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The spectral theorem allegedly says when a linear transform is diagonalizable. Well, an integral is a linear transform, how do you diagonalize an integral? What does that notion mean in a more general case?

PhiEarl
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    Some linear transformations are not diagonalizable. Not to toot my own horn, but https://math.stackexchange.com/questions/149842/understanding-the-symmetric-hermitian-matrix-operator/149845#149845 is a response relating to this issue (e.g. for operators on infinite dimensional spaces). The Halmos article linked there is still very readable and very good. The idea is that a "diagonalizable" linear transformation is one that looks like "multiplication" by something, where multiplication might be on a function space and not just coordinate-wise multiplication on $\mathbb{R}^n$ or $\mathbb{C}^n$. – leslie townes Feb 16 '21 at 21:37
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    Fredholm theory is a domain where integrals induce linear operators in function spaces and they can be diagonalized under suitable conditions. – Gribouillis Feb 16 '21 at 22:15
  • Is there perhaps an example you can go through of diagonalizing and integral or some other non-matrix linear transform over a Banach space or something idea? Or do you need an inner product space? – PhiEarl Feb 17 '21 at 12:51

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