Diagonalization of the Fourier transformation.

Question

Browsing the following link: https://webusers.imj-prg.fr/~bernard.maurey/agreg/Textes/ag001_a2.pdf , I was able to understand how to diagonalize the Fourier Transform $ \mathcal {F} $, but unfortunately, the article does not tackle the subject of knowing how to find the eigenspaces relative to the eigenvalues ​​of $ \mathcal {F} $. Can you please tell me how to find them ?

Thanks in advance for your help.

Answer 1

I'm not exactly sure kind of answer you're looking for, and I'm afraid that I don't know a very elegant way of going about this, but simply put we have the following situation.

The Fourier transform (when properly normalized) has eigenvalues $\{1,-1,i,-i\}$. Let us write $M_{a}$ for these eigenspaces for $a$ an eigenvalue, then we have \begin{equation} L^{2}(\mathbb{R}) = M_{1} \oplus M_{-1} \oplus M_{i} \oplus M_{-i}. \end{equation} According to this wikipedia article we have that the eigenfunctions of the Fourier transform are \begin{equation} \psi_{n}(x) = e^{-x^{2}/2} H_{n}(x), \end{equation} with $H_{n}$ the Hermite polynomials, and they have eigenvalues $(-i)^{n}$. The observation that the eigenvalue belonging to $\psi_{n}$ is equal to $1$ if and only if $n$ is divisible by $4$ tells us that \begin{equation} M_{1} = \langle \psi_{n} \mid \text{ n divisible by } 4 \rangle, \end{equation} where the brackets stand for closed linear span. Similar observations give similar expressions for the other $M_{a}$.

I'd recommend you also take a look at this other question and this MO question.

Answer 2

I'm going to write $F$ for the Fourier transform on $L^2$: $Ff=\hat f$.

The Plancherel theorem shows that $F^2f(x)=f(-x)$. Hence $F^4f=f$.

So in linear-algebra terms the minimal polynomial for $F$ is $x^4-1$; since this has four distinct roots it follows that $F$ is diagonalizable, and in fact a very simple procedure for writing $f$ as a sum of eigenfunctions also follows:

Suppose $f\in L^2$. Write $$f_0=\frac14(f+Ff+F^2f+F^3f).$$Since $F^4f=f$ it follows that $$Ff_0=f_0.$$Similarly, if $$f_2=\frac14(f-Ff+F^2f-F^3f)$$then $$Ff_2=-f_2.$$

I'll let you think about how to give similar definitions of $f_1$ and $f_3$ so that $Ff_1=if_1$, $Ff_3=-if_3$, and $$f=f_0+f_1+f_2+f_3.$$