Another thing i suddenly realized. If $F$ is the DFT matrix,
containing as columns the DFT base functions in terms of positive $i$,
we have $F^4 = I$, meaning $F$ represents a $90$ degree rotation
combined with a reflection.  The eigenvalues of $F$ are $4$--th roots
of unity $\in \{\pm 1, \pm i\}$.

Note that a DFT of matrix $A$ looks like $FAF^H$. For $2 \times 2$
matrices, the normalized $F$ looks like
$$F_2 = {1 \over \sqrt{2}}\matrix{1 & 1 \\  1 & -1}.$$
We can write this as 
$$\sqrt{2} F_2 = 
\matrix{1 & 0 \\  0 & -1} +
\matrix{0 & 1 \\  1 &  0} = E_1 + E_2,$$
where $E_1^2 = E_2^2 = I$ en $E_1E_2 = -I$. This is a $2^2$--dimensional
Clifford algebra. Note that for this case $F_2 ^2 = I$.
For $4 \times 4$ matrices we have
$$F_4 = {1 \over \sqrt{4}}
\matrix{
1 &  1  & 1 &  1 \\ 
1 &  i & -1 & -i \\
1 & -1  & 1 & -1 \\ 
1 & -i & -1 &  i}.$$

Something which has always bugged me is why no--one talks about
eigenvectors of a DFT. For the continuous case this is important, but
for DFT this doesn't really seem to make sense. Why?

[Edited] Maybe there is an answer in the recursive formulas for the
Hermite polynomials. Probably something needs to be said about the
Gaussian wave packet. On the real line, the exponential of $-x^2$
somehow makes sense, but on the fine cycle on which the DFT is
defined, it does not really.