The FFT processing gain comes from the fact that the DFT (of which FFT is simply a fast implementation) is a non-normalized linear transformation. This is a mouthful, so let's look at what this means.

I'm going to assume that you know what a [linear transformation][1] is. Namely, given vectors $\mathbf{x}$ and $\mathbf{y}$ and a matrix $A$ we have

$$\mathbf{y} = A \mathbf{x}.$$

DFT is just such a transformation. In fact, you can use MATLAB `dftmtx` command to generate this matrix for you based on the length of vector $\mathbf{x}$. This this case,

$$\mathbf{y} = \text{DFT}(\mathbf{x}).$$

This matrix $A$ has some properties. First of all, it's a square matrix, which means that it's probably invertible (and indeed it is!). It also tells us that we're essentially taking components of x and performing a [change of basis][2] given by the columns of $A$ to get its DFT. So far so good.

Now, let's get to some more important properties. Matrix $A$ is orthogonal. This means that every column of $A$ is perpendicular to every other column, or more mathematically, $A^TA$ is a diagonal matrix (you may have to think a little about why this is true). This is a *very* nice property, since simply transposing the matrix gives us something very close to its inverse.

To make this Transpose $\leftrightarrow$ Inverse relationship strict, we want the matrix $A$ to also be *normal*. This is a matrix whose every column vector has length 1. In other words, if $a$ is a column of $A$, then $\sqrt{a^Ta} = 1.$ If a matrix is both orthogonal and normal, we call it [orthonormal][3], and in this case $A^TA = I$, so $A^T$ is in fact the inverse of $A$. Neat!

The usual DFT matrix (or the usual DFT transform) is orthogonal, but *not* orthonormal. In fact, if $D$ is the DFT matrix, then $D^TD = N$ where $N$ is the number of columns (or rows, it's square!) in $D$. To make it orthonormal, we should use $\frac{D}{\sqrt{N}}$ instead. If you look at it long enough, you realize that is we scale both forward and inverse transforms by $\frac{1}{\sqrt{N}}$, we're doing extra work in performing the calculations, so we usually just scale by $\frac{1}{N}$ at the inverse.

There are better theoretical reasons for doing it on one side rather than of both. See [my answer here][4] for more information.


  [1]: http://en.wikibooks.org/wiki/Linear_Algebra/Linear_Transformations
  [2]: http://en.wikipedia.org/wiki/Change_of_basis
  [3]: http://en.wikipedia.org/wiki/Orthonormality
  [4]: http://dsp.stackexchange.com/questions/11376/why-are-magnitudes-normalised-during-synthesis-idft-not-analysis-dft/11379#11379