Simultaneous diagonalization

A finite family of pairwise commuting diagonalizable operators on a finite-dimensional complex vector space is simultaneously diagonalizable. In particular, commuting Hermitian operators admit an orthonormal common eigenbasis.

If linear operators $A,B$ have a common eigenbasis $(v_i)$, then $A+B$ has the corresponding eigenvalues ${\alpha }_i+{\beta }_i$, counting algebraic multiplicity. This follows directly from linearity on each basis vector. The pairing is determined by the common eigenvectors; it is not the collection of all pairwise sums of arbitrary eigenvalues.

For commuting real symmetric matrices $A,B$, each eigenspace of $A$ is invariant under $B$: $A(Bv)=B(Av)=\lambda Bv$. The restriction of $B$ to that eigenspace is again symmetric. The spectral theorem therefore supplies an orthonormal eigenbasis for $B$ inside each eigenspace of $A$. Combining these bases gives a common orthonormal eigenbasis of the whole space. Conversely, two symmetric matrices diagonal in the same orthonormal basis commute.

Suppose $D$ has $n$ distinct eigenvalues over a field and $C$ commutes with $D$. Each eigenspace of $D$ is one-dimensional. Since commuting maps preserve eigenspaces, $C$ acts as a scalar on each of them. The eigenvectors of $D$ give a common eigenbasis and hence simultaneous diagonalization. For real matrices, the result gives a real eigenbasis only if the eigenvalues of $D$ are real; otherwise it applies over $\mathbb{C}$.

For commuting diagonalizable operators $A_1,\dots ,A_r$ on a finite-dimensional vector space, the joint spectrum is the set of tuples $(a_1,\dots ,a_r)$ for which there is a common nonzero eigenvector $v$ with $A_{i}v=a_{i}v$. Commuting self-adjoint operators on a finite-dimensional inner product space admit a simultaneous orthonormal eigenbasis.