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

    Can someone point me in the right direction with this? Obviously I need to show that $\displaystyle \beta'$ is linearly independent and spans $\displaystyle V$. I know there are n vectors in $\displaystyle \beta'$ 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 $\displaystyle V$ be a finite-dimensional vector space over a field $\displaystyle F$, and let $\displaystyle \beta=\{x_1, ..., x_n\}$ be an ordered basis for $\displaystyle V$. Let $\displaystyle Q$ be an $\displaystyle n\times n$ invertible matrix with entries from $\displaystyle F$. Define $\displaystyle x_j'=\displaystyle\sum_{i=1}^n Q_{ij}x_i$ for $\displaystyle 1 \le j \le n$, and set $\displaystyle \beta'=\{x_1', ..., x_n'\}$. Prove that $\displaystyle \beta'$ is a basis for $\displaystyle V$ and hence that $\displaystyle Q$ is the change of coordinate matrix changing $\displaystyle \beta'$-coordinates into $\displaystyle \beta$-coordinates."

    Can someone point me in the right direction with this? Obviously I need to show that $\displaystyle \beta'$ is linearly independent and spans $\displaystyle V$. I know there are n vectors in $\displaystyle \beta'$ 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 $\displaystyle Q$ is invertible and thus $\displaystyle Q(\beta)$ is a basis, isn't $\displaystyle Q(\beta)=\beta'$?
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  3. #3
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    Re: Change of coordinate matrix, new basis

    How are you defining $\displaystyle Q(\beta)$?
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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 $\displaystyle Q(\beta)$?
    The image of the basis, i.e. if we take the matrix of the transformation $\displaystyle Q$ with respect to $\displaystyle Q$ then isn't $\displaystyle x'_j=Q(x_j)$?
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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 $\displaystyle Q(\beta)$?
    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, $\displaystyle Q(\beta) = \{Q(x_1),Q(x_2),\dots,Q(x_n)\}$. 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 $\displaystyle c_1x'_1 +\dots +c_nx'_n = 0$. then

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

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

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

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

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