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Linear space shilov

  1. Jul 9, 2011 #1
    Hi guys i just had a quick question re shilov linear algebra page 44 here he shows that for an 'n' dimentional linear(vector) space K, a subspace, say L, must have dimention no greater than 'n'.
    He then goed onto say that if a basis is chosen in the subspace L, say f1,f2,...,fr, then additional vectors can be chosen from the linear space K, say f(r+1),...,fn such that f1,f2,...,fr,...,fn form a basis for K. Which makes perfect sense. But here is the part that throws me:
    Suppose we have the relation:
    c1f1+c2f2+...+crfr+c(r+1)f(r+1)=0
    Where the c terms are constants and f(r+1) is an additional vector from K added to the basis of L.
    He says if c(r+1) does not equal zero then then L is K. Which is true. But then he says if c(r+1)=0 then the vectors f1,f2,...,fr are linearly dependent, which would be a contradiction. How can this be true? If this is the basis for L and we set it equal to zero then we have all the constants equal to zero, hence they will be linearly independent. I checked the errata and there was no mention of this.
    Any help will be appreciated, thanks guys.
     
  2. jcsd
  3. Jul 9, 2011 #2
    Read this line again, f(r+1) is not included in the basis for L:
     
  4. Jul 9, 2011 #3
    Yes but he say if you set the f(r+1)'s constant equal to zero then the remaining vectors would be linearly dependent? If you set this constant(c(r+1)) equal to zero wouldnt the f(r+1) dissapear as its coefficient is now zero and you are just left with the basis for L set equal to zero? Maybe im missing something but thats the way i see it...
     
  5. Jul 9, 2011 #4
    It is a bit hard to read your description, some details are missing, maybe I assumed some things that weren't. I don't see how the first statement is true with this information you gave, just pick a vector from L in K so there is something missing. Or I am just dumb and missing something you wrote.
     
  6. Jul 9, 2011 #5
    I think it may just be an error, the book was translated from russian so maybe thats an issue, he usually doesnt give much detail, but here he is trying to construct the basis for a vector space K from the basis of the subspace L by adding vectors that are not spanned by the basis of L. And he uses the reasoning above as a step in his proof, but no matter how i think of it it doesnt make sense.
     
  7. Jul 9, 2011 #6

    micromass

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    Is he trying to show that [itex]\{f_1,...,f_{n+1}\}[/itex] is linear independent?

    In that case he supposes that that

    [tex]c_1f_1+...+c_nf_n+c_{n+1}f_{n+1}=0[/tex]

    and that not all ck are 0. The aim is to get a contradiction.

    and suppose that cn+1=0, then

    [tex]c_1f_1+...+c_nf_n=0[/tex]

    Since the [itex]\{f_1,...,f_n\}[/itex] forms a basis, we see that [itex]c_1=...=c_n=0[/itex], so all the ck are 0, which is a contradiction.
     
  8. Jul 9, 2011 #7
    Ah yes that would make sense! Your a genius! He didnt mention explicitly that we should assume that not all the coefficients are zero, but its implict in there since he draws a contradiction, and your explaination is the only way this makes sense! Thanks alot! I cant believe you figured that out with so little information! Cheers.
     
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