Is a Fourier transform a change of basis, or is it a linear transformation? I've frequently heard that a Fourier transform is "just a change of basis".
However, I'm not sure whether that's correct, in terms of the terminology of "change of basis" versus "transformation" in linear algebra.
Is a Fourier transform of a function merely a change of basis (i.e. the identity transformation from the original vector space onto itself, with merely a change of basis vectors), or is indeed a linear transformation from the original vector space onto another vector space (the "Fourier" vector space, so to speak)?
Also: Does the same  answer hold for other similar transforms (e.g. Laplace transforms)? Or is there a different terminology for those?
 A: Calling the Fourier transformation "a change of basis" is misleading in the sense that the Fourier transformation is a unitary (linear) transformation between two different Hilbert spaces, namely $L^2(\mathbb R)$ and $L^2(\hat{\mathbb R})$.
Here $\hat{\mathbb R}$ is the dual group of $\mathbb R$. It turns out that $\hat{\mathbb R}\cong\mathbb R$, but there is no canonical isomorphism. So, only if you fix some arbitrary isomorphism $\hat{\mathbb R}\cong\mathbb R$, you can consider the Fourier transformation as a unitary transformation from some Hilbert space to itself, which really is essentially a change of basis.
A: Rasmus' answer is probably what you want. But remember that the Fourier transform can also be defined in a discrete domain, and we have the DTFT and the DFT. The latter, finite, maps a sequence of N complex numbers to other complex sequence of same length, via a linear unitary transform. In this case (and, perhaps, only in this case) we can confidently say that it's a "change of basis".
A: The way I think about it is that the Fourier transform defines the coefficients of the original function WRT a new basis (complex sinusoids). The coefficients are obtained simply by projecting the original function onto each (orthogonal) basis function (using the standard L^2-complex inner product, which is where the minus sign in the exponent comes from).
