Flip operator's equality to Double Fourier

[valjok]

Some authors say that applying Fourier transform twice flips the vector, F²[x(t)] = x(-t). Yet, the simple checks proves this wrong. For instance, take 2x2 DFT:

\left[\begin{array}{cc}1&amp;1\\ 1&amp;-1\end{array}\right]^2 = \left[\begin{array}{cc}1&amp;0\\ 0&amp;1\end{array}\right]<br />

The Identity is different from counter-identity. It cannot therefore flip the vector elements. Where is the mistake?

 

[adriank]

For the size-n DFT, (F²x)(t) = x(-t), where x is a periodic function with period n.

In the case n = 2, there is no mystery, since x(-0) = x(0) and x(-1) = x(-1 + 2) = x(1), since x has period 2.

 

[valjok]

This is a great answer!