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Math Help - Determinant of linear transform with ordered basis is independent of the choice

  1. #1
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    Determinant of linear transform with ordered basis is independent of the choice

    Given a basis \beta for V define \det(T) = \det([T]_\beta). Prove that the definition is independent of choice of an ordered basis for V (i.e. if \beta and \gamma are ordered basis for V then \det([T]_\beta) =  \det([T]_\gamma)).

    I don't have much in the form of a proof. I tried working out an example but it got me nowhere. So what I have as an example is:

    let T(a,b) = (a+b,a-2b) and \beta=\{(1,0), (0,1)\} \ \gamma = \{(1,2),(1,-1)\} then under the transformation we have

    [T]_\beta = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} where \det[T]_\beta = -3

    and for

    [T]_\gamma= \begin{pmatrix} 0 & 3 \\ 1 & -1 \end{pmatrix} where \det[T]_\gamma= -3

    where [T]_\gamma= \begin{pmatrix} 0 & 3 \\ 1 & -1 \end{pmatrix} was obtained by [T]_\gamma^\beta= \begin{pmatrix} 3 & -3 \\ 0 & 3 \end{pmatrix} and [\hat{T}] = \frac{1}{3} (a+b,2a-b) which gives the desired result.

    Now I'm thinking it has something with the eigenvalues and corresponding eigenvectors but that got nowhere, since I'm not to sure on how to proceeds.
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  2. #2
    Senior Member roninpro's Avatar
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    Actually, it is possible to show that if you have two matrix representations of a linear transformation, then they are similar. That is, if A and B are both representations of T, then A=SBS^{-1} for some invertible matrix S. Then, \det(B)=\det(SAS^{-1})=\det(S)\det(A)\det(S^{-1})=\det(A). This proves invariance of the determinant with respect to a basis.
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