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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?2012-07-03
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    @Arturo Magidin, anywhere I've seen coordinate transformation is used it is almost at once it is assumed that the transformation is reversible.2012-07-03
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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.2012-07-03
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    @Arturo Magidin, Do the two bases guarantee that the transformation will be reversible? I thought everything depends on the mapping function.2012-07-03
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    If you map each element in the basis $A$ *uniquely* into an element in the basis $B,$ then it's a bijection between $A$ and $B.$ Also, two bases for the same space will have same dimension. Everything else follows by linearity & definition.2012-07-03
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    @superM: I'm assuming you are talking about vector spaces (given the tag [linear-algebra]. A "change of coordinates" amounts to saying how to translate expressions using one basis to expressions using the other basis. Look at the image of a basis in your original coordinate system. This will be linearly independent (because your transformation is one-to-one), and by dimensionality considerations (assuming finite dimension) it must be a basis. So the map is also onto, because the image includes a basis. Hence, the map is invertible.2012-07-03
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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.2012-07-03
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    @Arturo Magidin, I think I got it. Coordinate transformation is invertible (thanks for the term, I didn't study math in English) by definition, because otherwise such transformation can't be useful.2012-07-03
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    @superM: Sorry... what do you mean, **exactly**, by "coordinate transformations"? I thought you meant change-of-coordinates transformations, but perhaps that is not what you meant? Linear transformations are very useful, and there are many of them that are *not* invertible (e.g., projections).2012-07-03
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    @Arturo Magidin, you understood everything right )) And I really think I got it at last!2012-07-03
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    @Arturo Magidin, Thanks2012-07-03
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    When you say reversible, you probably mean invertible. And coordinate transformation is bijective, so it's invertible.2012-07-03
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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!2012-07-03

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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.