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Shilov's Linear Algebra Subspace Question

  1. Mar 7, 2014 #1
    Hi,

    I'm reading Shilov's linear algebra and in part 2.44 he talks about linear independent vectors in a subspace L which is a subset of space K( he refers to it as K over L). I don't understand why he says that a linear combination of vectors of the subspace L and vectors of the subspace K over L is independent. Is it the same subspace or am I wrong? Shilov also says that the dimension of the subspace K over L n-l. Why?

    LINK to the page:

    http://books.google.com.mx/books?id=5U6loPxlvQkC&pg=PA44&lpg=PA44&dq=2.44+shilov+linear+algebra&source=bl&ots=bcYpdoyvx7&sig=3daYGhPuQKVDbO2nc1TGrXb75tA&hl=es&sa=X&ei=P1MaU9nrOYPj2AWKjYHQCw&ved=0CEkQ6AEwAw#v=onepage&q=2.44%20shilov%20linear%20algebra&f=false
     
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  3. Mar 7, 2014 #2

    micromass

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    He proves it in the text. He takes a linear independent set ##\{f_1,...,f_l\}\subseteq L## and a set ##\{g_1,...,g_k\}\subseteq K## such that they are linear indepent over L. Then Shilov explicitely shows that

    [tex]\alpha_1 f_1+...+\alpha_lf_l +\beta_1 g_1 +.... + \beta_k g_k = 0[/tex]

    implies that ##\alpha_1 = ... = \alpha_l = \beta_1 = ...=\beta_k = 0## which implies linear independence in K.

    He explicitely proves this in the book. What about the proof is unclear?
     
  4. Mar 7, 2014 #3
    So both f and g are bases of the subspace?
     
  5. Mar 7, 2014 #4

    micromass

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    The set ##\{f_1,...,f_l\}## is a linear independent set of ##L##. And ##L## is assumed to have dimension ##l##, so ##\{f_1,...,f_l\}## is indeed a basis.

    The ##\{g_1,...,g_k\}## do not (in general) form a basis of anything.

    However, we do know that if ##k = n-l## (where ##K## has been assumed to have dimension ##n##), then ##\{f_1,...,f_l,g_1,...,g_k\}## is a basis of ##K##.
     
  6. Mar 7, 2014 #5
    I still don't get it :(. The set of g's are just random vectors, or what? Because he says that vectors g_1....g_k are part of L.
     
  7. Mar 7, 2014 #6

    micromass

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    He never says that the ##g_1,...,g_k## are elements of ##L##.

    The ##g_1,...,g_k## are just random elements of ##K##. And you only know that they are linearly independent over ##L##.
     
  8. Mar 7, 2014 #7
    Ok... What does he mean then with:

    α1g1+...+αkgk belongs to L
     
  9. Mar 7, 2014 #8

    micromass

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    He means that if ##\alpha_1 g_1 + ... + \alpha_kg_k\in L##, then ##\alpha_1 = ...=\alpha_k = 0##.

    So the only linear combination of the ##\{g_1,...,g_k\}## belonging to ##L## is the zero vector. And it's not because some linear combination belongs to ##L##, that the ##g_j## separately belong to ##L##.
     
  10. Mar 8, 2014 #9
    Do you have an example? Because I still can't grasp the idea. Thank you
     
  11. Mar 8, 2014 #10

    micromass

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    Take ##K = \mathbb{R}^2## and ##L## the ##x##-axis, thus ##L = \mathrm{span}\{(1,0)\}##.

    Then ##(1,1)## and ##(1,-1)## are not independent over ##L## because

    [tex]1\cdot (1,1) + 1\cdot (1,-1) = (2,0)\in L[/tex]

    but the coefficients are not all ##0##.

    Consider however ##K=\mathbb{R}^3## and ##L = \mathrm\{(1,0,0)\}## the ##x##-axis again. Then ##(0,1,0)## and ##(0,0,1)## are independent over ##L##. Indeed, take any linear combination

    [tex]x:=\alpha(0,1,0) + \beta(0,0,1) = (0,\alpha,\beta)[/tex]

    if this were in ##L##, then there would exist some scalar ##c## such that ##x = c(1,0,0)##. But then

    [tex](c,0,0) = (0,\alpha,\beta)[/tex]

    which is clearly only possible if ##c=\alpha=\beta=0##.
     
  12. Mar 8, 2014 #11
    Ok now I get it. Thanks a lot for taking the time to answer my questions and having the patience.
     
  13. Mar 8, 2014 #12

    micromass

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    I think that these things will make more sense to you once you get to quotient spaces. Right now, I think the concept is pulled a bit out of the air.
     
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