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Math Help - Change of coordinate matrix, new basis

  1. #1
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    Change of coordinate matrix, new basis

    I've been working on the following problem:

    "Let V be a finite-dimensional vector space over a field F, and let \beta=\{x_1, ..., x_n\} be an ordered basis for V. Let Q be an n\times n invertible matrix with entries from F. Define x_j'=\displaystyle\sum_{i=1}^n Q_{ij}x_i for 1 \le j \le n, and set \beta'=\{x_1', ..., x_n'\}. Prove that \beta' is a basis for V and hence that Q is the change of coordinate matrix changing \beta'-coordinates into \beta-coordinates."

    Can someone point me in the right direction with this? Obviously I need to show that  \beta' is linearly independent and spans V. I know there are n vectors in \beta' so it only remains to show that its elements are linearly independent, correct? I can't seem to figure it out though.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Change of coordinate matrix, new basis

    Quote Originally Posted by AlexP View Post
    I've been working on the following problem:

    "Let V be a finite-dimensional vector space over a field F, and let \beta=\{x_1, ..., x_n\} be an ordered basis for V. Let Q be an n\times n invertible matrix with entries from F. Define x_j'=\displaystyle\sum_{i=1}^n Q_{ij}x_i for 1 \le j \le n, and set \beta'=\{x_1', ..., x_n'\}. Prove that \beta' is a basis for V and hence that Q is the change of coordinate matrix changing \beta'-coordinates into \beta-coordinates."

    Can someone point me in the right direction with this? Obviously I need to show that  \beta' is linearly independent and spans V. I know there are n vectors in \beta' so it only remains to show that its elements are linearly independent, correct? I can't seem to figure it out though.
    Well, you know that Q is invertible and thus Q(\beta) is a basis, isn't Q(\beta)=\beta'?
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  3. #3
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    Re: Change of coordinate matrix, new basis

    How are you defining Q(\beta)?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Change of coordinate matrix, new basis

    Quote Originally Posted by AlexP View Post
    How are you defining Q(\beta)?
    The image of the basis, i.e. if we take the matrix of the transformation Q with respect to Q then isn't x'_j=Q(x_j)?
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  5. #5
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    Re: Change of coordinate matrix, new basis

    Quote Originally Posted by AlexP View Post
    How are you defining Q(\beta)?
    in general, if A is a set, and T is a linear transformation (which every matrix is), then T(A) is the set of all T(a) where a is an element of A.

    so in this case, Q(\beta) = \{Q(x_1),Q(x_2),\dots,Q(x_n)\}. the trick is to show all of these elements are distinct (which you

    should be able to do just from the fact that Q is invertible, and hence 1-1). do these n vectors span V? well, they have to, since Q

    is invertible, and thus onto (so we can write any vector w in V as Q(v) for some other vector v in V, and we can write v as a linear

    combination of the xj, and thus write w = Q(v) as the same linear combination of the Q(xj), since Q is linear).

    but, if Q(β) spans V, and has the same cardinality as a basis (this is where the distinctness comes in), it has to be a basis, too,

    because it is a minimal spanning set.

    **********************

    you can also show the linear independence directly:

    suppose c_1x'_1 +\dots +c_nx'_n = 0. then

    c_1x'_1 +\dots +c_nx'_n = c_1Q(x_1)+\dots +c_nQ(x_n)=

    Q(c_1x_1)+\dots +Q(c_nx_n) = Q(c_1x_1+\dots +c_nx_n) = 0.

    since Q is 1-1, c_1x_1+\dots +c_nx_n = 0, which means by the

    linear independence of the xj that c_1 = \dots = c_n = 0.

    this is a recurring theme in linear algebra:

    injective/1-1 ~~ linear independence
    surjective/onto ~~ spanning
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  6. #6
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    Re: Change of coordinate matrix, new basis

    I completely understand how the invertibility of Q results in a new basis by taking Q(\beta), but I didn't think that Q was acting as left multiplication on x_i. I feel this is somewhat dumb, but am I really interpreting x'_j=\displaystyle\sum_{i=1}^nQ_{ij}x_i for 1 \le j \le n incorrectly? I don't see how it amounts to left multiplication by Q on an element of \beta. This must be where my real confusion lies.
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