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?


3 Answers 3


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.

  • 1
    $\begingroup$ When is a morphism canonical? The construction of $\hat{\mathbb{R}}$ relies on $\text{Hom}(\mathbb{R},T)$, right? Does this miss any of the structure of $\mathbb{R}$? Don't both not even have a metric which can be matched etc.? $\endgroup$
    – Nikolaj-K
    Commented Nov 3, 2012 at 1:26
  • $\begingroup$ inst taking the Fourier transform is like "picking a basis" better? to beginwith we only have the vector $f$ w.o having coordiates! Change implies we had one to begin with, $\endgroup$
    – user415535
    Commented Mar 16, 2018 at 6:53
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    $\begingroup$ Why can't we say that the Fourier transform is a change of basis? Isn't a change from some basis for $L^2$ to the $\sin$ and $\cos$ basis? Naively looking at the Fourier transform it also looks like a change of basis in the sense that $\hat{f}$ is written as a weighted sum of basis "vectors (functions)" where the weights are given by the "transformed" function. $\endgroup$
    – Cuhrazatee
    Commented Feb 21, 2019 at 0:27
  • $\begingroup$ @Rasmus Can you clarify what you mean with $\widehat{\mathbb{R}}\cong \mathbb{R}$ non-canonically? The map $\mathbb{R}\to \widehat{\mathbb{R}}: s \mapsto (t \mapsto \exp(2\pi i st))$ looks pretty canonical to me! $\endgroup$
    – J. De Ro
    Commented May 16, 2022 at 13:50
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    $\begingroup$ The question you might ask yourself is what makes $s \mapsto (t \mapsto \exp(2\pi i s t))$ is more or less 'canonical' than $s \mapsto (t \mapsto \exp(i s t))$. The first one chooses $1$ has a special element, the other one chooses $2\pi$. This might serve as a hint that this is non-canonical $\endgroup$ Commented Jun 29, 2023 at 16:47

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".

  • $\begingroup$ +1 The context was certainly also related to DTFTs and DFTs/DFSs, so this answer points out something important; thanks. $\endgroup$
    – user541686
    Commented Mar 24, 2012 at 0:59

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).


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