Spectral resolution and functional calculus

From Q D Q^T to a sum of projectors

The factorization A = Q D Q^T is basis-bound; the spectral resolution frees it. Group the orthonormal eigenvectors by their eigenvalue, and for each distinct eigenvalue lambda_i let P_i be the orthogonal projection onto its eigenspace. Then A = lambda_1 P_1 + ... + lambda_k P_k. Each P_i is built coordinate-free as a sum of outer products q q^T over the eigenvectors q sharing that eigenvalue.

Spectral resolution of a normal operator with eigenvalues lambda_1..lambda_k:

  P_i = sum over orthonormal eigenvectors q of lambda_i  of  q q^*

  (1)  P_i = P_i^*       each projector is self-adjoint (ORTHOGONAL projection)
  (2)  P_i^2 = P_i       idempotent
  (3)  P_i P_j = 0  (i!=j)   eigenspaces are mutually orthogonal
  (4)  P_1 + ... + P_k = I    resolution of the identity

Spectral form:
  A = lambda_1 P_1 + lambda_2 P_2 + ... + lambda_k P_k

Example  A = [2, 1; 1, 2]  (eigenvalues 3, 1):
  q_1 = (1, 1)/sqrt2,  q_2 = (1,-1)/sqrt2
  P_1 = q_1 q_1^T = [0.5, 0.5; 0.5, 0.5]
  P_2 = q_2 q_2^T = [0.5,-0.5;-0.5, 0.5]
  check  3 P_1 + 1 P_2 = [2, 1; 1, 2] = A   and  P_1 + P_2 = I.

Orthogonal projectors and the spectral measure

The key upgrade over Vol I's spectral projections is property (1): here the projectors are *orthogonal*, P_i = P_i*, because the eigenspaces are mutually orthogonal — a gift of the spectral theorem that does not hold for a general diagonalizable operator. This family {P_i} is the finite-dimensional projection-valued measure: it assigns to each eigenvalue the projection that 'measures how much of a vector lives there.'

Functional calculus: applying f to an operator

The resolution makes functional calculus effortless. To apply a function f to A, apply it to each eigenvalue and keep the same projectors: f(A) = f(lambda_1) P_1 + ... + f(lambda_k) P_k. This *defines* f(A) for any f on the spectrum — not just polynomials, but sqrt, exp, log, 1/x, even a step function — because A acts as plain multiplication by lambda_i on each eigenspace.

1. Diagonalize the normal operator A = U D U* and read its spectrum {lambda_i} off the diagonal of D.
2. Apply f scalar-wise: form the diagonal matrix f(D) = diag(f(lambda_1), ..., f(lambda_n)).
3. Conjugate back: f(A) = U f(D) U*, equivalently sum f(lambda_i) P_i over the spectral resolution.