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eigenvalues of matrices or operators

4 votes
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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 $\mu$. …
Felix Benning's user avatar
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 …
Felix Benning's user avatar