The finite-dimensional version of the theorem states if $A$ is a Hermitian matrix, then it may be decomposed into $A = Q\Lambda Q^*$. In other words, the eigenvectors of $A$ form an orthonormal basis for $A$.
Now, let $H$ be an infinite-dimensional Hilbert space and $B(H)$ the set of all bounded linear operators from $H$ to $H$. Let $\sigma(A)$ denote the spectrum of an operator $A$. With this notation, the infinite dimensional version of the Spectral theorem is (taken from "Quantum Theory for Mathematicians" by Brian Hall):
If $A \in B(H)$ is self-adjoint, then there exists a unique projection-valued measure $\mu^A$ on the Borel $\sigma$-algebra in $\sigma(A)$, with values in projections on $H$, such that
$$\int_{\sigma(A)} \lambda\,d\mu^A(\lambda) = A.$$
I am having trouble seeing how the above has any relation to the conclusion drawn for the finite-dimensional case. How does it imply an orthogonal basis of (perhaps generalized) eigenvectors for a self-adjoint operator? I suppose a more general way to put it is, how does the infinite-dimensional spectral theorem really help us? It is easy to see the significance of the theorem in finite dimensions, but what makes the infinite-dimensional version so powerful?
Best Answer
Your $A$ can be written as $\sum \lambda_i P_i,$ where $P_i$ is the projection onto the eigenspace of $\lambda_i.$ A sum is a finite version of an integral.