits_{1leq ileq n}|beta_i|=| x|_0+|y|_0,;forall;x,yin Eend{align}






      share|cite|improve this answer
























        0












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        0






        $a.qquad$ Let $xin E,$ then begin{align}|x|_0=0&iff maxlimits_{1leq ileq n}|alpha_i|=0 iff |alpha_i|=0,,forall ,{1leq ileq n}\&iff alpha_i=0,,forall ,{1leq ileq n}\&iff x=0,;forall;xin Eend{align}
        $b.qquad$ Let $lambda in K$, then $lambda x=sum_{i=1}^{n}lambda, alpha_i e_i$ and



        begin{align}|lambda x|_0&= maxlimits_{1leq ileq n}|lambda,alpha_i|=|lambda,|maxlimits_{1leq ileq n}|alpha_i|=|lambda,|| x|_0,;forall;xin Eend{align}



        $c.qquad$ Let $x,yin E$, then $x=sum_{i=1}^{n}alpha_i e_i$ and $y=sum_{i=1}^{n}beta_i e_i,$ for some $alpha_i,beta_i,;iin{1,2,cdots,n}.$ Thus,



        begin{align}| x+y|_0&= maxlimits_{1leq ileq n}|alpha_i+beta_i|leq maxlimits_{1leq ileq n}|alpha_i|+maxlimits_{1leq ileq n}|beta_i|=| x|_0+|y|_0,;forall;x,yin Eend{align}






        share|cite|improve this answer












        $a.qquad$ Let $xin E,$ then begin{align}|x|_0=0&iff maxlimits_{1leq ileq n}|alpha_i|=0 iff |alpha_i|=0,,forall ,{1leq ileq n}\&iff alpha_i=0,,forall ,{1leq ileq n}\&iff x=0,;forall;xin Eend{align}
        $b.qquad$ Let $lambda in K$, then $lambda x=sum_{i=1}^{n}lambda, alpha_i e_i$ and



        begin{align}|lambda x|_0&= maxlimits_{1leq ileq n}|lambda,alpha_i|=|lambda,|maxlimits_{1leq ileq n}|alpha_i|=|lambda,|| x|_0,;forall;xin Eend{align}



        $c.qquad$ Let $x,yin E$, then $x=sum_{i=1}^{n}alpha_i e_i$ and $y=sum_{i=1}^{n}beta_i e_i,$ for some $alpha_i,beta_i,;iin{1,2,cdots,n}.$ Thus,



        begin{align}| x+y|_0&= maxlimits_{1leq ileq n}|alpha_i+beta_i|leq maxlimits_{1leq ileq n}|alpha_i|+maxlimits_{1leq ileq n}|beta_i|=| x|_0+|y|_0,;forall;x,yin Eend{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 at 14:18









        Mike

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        1,278220






























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