# Matrix Notation of Inverse Discrete Fourier Transform

Notation: I will write $a_k$ to denote a sequence/vector of scalar, non-complex elements.

The discrete Fourier transform $\hat a_k$ of a spatial signal $a_k$ is defined as:

$$\hat a_k = \frac{1}{\sqrt n} \sum_{w=0}^{n-1} a_w e^{2\pi j w/n}$$

The inverse transform of the frequency-domain signal $\hat a_k$ only differs in the sign of the exponent:

$$a_k = \frac{1}{\sqrt n} \sum_{w=0}^{n-1} \hat a_w e^{-2\pi j w/n}$$

So obviously, $a_k$ the inverse discrete Fourier transform is just the complex-conjugate of the regular discrete Fourier transform.

Since both operations are linear, they often also are written as matrix expressions:

$$\hat a_k = F a_k, \quad a_k = F^\mathrm H \hat a_k$$

Here $F^\mathrm H$ denotes the Hermitian of $F$, that is its conjugate-transpose. But why is $a_k = F^\mathrm H \hat a_k$ and not just $a_k = \overline F \hat a_k$, i.e. why is here the conjugate-transpose involved and not only the conjugate?

Update: According to Wikipedia, the matrix $F$ is symmetrical, i.e. its conjugate and its conjugate-transpose are the same. Doesn't that mean that $F^\mathrm H = \overline F$?

You are correct. $F^H = \bar{F}$ since $F^T = F$. Thus $F^{-1} = F^H = \bar{F}$, and so:

$$\hat{\mathbf{a}} = F\mathbf{a} \Leftrightarrow F^H\hat{\mathbf{a}} = \mathbf{a} \Leftrightarrow \bar{F}\hat{\mathbf{a}} = \mathbf{a}$$

where $\mathbf{a} = \begin{bmatrix} a_0 & a_1 & \cdots & a_{n-1} \end{bmatrix}^T$ and similar for $\mathbf{\hat{a}}$

Why is $a_k=F^H\hat{a}_k$ and not just $a_k=\bar{F}\hat{a}_k$, i.e. why is here the conjugate-transpose involved and not only the conjugate?

You're notation is a little off here. You have a matrix operating on a scalar. I believe what you intended is the DFT matrix acting on the vector of $a_k$'s as I've written above.

As you can see from above, they are equivalent. So it is perfectly acceptable to write it either way.

Now, the question of why is $F^{-1}$ typically written as $F^H$ instead of $\bar{F}$, is a different question all together. I can't say for certain, but I would conjecture that this is because this is because $F$ is a type of matrix from a larger class, called unitary matrices. For a unitary matrix, $P$, it is the case that:

$$P^{-1} = P^H$$

Writing $F^H$ is perhaps more obvious a reminder that the DFT matrix is unitary, and unitary matrices have useful properties which are often used in derivations (such as norm preservation, i.e., Parseval's theorem).

• Thanks, but I'm not quite sure how this relates to my question. Could you please add some details on this to your answer? – theV0ID Sep 11 '15 at 14:24
• @theV0ID that should clarify... let me know if not – Chester Sep 11 '15 at 14:58