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Math Help - Linear Algebra- Basis

  1. #1
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    Linear Algebra- Basis

    Let V be a finite n-dim vector space

    v1,v2,....,vn
    w1,w2,...,wn

    be two set of basis

    Let v be a vector in V be represented in this basis as follows:
    v = a1.v1 + a2.v2 + .... + an.vn
    also
    v = a1.w1 + a2.w2 + ... + an.wn
    (note the coefficients a1,a2,...,an are same for both basis)

    Can we conclude that v1=w1; v2 = w2' ....vn = wn from this?

    Is yes - why?
    (obviously I assume v is not equal to 0)
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    Let V be a finite n-dim vector space

    v1,v2,....,vn
    w1,w2,...,wn

    be two set of basis

    Let v be a vector in V be represented in this basis as follows:
    v = a1.v1 + a2.v2 + .... + an.vn
    also
    v = a1.w1 + a2.w2 + ... + an.wn
    (note the coefficients a1,a2,...,an are same for both basis)

    Can we conclude that v1=w1; v2 = w2' ....vn = wn from this?

    Is yes - why?
    (obviously I assume v is not equal to 0)
    no! for example in V=\mathbb{R}^2 let \{e_1,e_2 \} be the standard basis. then \{e_1-e_2, 2e_2 \} is also a basis for V and e_1+e_2=e_1-e_2 + 2e_2.
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  3. #3
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    Thanks - a very neat counter example. But does my equation put some kind of constraint on the basis - I mean is there a special relation between v1,v2...vn and w1,w2,...,wn for such a thing to happen? I am kind of confused here
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  4. #4
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    Quote Originally Posted by aman_cc View Post
    Thanks - a very neat counter example. But does my equation put some kind of constraint on the basis - I mean is there a special relation between v1,v2...vn and w1,w2,...,wn for such a thing to happen? I am kind of confused here
    what i can say now is that we'll have this situation whenever we have an n \times n matrix C=[c_{ij}] with c_{ij} \in F (the base field) with these properties that \det C \neq 0 and \det (C - I) = 0.

    to see that let \{v_1, \cdots , v_n \} be any basis for V and define w_j=c_{1j}v_1 + \cdots + c_{nj}v_n, \ 1 \leq j \leq n. since \det C \neq 0, the set \{w_1, \cdots , w_n \} is a basis for V. also since \det(C-I)=0, there

    exists a vector \bold{0} \neq \bold{x}=\begin{bmatrix}a_1 & . & . & . & a_n \end{bmatrix}^T \in F^n such that C \bold{x}=\bold{x}, which is equivalent to a_1v_1 + \cdots + a_nv_n = a_1w_1 + \cdots + a_nw_n.
    Last edited by NonCommAlg; September 18th 2009 at 01:48 PM.
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