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This is a question about an off-hand remark from a lecturer a few weeks ago. He was talking about eigenvectors/values of a matrix, and rhetorically asked us if we'd seen the interpretation of eigenvalues as frequencies.

I haven't been able to find a clear explanation for this, and I'd be interested if anyone could enlighten me or (perhaps even more usefully) point me in the direction of a good text which covers this interpretation/application.

Thankyou very much!

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It only makes sense if your vector space can reasonably be interpreted as modelling oscillations. Otherwise, I don't see why you would gain something by speaking about frequencies instead of eigenvalues. Besides, even when they can be interpreted as frequencies, we still speak about eigenfrequencies of the system to distinguish them from frequencies which are no eigenvalues of the operator under consideration. – Raskolnikov Apr 2 '12 at 10:11

Standard Fourier series. The linear transformation (instead of a matrix) is $$ T = \frac{-d^2}{dx^2}. $$ The vector space is smooth functions of $x$ with period $2\pi.$ And we get $$ T (\cos n x ) = n^2 \cos n x,$$ so $n^2$ is an eigenvalue of $T.$ However, $$ T (\sin n x ) = n^2 \sin n x$$ as well, so we get two different eigenvectors for that eigenvalues. How different are they? Sticking with real functions, we have an inner product $\langle, \rangle$ on pairs of periodic functions given by $$ \langle f,g \rangle = \int_0^{2 \pi} \; f(x) g(x) dx. $$ And the pair of functions we gave are orthogonal under the given inner product. Also other pairs, such as $\cos nx, \cos mx$ give $0$ when $m \neq n,$ same for sines, same for sin and cosine, one $m$ the other $n.$ Finally, integration by parts tells us that $T$ is self-adjoint with respect to the inner product, as $$ \int_0^{2 \pi} \; u'(x) v'(x) dx = \langle u,Tv \rangle = \langle Tu,v \rangle. $$

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