Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
"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
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
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
@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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.