Commuting maps preserve eigenspaces ($C(E_{l}ambda(D))subseteq{E}_{l}ambda(D)$)

If $CD=DC$ and $Dx=\lambda x$, then $D(Cx)=C(Dx)=\lambda Cx$. Therefore each eigenspace of $D$ is invariant under $C$. The vector $Cx$ may be zero; invariance does not require it to be a nonzero eigenvector. This observation is the starting point for simultaneous diagonalization.