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Math Help - Identity map between bases

  1. #1
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    Identity map between bases

    My book (Tensor Geometry - Poston & Dodson) says the following:

    If  \beta = (b_1,..., b_n) is a basis for X, and A : X \rightarrow Y is an isomorphism, then  A\beta = (Ab_1,..., Ab_n) is a basis for Y.
    If \beta is a basis for X and A : X \rightarrow X is an isomorphism, the change of basis matrix [I]_\beta^{A\beta} is exactly the matrix ([A]_\beta^\beta)^{\leftarrow}.
    I just can't seem to agree with this result!

    After hours of tearing my hair out I have come up with the following argument...Please point out where I've gone wrong...

    For some basis \beta, some vector \mathbf{x} and its representation x^\beta in the \beta coords.

    \beta x^\beta=\mathbf{x}
    \Rightarrow x^\beta=\beta^{\leftarrow}\mathbf{x}
    \Rightarrow [I]_\beta^{\beta'} x^\beta=[I]_\beta^{\beta'} \beta^{\leftarrow}\mathbf{x}

    for some other basis \beta' where [I]_\beta^{\beta'} is the change of basis matrix from \beta to \beta' coordinates.
    so
     [I]_\beta^{\beta'} x^\beta=[I]_\beta^{\beta'} \beta^{\leftarrow}\mathbf{x}= x^{\beta'} --(*)


    We also know the coordinates of \mathbf{x} in the \beta' coords using the \beta' basis:

    \beta' x^{\beta'}=\mathbf{x}
    \Rightarrow x^{\beta'}=\beta'^{\leftarrow}\mathbf{x} --(**)

    (*) and (**) combine to give

    [I]_\beta^{\beta'} \beta^{\leftarrow}\mathbf{x}=\beta'^{\leftarrow}\m  athbf{x}
    \Rightarrow [I]_\beta^{\beta'} \beta^{\leftarrow}=\beta'^{\leftarrow}
    \Rightarrow [I]_\beta^{\beta'}=\beta'^{\leftarrow}\beta

    This seems like a nice neat result to me, but if \beta'=A\beta as it is in the book, we have

     [I]_\beta^{\beta'}=\beta'^{\leftarrow}\beta
    \Rightarrow [I]_\beta^{A \beta}=(A\beta)^{\leftarrow}\beta
    \Rightarrow [I]_\beta^{A \beta}=\beta^{\leftarrow}A^{\leftarrow}\beta
    \not=A^{\leftarrow}

    However, if \beta'=\beta A
     [I]_\beta^{\beta A}=\beta'^{\leftarrow}\beta
    \Rightarrow [I]_\beta^{\beta A}=(\beta A)^{\leftarrow}\beta
    \Rightarrow [I]_\beta^{\beta A}=A^{\leftarrow}\beta^{\leftarrow}\beta
    =A^{\leftarrow}

    which is the required result.....

    I have tried some basic examples with actual numbers and the results support what I have here... Unless I have some fundamental misunderstanding of all this and what it is supposed to mean, which is quite possible...
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Mmmm View Post
    My book (Tensor Geometry - Poston & Dodson) says the following:



    I just can't seem to agree with this result!

    After hours of tearing my hair out I have come up with the following argument...Please point out where I've gone wrong...

    For some basis \beta, some vector \mathbf{x} and its representation x^\beta in the \beta coords.

    \beta x^\beta=\mathbf{x}
    \Rightarrow x^\beta=\beta^{\leftarrow}\mathbf{x}
    \Rightarrow [I]_\beta^{\beta'} x^\beta=[I]_\beta^{\beta'} \beta^{\leftarrow}\mathbf{x}

    for some other basis \beta' where [I]_\beta^{\beta'} is the change of basis matrix from \beta to \beta' coordinates.
    so
     [I]_\beta^{\beta'} x^\beta=[I]_\beta^{\beta'} \beta^{\leftarrow}\mathbf{x}= x^{\beta'} --(*)


    We also know the coordinates of \mathbf{x} in the \beta' coords using the \beta' basis:

    \beta' x^{\beta'}=\mathbf{x}
    \Rightarrow x^{\beta'}=\beta'^{\leftarrow}\mathbf{x} --(**)

    (*) and (**) combine to give

    [I]_\beta^{\beta'} \beta^{\leftarrow}\mathbf{x}=\beta'^{\leftarrow}\m  athbf{x}
    \Rightarrow [I]_\beta^{\beta'} \beta^{\leftarrow}=\beta'^{\leftarrow}
    \Rightarrow [I]_\beta^{\beta'}=\beta'^{\leftarrow}\beta

    This seems like a nice neat result to me, but if \beta'=A\beta as it is in the book, we have

     [I]_\beta^{\beta'}=\beta'^{\leftarrow}\beta
    \Rightarrow [I]_\beta^{A \beta}=(A\beta)^{\leftarrow}\beta
    \Rightarrow [I]_\beta^{A \beta}=\beta^{\leftarrow}A^{\leftarrow}\beta
    \not=A^{\leftarrow}

    However, if \beta'=\beta A
     [I]_\beta^{\beta A}=\beta'^{\leftarrow}\beta
    \Rightarrow [I]_\beta^{\beta A}=(\beta A)^{\leftarrow}\beta
    \Rightarrow [I]_\beta^{\beta A}=A^{\leftarrow}\beta^{\leftarrow}\beta
    =A^{\leftarrow}

    which is the required result.....

    I have tried some basic examples with actual numbers and the results support what I have here... Unless I have some fundamental misunderstanding of all this and what it is supposed to mean, which is quite possible...
    I hate the notation tensor analysts use. What does the arrow in A^{\leftarrow} mean?
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  3. #3
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    It is just the inverse of A, ie A^{-1}
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  4. #4
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    An Example

    Just to make things a little simpler I'll give an example and hopefully
    somebody will be able to tell me what I'm doing wrong.

    Given a basis, \beta for X

    \[<br />
\beta=\left(\begin{array}{cc}<br />
1 & \frac{1}{2}\\<br />
1 & 1\end{array}\right)\text{ so that }\beta^{-1}=\left(\begin{array}{rr}<br />
2 & -1\\<br />
-2 & 2\end{array}\right)\]<br />

    and a linear isomorphic map A:X\rightarrow X

    \[<br />
A=\left(\begin{array}{rr}<br />
1 & 2\\<br />
1 & 1\end{array}\right)\text{ so that }A^{-1}=\left(\begin{array}{rr}<br />
-1 & 2\\<br />
1 & -1\end{array}\right)\]


    Case (1):

    Given a new basis \beta'=A\beta

    \[<br />
\beta'=\left(\begin{array}{rr}<br />
1 & 2\\<br />
1 & 1\end{array}\right)\left(\begin{array}{rr}<br />
1 & \frac{1}{2}\\<br />
1 & 1\end{array}\right)=\left(\begin{array}{rr}<br />
3 & \frac{5}{2}\\<br />
2 & \frac{3}{2}\end{array}\right)\]


    and a vector $\mathbf{x}=\left(\begin{array}{c}<br />
3\\<br />
2\end{array}\right)$ chosen so that we know that $x^{\beta'}=\left(\begin{array}{c}<br />
1\\<br />
0\end{array}\right)$

    Now, according to the book, $A^{-1}x^{\beta}=x^{\beta'}$, so I'll try it and see if it works

    First I need to find $x^{\beta}$

    \[<br />
x^{\beta}=\beta^{-1}\mathbf{x}=\left(\begin{array}{rr}<br />
2 & -1\\<br />
-2 & 2\end{array}\right)\left(\begin{array}{c}<br />
3\\<br />
2\end{array}\right)=\left(\begin{array}{r}<br />
4\\<br />
-2\end{array}\right)\]


    now,

    \[<br />
A^{-1}x^{\beta}=\left(\begin{array}{rr}<br />
-1 & 2\\<br />
1 & -1\end{array}\right)\left(\begin{array}{r}<br />
4\\<br />
-2\end{array}\right)=\left(\begin{array}{r}<br />
-8\\<br />
6\end{array}\right)\neq\left(\begin{array}{r}<br />
1\\<br />
0\end{array}\right)\]


    so it doesn't work!



    Case (2):

    Do the same but with $\beta'=\beta A$

    \[<br />
\beta'=\left(\begin{array}{rr}<br />
1 & \frac{1}{2}\\<br />
1 & 1\end{array}\right)\left(\begin{array}{rr}<br />
1 & 2\\<br />
1 & 1\end{array}\right)=\left(\begin{array}{rr}<br />
\frac{3}{2} & \frac{5}{2}\\<br />
2 & 3\end{array}\right)\]


    and a vector $\mathbf{x}=\left(\begin{array}{c}<br />
\frac{3}{2}\\<br />
2\end{array}\right)$ chosen so that we know that $x^{\beta'}=\left(\begin{array}{c}<br />
1\\<br />
0\end{array}\right)$

    Finding $x^{\beta}$

    \[<br />
x^{\beta}=\beta^{-1}\mathbf{x}=\left(\begin{array}{rr}<br />
2 & -1\\<br />
-2 & 2\end{array}\right)\left(\begin{array}{c}<br />
\frac{3}{2}\\<br />
2\end{array}\right)=\left(\begin{array}{r}<br />
1\\<br />
1\end{array}\right)\]


    now,

    \[<br />
A^{-1}x^{\beta}=\left(\begin{array}{rr}<br />
-1 & 2\\<br />
1 & -1\end{array}\right)\left(\begin{array}{r}<br />
1\\<br />
1\end{array}\right)=\left(\begin{array}{r}<br />
1\\<br />
0\end{array}\right)\]


    It seems to work for $\beta'=\beta A$ but not $\beta'=A\beta$.
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  5. #5
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    I've figured it out...
    So just for the sake of completeness and if anyone is interested I'll post the conclusion to this problem.

    My mistake was in thinking that A=\left[A\right]_{\beta}^{\beta}.

    The map A:X\rightarrow X maps a vector in the vector space X to a new vector in X.

    Wheras \left[A\right]_{\beta}^{\beta} maps the components of a vector in the \beta basis to new components in the \beta basis.

    The result is the same but the maps are different.

    You can do the map \left[A\right]_{\beta}^{\beta} in terms of A by first converting components into a vector ( \beta(x^{\beta})), then applying A ( A(\beta x^{\beta})) and then converting back into components ( \beta^{-1}(A\beta x^{\beta})).

    ie

    \left[A\right]_{\beta}^{\beta}=\beta^{-1}A\beta

    My result in my first post was \left[I\right]_{\beta}^{A\beta}=\beta^{-1}A^{-1}\beta which completely confused me (I was expecting \left[I\right]_{\beta}^{A\beta}=A^{-1})

    But this was quite correct and can be taken further:

    \begin{aligned} \left[I\right]_{\beta}^{A\beta} & =\beta^{-1}A^{-1}\beta\\ & =(A\beta)^{-1}\beta\\ & =(\beta^{-1}A\beta)^{-1}\\ & =(\left[A\right]_{\beta}^{\beta})^{-1}\end{aligned}
    Which is the required result.

    So there you go...
    Mystery solved!
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