# Could we write Fourier transform as a matrix?

I have heard that Fourier transform is a linear transformation.

I have also heard that any linear transformation can be written as a matrix multiplication.

(probably I'm missing some details in the above two statements)

So my guess is, the above two notions are related (may be not). Carrying on with the question, I also know that Fourier transform acts on the space of $L^2$ functions (again correct me please if wrong).

So my questions are:

How can we write the Fourier transform as a matrix operator say $\mathcal{F}$?

What are the elements of that matrix $\mathcal{F}$?

What is the dimension of that matrix $\mathcal{F}$?

When $\mathcal{F}$ acts on a function $g$, how do we write $g$ as a vector, to apply the matrix $\mathcal{F}$?

I know these can be done in DFT, but can it be done in continuous case is my question?

I read the following posts and they are related but not the same.

Is a Fourier transform a change of basis, or is it a linear transformation?

How is the Fourier transform "linear"?

• Well, on the real line, the Fourier transform is a diagonalizable unitary operator on the Hilbert space $L^2$. The eigenvectors are the Hermite polynomials. On the integers, the Fourier transform is a unitary from $l^2(\mathbb{Z})$ to $L^2(\mathbb{T})$. – Michael Oct 16 '13 at 10:28
• @Michael thanks a lot. Could you please elaborate on your answer? It's too dense for me to understand. – triomphe Oct 16 '13 at 12:06
• @Michael You mean Hermite functions, not Hermite polynomials. – user100000 Oct 16 '13 at 12:35
• My bad, thanks for the correction. – Michael Oct 17 '13 at 3:56

What are the elements of that matrix $\mathcal F$?

Depends on the basis we choose. The same linear operator is represented by different matrices in different bases.

What is the dimension of that matrix $\mathcal F$?

It is an infinite matrix (so, strictly speaking, not a matrix).

When $\mathcal F$ acts on a function $g$, how do we write $g$ as a vector, to apply the matrix $F$?

Complex-valued functions are vectors, if a vector is understood abstractly, as an element of a vector space. (That is to say, functions form a vector space). If you mean the concrete representation of a vector as a row or column of numbers, then that can be obtained by expanding $g$ in a basis. The row/column will be infinite.

More details below the cut.

To represent a linear operator as a matrix, we need to choose a basis for our space. The most convenient space on which to study $\mathcal F$ is $L^2(\mathbb R)$, the space of square-integrable functions. One convenient basis of that space is given by Hermite functions $$\Phi_n(x)= (-1)^n (2^{n}n! \sqrt{\pi})^{-1/2} e^{x^2/2}\frac{d^n(e^{-x^2})}{dx^n}, \quad n=0,1,2,\dots$$ This basis is orthonormal (sketch here), which makes it easy to expand a given function $g\in L^2(\mathbb R)$ in this basis: $$g=\sum_{n=0}^\infty g_n \Phi_n,\quad g_n = \langle g,\Phi_n\rangle = \int_{-\infty}^\infty g(x)\Phi_n(x)\,dx$$ (I did not need conjugation over $\Phi_n$, since it happens to be real-valued.)

Each $\Phi_n$ is an eigenvector (eigenfunction) of $\mathcal F$, with eigenvalue $(-i)^n$, see Wikipedia. Therefore, the matrix of $\mathcal F$ in this basis is diagonal with the periodic sequence $(-i)^n$ along the diagonal: $$\begin{pmatrix} 1 & 0& 0 & 0 &0 &\dots \\ 0 & -i & 0 & 0 &0 &\dots \\ 0 & 0 & -1 & 0 &0 &\dots \\ 0 & 0 & 0 & i &0 &\dots \\ 0 & 0 & 0 & 0 &1 &\dots \\ \vdots & \vdots &\vdots &\vdots &\vdots &\ddots \end{pmatrix}$$

• Thanks a lot for the detailed answer. So when "they" say any linear transformation can be written as a matrix does that include infinite matrices as well? – triomphe Oct 16 '13 at 13:50
• @MLT Impossible to know without context in which the statement was made. – user100000 Oct 16 '13 at 13:59
• I think it was in vector spaces I heard it. – triomphe Oct 16 '13 at 14:00