Spectral Decomposition

Let there be a Hermitian operator A^\hat{A} on a finite-dimensional Hilbert space. Spectral theorem says that an operator A^\hat{A} can be spectrally decomposed if it is unitarily diagonalizable i.e., normal. There exists real eigenvalues {an}\{a_n\}, orthonormal eigenvectors {∣k⟩}\{\left\lvert k \right\rangle\} which form orthogonal Projector {Π^n}\{\hat{\Pi}_n\} such that

A^=∑n=1NanΠ^n=∑n=1Nan∣k⟩⟨k∣\hat{A}=\sum_{n=1}^N a_n\hat{\Pi}_n=\sum_{n=1}^N a_n\left\lvert k \right\rangle\left\langle k \right\rvert

where projectors satisfy

  1. Π^nΠ^m=0\hat{\Pi}_n\hat{\Pi}_m=0 for n≠mn\neq m which means they are orthogonal

  2. ∑n=1NΠ^n=I^\sum_{n=1}^N \hat{\Pi}_n=\hat{I} which means they are a complete Set of projectors

  3. this projector projects onto the ana_n-eigenspace of A^\hat{A}

So {an}\{a_n\} is the spectrum of A^\hat{A}.

And that means you can think A^\hat{A} as scaling each eigenspace by its own eigenvalue.