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Homework Help: Linear algebra

  1. Jun 7, 2006 #1
    Let b = {x_1,...,x_n} be a basis for a vector space V over a field F.
    let x'_j = Sum i=1 to n[Bij x_j] where Bij are the entries of any matrix (n x n). Prove that b'={x'_1,...,x'_n} is a basis for V and therefore B is the change of coordinate matrix.

    Ok, the final connection between the proof and the coordinate matrix is obvious (that's how its defined). But im not sure how to prove that b' is a basis. I know that
    Sum j=1 to n[a_j x'_j]
    = Sum j=1 to n[a_j Sum i=1 to n[Bij x_i]]
    = Sum j=1 to n[Sum i=1 to n[a_j Bij] x_i]

    So you can consider the Sum i=1 to n[a_j Bij] as the coefficient, call it c_i such that each element of the vector space v = Sum i=1 to n[c_i x_i] for unique scalars c_1,...,c_n.
    But this would mean b' is a basis only if for each combination c_1,...,c_n there exists a a_1,...,a_n that generates this c_1,...,c_n. Im having trouble showing this.
    I know theres another way to look at it. For example, what i just said is equivalent to having:
    c_1 = Sum j=1 to n[a_j B1j]
    c_2 = Sum j=1 to n[a_j B2j]
    c_n = Sum j=1 to n[a_j Bnj]

    so if i can prove that given these n equations and n unknows, there exists for every combination, c_1,...c_n a solution for j_1,...,j_n then i would be done.
    But the book im reading still didnt get to the chapter about solving equations so there must be a more elementary way of showing this.
    So ya, if anyone could help that would be cool.

    edit: the matrix is invertible
    Last edited: Jun 7, 2006
  2. jcsd
  3. Jun 7, 2006 #2
    maybe this is more suited as a homework question.
  4. Jun 8, 2006 #3


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    Homework Helper

    Any matrix (n x n)? Then this is wrong. Take B = 0. Then x'j = 0 for all j, and b' = {0} is not a basis for V. You can prove that:

    b' is a basis if and only if B is a change of coordinate matrix

    In other words, you can prove that:

    1) if b' is a basis, then B is a change of coordinate matrix, and 2) if B is a change of coordinate matrix, then b' is a basis

    Do you want to prove, 1), 2), or both? Also, how does your book define "change of coordinate matrix"? Finally, do you know that since {x1, ..., xn} is a basis of V, then

    a) any set of n vectors which span V are a basis of V
    b) any set of n linearly independent vectors are a basis of V
  5. Jun 8, 2006 #4


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    More generally, if B is an invertible matrix, then B is a "change of coordinate matrix".
  6. Jun 8, 2006 #5
    Ya i wrote under "edit" that B is invertible. But how do I prove, given that definition for x'_j that {x'_1,...,x'_n} is a basis for V. Once I know that then the change of coordinate matrix statement immediately follows.
  7. Jun 8, 2006 #6
    Actually, i think i have the answer. Tell me if this is right.
    Remember from my first post I said that if i can choose for every combination (c_1,...,c_n) a combination (a_1,...,a_n) that would generate that (c_1,...,c_n) then I would be done. Well i can. Knowing that B is invertible, this means there exists a B^-1 s.t B B^-1 = I. Then this means, for each i there exists a (B B^-1)ii = Sum j=1 to n[Bij B^-1ji] = 1. So for each c_i, choose a_j = B^-1ji * c_i and were done. Right?
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