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## Main Question or Discussion Point

Some authors say that applying Fourier transform twice flips the vector, F

[tex]\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]^2 = \left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]

[/tex]

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

^{2}[x(t)] = x(-t). Yet, the simple checks proves this wrong. For instance, take 2x2 DFT:[tex]\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]^2 = \left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]

[/tex]

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