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Intuitively I understand why coordinate transformation should be reversible. New coordinates should cover the same area covered by the initial coordinates, i.e. there should be one-to-one mapping.

But still, are reversible transformations used only because it is convenient or is there any theoretical background?

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"used" for what purpose? –  Arturo Magidin Jul 3 '12 at 19:21
    
@Arturo Magidin, anywhere I've seen coordinate transformation is used it is almost at once it is assumed that the transformation is reversible. –  superM Jul 3 '12 at 19:23
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Ah, so you mean "used" for "coordinate transformations"? (It wasn't clear to me) The fact that you are dealing with two bases guarantees that the transformation must be reversible: you can express either basis in terms of the other basis. –  Arturo Magidin Jul 3 '12 at 19:25
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In other words, "coordinate transformations" are necessarily invertible; otherwise, they would not be "coordinate transformations". It's not a matter of convenience. –  Arturo Magidin Jul 3 '12 at 19:36
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@all When I read his question I reallly get the feeling the OP is thinking about coordinate maps on a manifold, but he swears he's happy so I guess I shouldn't complain! –  rschwieb Jul 3 '12 at 20:02
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up vote 2 down vote accepted

In a "coordinate transformation" (change-of-coordinates transformation), if you start with a basis $\beta=[\mathbf{v}_1,\ldots,\mathbf{v}_n]$, then because the transformation is one-to-one, the image of $\beta$ must be linearly independent, and hence a basis. Thus, the image of the transformation contains a basis, and so will necessarily be onto. Thus, by virtue of being one-to-one on a finite dimensional space, it must necessarily be onto as well, and thus will be invertible.

Or you can view a "change-of-coordinates" transformation as a way of expressing vectors in one basis, $\beta$, in terms of vectors in another basis, $\gamma$; if you then write out what it would mean to express the vectors in $\gamma$ in terms of $\beta$, and you compose the resulting two transformations, you get the elements of $\beta$ expressed in terms of $\beta$ (and composing the other way, the elements of $\gamma$ in terms of the vectors of $\gamma$). Because each vector can be expressed in a unique way in terms of the vectors of a basis, the only way to express the vectors of $\beta$ in terms of the vectors of $\beta$ is by the identity transformation: $$\begin{align*} \mathbf{v}_1 &= 1\mathbf{v}_1 + 0\mathbf{v}_2 + \cdots + 0\mathbf{v}_n\\ \mathbf{v}_2 &= 0\mathbf{v}_1 + 1\mathbf{v}_2 + \cdots + 0\mathbf{v}_n\\ &\vdots\\ \mathbf{v}_n &= 0\mathbf{v}_1 + 0\mathbf{v}_2 + \cdots + 1\mathbf{v}_n. \end{align*}$$ so that the composition is the identity. Composing them the other way also gives the identity, so that shows the original change-of-coordinates transformation must be invertible.

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