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eigenvalues of matrices or operators
4
votes
2
answers
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views
Is it possible to use the Laplace Transform to calculate eigenvalues?
The relationship of Eigenvalues with Gradient Descent
Let $A$ be a symmetric (and thus diagonalizable) matrix, with diagonalization
$$A=VDV^T.$$
Let us define the quadratic function
$$f(x) = x^T A x.$$ … If we could invert our Laplace transform we would of course immediately get the eigenvalues as the non-zero "frequencies" in our discrete measure $\mu$. …
2
votes
Is it possible to use the Laplace Transform to calculate eigenvalues?
A very unsatisfying answer is matrix exponentiation (which is very expensive itself):
$$x_n=\exp(ihA)x_{n-1}$$
results in
$$
x_n^{(j)} = \exp(ih\lambda_jn) x_0^{(j)}
$$
and thus
$$
\|x_n\|^2 = \sum_{j …