# Relation between spectra of operator and spectrum of a fourier transfo

1. Jun 23, 2013

### Damidami

Hello,

Something I have some time wondering and still couldn't find the answer is to this question: if there is some relation between the Spectrum (functional analysis) and the Frequency spectrum in Fourier Analysis.

Now that I think about it there seems to be a casuality the use of the same word spectrum, without any relation of one concept to the other, but when I asked my linear algebra teacher if there was some relation between (eigenvectors/eigenvalues) and the fourier transform she told me yes, without any more info as to what was that relation.

Any ideas? Pure coincidence? Heavily related? Hints?
Thanks!

2. Jun 23, 2013

### Tenshou

inner product space, is the only thing I can find , but other than that I am not completely sure

Also nothing in mathematics is a coincidence :D new discoveries are always ready to be found, new connections ready to be made, from now perspectives!

3. Jun 23, 2013

### Damidami

Hi Tenshou,
I agree, but my question was pointing more to if there was or not any reason why the word spectrum or spectra to be used for these two apparently completly different things: the spectra of a linear operator, and the spectra obtained by a fourier transformation.

The use of the same word is pure coincidence? Historically speaking at least?

Thanks,
Damián.

4. Jun 25, 2013

### Bacle2

For an extreme case of an overuse of a word in Mathematics, look up the term "Normal", which has a lot of very different uses. Normal in normal subgroup, Normal Topological Space, Normal Line, Normal Subspace....

5. Jun 28, 2013

### jbunniii

A linear, time-invariant operator can be represented in terms of its impulse response, which is a time-domain function $h$ which relates an input $x$ to the corresponding $y$ by means of a convolution:
$$y(t) = \int_{-\infty}^{\infty} h(\tau) x(t - \tau) d\tau$$
If we choose $x$ to be a complex exponential function, of the form $x(t) = a e^{i \omega t}$, then we obtain the result
$$y(t) = \int_{-\infty}^{\infty} h(\tau) a e^{i \omega (t - \tau)} d\tau = a e^{i \omega t} \int_{-\infty}^{\infty} h(\tau) e^{-i \omega \tau} d\tau = a e^{i \omega t} \hat{h}(\omega)$$
where $\hat{h}$ is the Fourier transform of $h$. (Assuming that $h$ is integrable.) Thus, we have shown that a complex exponential is an eigenfunction of the operator corresponding to $h$: if we apply the operator to a complex exponential, the result is the same complex exponential, scaled by the factor $\hat{h}(\omega)$. Thus we may consider $\hat{h}(\omega)$ to be the eigenvalue associated with the eigenfunction $e^{i\omega t}$. In Fourier analysis, we may define the spectrum to be the function $\hat{h}$. The image of this function is simply the set of all eigenvalues, which corresponds to the definition of the spectrum in functional analysis.