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Spectral projections and simultaneous diagonalization

Once you can diagonalize, you can split the space into eigenspaces with explicit projectors — and ask when two operators share one eigenbasis. The answer hinges on a single algebraic condition: commuting.

Decomposing the identity by eigenvalue

A diagonalizable A splits the space as a direct sum of its eigenspaces. For each eigenvalue lambda there is a spectral projection P_lambda — a projection onto E_lambda along all the other eigenspaces. These projectors are the spectral skeleton of A: they are idempotent, they multiply to zero across distinct eigenvalues, and they sum to the identity.

If A is diagonalizable with distinct eigenvalues lambda_1, ..., lambda_k:

  P_i^2 = P_i              (each is a projection)
  P_i P_j = 0  for i != j  (orthogonal idempotents)
  P_1 + ... + P_k = I      (resolution of the identity)

Spectral decomposition of A:
  A = lambda_1 P_1 + ... + lambda_k P_k

And then ANY power / polynomial is read off the same projectors:
  A^m   = lambda_1^m P_1 + ... + lambda_k^m P_k
  f(A)  = f(lambda_1) P_1 + ... + f(lambda_k) P_k
Spectral resolution: A is a weighted sum of its eigenspace projectors.

This is the operator-level upgrade of A = P D P^-1: instead of one change-of-basis, you get k standalone pieces, each carrying one eigenvalue. Functions of A — powers, exponentials, inverses — become functions applied to scalars, eigenvalue by eigenvalue.

Building the projectors concretely

You don't have to guess the spectral projections — they read straight off A = P D P^-1. Group the columns of P by eigenvalue; the projector P_lambda keeps the coordinates belonging to lambda's eigenspace and zeroes the rest, expressed back in the standard basis. Equivalently, P_lambda = P E_lambda P^-1, where E_lambda is the diagonal indicator that is 1 on lambda's slots and 0 elsewhere.

A = [3, 1; 0, 2]      eigenvalues 3 and 2 (distinct -> diagonalizable)

  eigenvector for 3:  v1 = (1, 0)
  eigenvector for 2:  v2 = (1, -1)
  P = [1, 1; 0, -1]     P^-1 = [1, 1; 0, -1]   (P is its own inverse here)

  D = diag(3, 2)

  P_3 = P diag(1,0) P^-1 = [1, 1; 0, 0]    (onto E_3 along E_2)
  P_2 = P diag(0,1) P^-1 = [0, -1; 0, 1]   (onto E_2 along E_3)

check:  P_3 + P_2 = I,  P_3 P_2 = 0,  3 P_3 + 2 P_2 = A   OK
Reading the two spectral projectors of a 2x2 directly off its eigenvectors.

Two operators, one basis

Now ask: can two diagonalizable operators A and B be diagonalized by the same P? That's simultaneous diagonalization. The clean theorem: two diagonalizable operators are simultaneously diagonalizable iff they are commuting operators, AB = BA. Commuting is exactly the algebraic shadow of sharing an eigenbasis.

The intuition for one direction: if B commutes with A, then B maps each eigenspace of A into itself (an A-eigenspace is invariant under B). Restrict B to each eigenspace, diagonalize it there, and stitch the local eigenbases together — the result diagonalizes both at once.