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Relation between spectra of operator and spectrum of a fourier transfo

  1. Jun 23, 2013 #1
    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. jcsd
  3. Jun 23, 2013 #2
    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!
     
  4. Jun 23, 2013 #3
    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.
     
  5. Jun 25, 2013 #4

    Bacle2

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    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....
     
  6. Jun 28, 2013 #5

    jbunniii

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    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.
     
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