interpolation and spectral projections
Suppose T is diagonalizable with distinct eigenvalues lambda_1, ..., lambda_k. We would love to build, for each eigenvalue, an operator P_i that projects onto its eigenspace along the others. Remarkably, you can write each P_i explicitly as a POLYNOMIAL in T — and the recipe is classical Lagrange interpolation, repurposed from numbers to operators.
Recall the Lagrange basis polynomial L_i(x) = prod_(j != i) (x - lambda_j) / (lambda_i - lambda_j). It is built so that L_i(lambda_i) = 1 and L_i(lambda_j) = 0 for j != i. Now evaluate it at the operator: set P_i = L_i(T). Because a polynomial acts on each eigenvector by its value at that eigenvalue, P_i acts as 1 on the lambda_i-eigenspace and as 0 on every other eigenspace — exactly the spectral projection.
These projections satisfy the spectral identities for free: they are idempotent (P_i^2 = P_i), mutually orthogonal as operators (P_i P_j = 0 for i != j), and they sum to the identity (sum P_i = I, which is the interpolation statement that sum L_i(x) = 1). Moreover T itself rebuilds as T = sum lambda_i P_i — the spectral decomposition — and any function f of T becomes f(T) = sum f(lambda_i) P_i.
Why this is beautiful: it shows the spectral projections are not extra data bolted onto T but live INSIDE the algebra F[T]; they are polynomials in the operator. It also gives a slick, computation-free proof that diagonalizable operators have a clean functional calculus, and it generalizes — for non-diagonalizable T, Hermite interpolation (matching derivatives too) produces the projections plus nilpotent parts.
For a two-eigenvalue operator, the Lagrange polynomials are linear; evaluated at T they become the spectral projections that reconstruct T and any function of it.
Because each P_i is a polynomial in T, every operator that commutes with T also commutes with every spectral projection — and hence preserves each eigenspace. This single observation drives many simultaneous-diagonalization and commutant arguments.