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Math Help - Vector

  1. #16
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    Quote Originally Posted by Gamma View Post
    Are you talking about the other question on the other Thread? I am really confused as to what you are talking about? Again I cannot stress enough, there is no "THE basis" there is no preferred basis, it is just a basis.

    If you need to find a basis for \mathbb{R}^n,just pick the standard basis for \mathbb{R}^n.
    {(1,0,...,0), (0,1,0,...,0),... (0,0,...,1)}
    I was talking about this post. What i wanted to understand is how can i derive a standard basis for this, how can i find the rank of every row how can i know which rows have pivot?
    Thanks
    KC
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  2. #17
    Super Member Gamma's Avatar
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    Definition

    Man, I am really lost with what you are asking now. I think you need to review some of your definitions, it is getting to the point that these posts are becoming incoherent to me.

    You do not find a basis for a matrix.

    A matrix represents a linear transformation, a basis is for a vector space, not a function. Matrices have ranks, it does not make sense to talk about the rank of a row.

    The dual space to \mathbb{R}^n does not have a "standard" basis. It is not canonically isomorphic to \mathbb{R}^n. It is isomorphic, but only in terms of a basis. If you give me a basis for \mathbb{R}^n, I will give you a basis for (\mathbb{R}^n)^*.

    I already wrote out the basis for the dual space in a previous post on here titled "bassis(I can't type apparently) for the Dual Space".

    Look at that. If you want the basis for the dual space for \mathbb{R}^n subordinant to the standard basis, then e_1=(1,0,0,...,0)
    e_2=(0,1,0,...,0)
    .
    .
    .
    e_n=(0,0,0,...,1)

    and this yields the basis for (\mathbb{R}^n)^*
    \epsilon_1(a_1e_1+a_2e_2+...+a_ne_n)=a_1
    \epsilon_2(a_1e_1+a_2e_2+...+a_ne_n)=a_2
    .
    .
    .
    \epsilon_n(a_1e_1+a_2e_2+...+a_ne_n)=a_n
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  3. #18
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    Quote Originally Posted by Gamma View Post
    This is the point, it depends on the basis, there are an infinite number of possible choices for basis of \mathbb{R}^n each of these yields a different basis for (\mathbb{R}^n)^*.

    To give you the idea though for any given basis for \mathbb{R}^n call it \{e_1,e_2,...,e_n\}, you get n linear functionals in (\mathbb{R}^n)^* that will form the basis, say they are \{\epsilon^1,\epsilon^2,...,\epsilon^n \}.

    Let the n functionals be defined as follows.

    \epsilon^i(a_1e_1 +a_2e_2 + ... + a_ie_i + ... +a_ne_n)=a_i

    Yeah, there is your basis. If you took the standard basis for \mathbb{R}^n the ith coordinate function would simply pick out the ith coefficient.

    could you please explain to me in a little more detail on how i can find the basis for this.
    thanks
    Last edited by mr fantastic; May 7th 2009 at 11:58 PM. Reason: Restored post deleted by OP
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