Projections in spectral decomposition.

Question

In my quantum mechanics book the spectral decomposition of operator $A$ is given as

$A=\sum\limits_j\lambda_jP_j$

where $\lambda_j$ are the eigenvalues of matrix $A$ and $P_j$ is the orthogonal projection of the particular eigenvalue. Now I'm doing an example in $\mathbb{C}^2$ where I got this spectral decomposition

$A=\frac{1}{2}(a+\lambda)\frac{1}{2}(I+\mathbf{a\cdot\sigma})+\frac{1}{2}(a-\lambda)\frac{1}{2}(I-\mathbf{a\cdot\sigma}),\quad a,\lambda\in \mathbb{R},\|a\|=1$

for $A=\frac{1}{2}(aI+\lambda\mathbf{a\cdot\sigma}),\quad a,\lambda\in \mathbb{R},\|a\|=1$

where $\sigma$ are Pauli matrices, $I$ identity matrix, $\frac{1}{2}(a \pm\lambda)$ are the eigenvalues and the projections are $T_{\pm\mathbf{a}}=\frac{1}{2}(I+\mathbf{a\cdot\sigma})$

Now my question is that when I did some math I got that the projections $T_{\pm\mathbf{a}}$ are actually the eigenvectors corresponding to their respectful eigenvalues. Is this true in general that the projections are eigenvectors of the corresponding eigenvalue or is this just some special case?

Answer 1

For self-adjoint (more generally, normal) matrices, spectral projections are by definition projections onto eigenspaces. They exist by the spectral theorem. Spectral projections corresponding to distinct eigenvalues are pairwise orthogonal.

For general matrices, what you call spectral decomposition may not exist. Quantum mechanical observables are self-adjoint operators.