# DFT's Unitary Property

A transform is said to be unitary transform when the product of the transform matrix and its conjugate transpose (Hermitian) matrix is equal to Identity matrix.

DFT for N= 4 gives

c =

4     0     0     0
0     4     0     0
0     0     4     0
0     0     0     4

Thus, it satisfies the Unitary property

But when N=2, we have

2      1+j
1-j     2

This doesn't satisfy unitary property right? Then how can we say the DFT is a Unitary Transform? Where I have misunderstood the concept? Help

• Could you edit your answer to include your work in finding that for $N=2$, the DFT is not a unitary transform? – Dilip Sarwate Dec 15 '13 at 14:46
• I calculated wrongly.. But DFT is a Unitary transform – Premnath D Dec 15 '13 at 15:34

Just so we're on the same page: given some sequence $x[n]$ for $n=0, \dots, N-1$, the $k$'th DFT coefficient is (forgetting the scaling factor)
$$X[k] = \sum_{n=0}^{N-1} x[n] \exp(-j 2\pi k n/N)$$
We can get all the coefficients by writing this as $X = Dx$, where $D$ is the $N \times N$ DFT matrix and $D_{kn} = e^{-j 2\pi k n /N}$. For the N = 2 case,
$$D = \begin{pmatrix} e^{-j 2\pi (0)(0) /2} & e^{-j 2\pi (0)(1)/2}\\ e^{-j 2\pi (1)(0) /2} & e^{-j 2\pi (1)(1)/2} \end{pmatrix} = \begin{pmatrix} e^{-j 0} & e^{-j 0}\\ e^{-j 0} & e^{-j \pi} \end{pmatrix} = \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix},$$ for which $D^* D = D^T D = D^2 = 2I$, so (minus the scaling factor), unitary.