Spectral Theorems

functional calculus

Once you can diagonalize an operator, you can do far more than add and multiply it — you can feed it into a function. The idea is disarmingly simple: if T scales the eigendirection q_i by lambda_i, then a function f of T should scale that same direction by f(lambda_i). The functional calculus makes this precise. Given the spectral resolution T = sum lambda_i P_i, define f(T) = sum f(lambda_i) P_i. You just apply f to each eigenvalue and keep the same projections.

This single definition unifies a pile of constructions you have seen piecemeal. Polynomials match the obvious meaning: if f(x) = x^2 then f(T) = T^2. Taking f(x) = 1/x (defined when 0 is not an eigenvalue) gives T^-1. Taking f(x) = sqrt(x) on a positive operator gives a square root. Taking f(x) = exp(x) gives the matrix exponential exp(T), the engine of linear differential equations. All of these are the one rule f acts eigenvalue by eigenvalue.

Two things make this rigorous and well-behaved. First, it only needs f to be defined on the spectrum of T, a finite set of numbers — so even a function with wild behavior elsewhere is fine. Second, because the P_i are orthogonal projections summing to I, the map f -> f(T) respects sums and products: (f g)(T) = f(T) g(T), and f(T) commutes with T. For self-adjoint T with real f the result is self-adjoint; for general normal T any complex-valued f works on its complex spectrum.

T = sum lambda_i P_i -> exp(T) = sum exp(lambda_i) P_i, T^-1 = sum (1/lambda_i) P_i

Apply any function f eigenvalue by eigenvalue, reusing the very same spectral projections.

The functional calculus only sees the eigenvalues, so two functions that agree on the spectrum of T give the same operator — even if they differ everywhere else. This is why f need only be defined on a finite set.

Also called
functions of an operatorf(T)