Simultaneous diagonalisation and a complete set of commuting observables

Both matrices are real and symmetric, hence self-adjoint. Both are block diagonal, with a $2\times2$ block acting on $\ket{e_1},\ket{e_2}$ and a $1\times1$ block acting on $\ket{e_3}$. On the first block, $A$ is the identity, which commutes with everything; on the second, they are numbers. Thus $AB=BA$, as may also be checked by direct calculation: $AB=BA=\begin{pmatrix}0&1&0\\1&0&0\\0&0&-1\end{pmatrix}$.

The matrix $A$ is diagonal: the eigenspace for eigenvalue $1$ is the plane spanned by $\ket{e_1}$ and $\ket{e_2}$, and the eigenspace for eigenvalue $-1$ is the line spanned by $\ket{e_3}$. For $B$, the $2\times2$ block is the matrix $\sigma_x$, with eigenvalues $\pm1$ and eigenvectors $\frac{1}{\sqrt2}(\ket{e_1}\pm\ket{e_2})$, while $\ket{e_3}$ is an eigenvector with eigenvalue $1$. Thus the eigenspace of $B$ for eigenvalue $1$ is the plane spanned by $\frac{1}{\sqrt2}(\ket{e_1}+\ket{e_2})$ and $\ket{e_3}$, and the eigenspace for eigenvalue $-1$ is the line spanned by $\frac{1}{\sqrt2}(\ket{e_1}-\ket{e_2})$.

Each of the two observables has a degenerate eigenvalue. Neither is therefore sufficient to define a unique eigenbasis: in an eigenspace of dimension $2$, any orthonormal basis is suitable. For example, $\ket{e_1}$ and $\ket{e_2}$ form an eigenbasis of $A$, but are not eigenvectors of $B$.

If $A\ket v=a\ket v$, then $A(B\ket v)=BA\ket v=a\,B\ket v$: the vector $B\ket v$ still belongs to the eigenspace of $A$ associated with $a$. The subspace $E_A(1)$, spanned by $\ket{e_1}$ and $\ket{e_2}$, is therefore invariant under $B$, and the restriction of $B$ to this plane has matrix $\sigma_x$ in the basis $(\ket{e_1},\ket{e_2})$. Diagonalising it gives the eigenvectors $\ket{u_\pm}=\frac{1}{\sqrt2}(\ket{e_1}\pm\ket{e_2})$, with eigenvalues $\pm1$. The subspace $E_A(-1)$ is the line spanned by $\ket{e_3}$, which is automatically an eigenvector of $B$. We obtain the orthonormal basis of common eigenvectors

$$\ket{u_+}=\frac{\ket{e_1}+\ket{e_2}}{\sqrt2}\ (a=1,\,b=1),\qquad \ket{u_-}=\frac{\ket{e_1}-\ket{e_2}}{\sqrt2}\ (a=1,\,b=-1),\qquad \ket{e_3}\ (a=-1,\,b=1).$$

This is the general method of simultaneous diagonalisation: diagonalise the second observable within each eigenspace of the first.

The three pairs $(1,1)$, $(1,-1)$ and $(-1,1)$ are distinct. Each corresponds to a common eigenspace of dimension $1$: for example, the vectors satisfying both $A\ket v=\ket v$ and $B\ket v=\ket v$ form the line spanned by $\ket{u_+}$. A common eigenvector is therefore determined, up to a phase, by its two eigenvalues, and may be denoted by $\ket{a,b}$. In quantum mechanics, if $A$ and $B$, which are compatible, are measured successively and the results $a$ and $b$ are obtained, the state after the two measurements is completely determined: it is $\ket{a,b}$, irrespective of the initial state. A complete set of commuting observables thus makes it possible to prepare a well-defined state and to label a basis of the state space by quantum numbers. For example, this is the role of $\hat H$, $\hat L^2$ and $\hat L_z$ for the hydrogen atom, whose common eigenstates are denoted by $\ket{n,\ell,m}$.