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Schrodinger operators, operators on manifolds, general differential operators, numerical studies, integral operators, discrete models, resonances, non-self-adjoint operators, random operators/matrices

4 votes

What does multiplying a matrix by its transpose have to do with spectral theorem?

In the finite dimensional case the Spectral Theorem says that we can decompose a self-adjoint operator into a sum of projection operators: if $A$ is self-adjoint then we can write $A=\lambda_{1}P_{1} …
George Melvin's user avatar