TITLE: How is the Fourier transform "linear"?
QUESTION [23 upvotes]: A "linear" function usually means one who's graph is a straight line, or that involves no powers higher than 1. And yet, many sources will tell you that the Fourier transform is a "linear transform".
Both the discrete and continuous Fourier transforms fundamentally involve the sine and cosine functions. These functions are about as non-linear as you can get. They are transcendental. They are solutions to differential equations. They are infinite-degree polynomials. They have infinity turning points. (So, not straight lines at all.) In the discrete case, you take a finite sum with them, but in the continuous case you compute an integral with them! This all sounds highly non-linear to me...
Then again, apparently "linear" is one of those mathematical terms that has many, many meanings. (Much like "normal" means a few trillion different things depending on the exact context in question.) And perhaps not all of those meanings are actually related.
So, my question is, in which exact sense is the Fourier transform considered "linear"?

REPLY [4 votes]: While the previous answers correctly made it clear that straight lines are not was defines a linear transformation, it is actually possible to visualise more general linear transformations with such lines. However, you need to come clear about where these lines lie: it has nothing to do with the trigonometric functions' graphs. After all, these functions are just special elements of the domain of the Fourier transform.
What matters is the graph of the transform itself. As was already said, the Fourier transform maps functions to functions, which is considerably more complicated than mapping numbers to numbers. However, functions form a vector space: naïvely, you can write a tuple something like
$$
  \begin{pmatrix} \vdots \\ f(0) \\ f(0.000\ldots0001) \\ f(0.000\ldots0002) \\ \vdots \\ f(1) \\ \vdots \end{pmatrix}
$$
of all the function values. Strictly this space is infinite-dimensional, but in many applications you're only interested in some finite range and resolution anyway, making the space finite-dimensional. Still rather too big to visualise, but for the extreme case $n=2$ it's actually almost possible: our "functions" $f(x)$ can now be seen as points in a plane with coordinate $(f(x_1),f(x_2)) =: (f_0,f_1)$. The Fourier transform reduces to just two different modes as well: "in phase" and "anti-phase" or "frequency $0$" and "frequency $1$", i.e.
$$\begin{aligned}
  F_0 &= \left(\sqrt{\tfrac12},\sqrt{\tfrac12}\right) \\
  F_1 &= \left(-\sqrt{\tfrac12},\sqrt{\tfrac12}\right)
\end{aligned}$$
That shows already a matrix representation of the (inverse) Fourier transform, and a matrix representation is enough to show that the transformation is linear. But you still don't see the lines, all right...
The graph of this Fourier-transform function (which maps now $\mathbb{R}^2\to\mathbb{R}^2$) lives in $\mathbb{R}^4$. Still a bit bulky that space, but let's look a 3-dimensional subspace: we put in only values where the coefficient of $F_1$ is zero, i.e. the part of the graph we look at consists of triples
$$
  \begin{pmatrix} a_0 \\ a_0\cdot(F_0)_0 + 0 \\ a_0\cdot(F_0)_1 + 0 \end{pmatrix}
 = \begin{pmatrix} a_0 \\ \sqrt{\tfrac12}a_0 \\ \sqrt{\tfrac12}a_0 \end{pmatrix}
$$
and if we follow the path of these points take through $\mathbb{R}^3$ while varying $a_0$ we see – exactly: a straight line, leading away from the origin in an oblique angle!
The actual graph is not a line, but a plane in $\mathbb{R}^4$.