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Math Help - Linear independence

  1. #1
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    Linear independence

    In a vector space (F(\mathbb{R},\mathbb{R});+; .), we consider the next functions defined as :

    \forall n \in \mathbb{N}\  e_n:x \to |x+n|,

    we must show that \forall n \in \mathbb{N} the indexed family (e_0;e_1;...;e_n) is linear independant.

    How to show it???
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  2. #2
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    Quote Originally Posted by lehder View Post
    In a vector space (F(\mathbb{R},\mathbb{R});+; .), we consider the next functions defined as :

    \forall n \in \mathbb{N}\ e_n:x \to |x+n|,

    we must show that \forall n \in \mathbb{N} the indexed family (e_0;e_1;...;e_n) is linear independant.

    How to show it???

    Ok, so it seems to be the vector space you're talking about is the space of all real valued real functions, and you want to show \{e_0,\ldots,e_n\} is a l.i. set for all n.

    So assume \sum^n_{k=0}a_ke_k=0\,,\,\,a_i\in\mathbb{R} \Longrightarrow \,\,\forall\,x\in\mathbb{R}\,,\,0=\sum^n_{k=0}a_ke  _k(x) =\sum^n_{k=0}a_k|x+k|\Longrightarrow take now succesively the values x=0,-1,-2,\ldots,-n and you'll get the equations:

    0=\sum^n_{k=0}a_k|0+k|=\sum^n_{k=1}a_kk =a_1+2a_2+3a_3+\ldots+na_n

    0=\sum^n_{k=0}a_k|-1+k|=\sum^n_{k=0\,k\neq 1}a_k|-1+k| =a_0+a_2+2a_3+\ldots+(n-1)a_n

    0=\sum^n_{k=0}a_k|-2+k|=2a_0+a_1+a_3+2a_4+\ldots+(n-2)a_n

    -------------------------------------------------------------

    0=\sum^n_{k=0}a_k|-n+k|=na_0+(n-1)a_1+\ldots+2a_{n-2}+a_{n-1}


    Now the trick that you'll generalize: substract the 2nd eq. from the 1st one and you get

    -a_0+a_1+\ldots+a_n=0 , and now substract the 3rd eq. from the 1st one and you get -2a_0+2a_2+2a_3+\ldots+2a_n=2(-a_0+a_2+a_3+\ldots+a_n)=0\Longrightarrow

     -a_0+a_2+a_3+\ldots+a_n=0.

    Compare these two equations we just got (ok, substract them!) and get that a_1=0 ...now you continue.

    Tonio
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  3. #3
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    Smile

    i might have another approach,
    for  n=1 the system \left \{ e_1 \right \} is independent since the function e_1 isn't zero.
    now suppose that \left \{ e_0,e_1,...,e_ n\right \} is independent and try to prove that \left \{ e_0,e_1,...,e_n,e_{n+1} \right \} is also independent.
    for that reason we must prove that e_{n+1}\notin \texttt{Span}(e_0,e_1,...,e_n).
    you can see that each element of e_n is differentiable at x_0=-n-1,since it's a linear combination of differentiable functions at x_0.
    but \lim_{x\to x_0^{+}}\frac{e_{n+1}(x)-e_{n+1}(x_0)}{x-x_0}=1 and \lim_{x\to x_0^{-}}\frac{e_{n+1}(x)-e_{n+1}(x_0)}{x-x_0}=-1 which means e_{n+1} isn't differentiable at x_0=-n-1 and it follows that e_{n+1}\notin \texttt{Span}(e_0,e_1,...,e_n).
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  4. #4
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    independence versus linear independence

    Hi, this thread is related to independence but unrelated to the initial thread. I hope this will bother no one.

    I have had a lively discussion with one of my comrades about linear independence. I stated that independence is the same as linear independence in the context of linear algebra. That is, a set of vectors is independent (in the set theoretical sense) iff it is linearly independent. I still find this very confusing and would greatly appreciate if someone could clarify the matter.

    Thanks in advance.
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  5. #5
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    Quote Originally Posted by arta View Post
    Hi, this thread is related to independence but unrelated to the initial thread. I hope this will bother no one.


    Well, it does. First, you're kidnapping another person's thread for your own question, but perhaps more important: your own question can get lost into the thread.
    Please do start a new one...oh, and by the way: what does it mean "independence" in the "set theoretical sense"? Whose independence, to begin with?

    Tonio


    I have had a lively discussion with one of my comrades about linear independence. I stated that independence is the same as linear independence in the context of linear algebra. That is, a set of vectors is independent (in the set theoretical sense) iff it is linearly independent. I still find this very confusing and would greatly appreciate if someone could clarify the matter.

    Thanks in advance.
    .
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  6. #6
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    Sorry then.
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