Eigenspace invariance for commuting linear maps ($AB=BA⟹B(E_{\lambda }(A))\subseteq E_{\lambda }(A)$)

If two linear maps commute and $Av=\lambda v$, then $A(Bv)=B(Av)=\lambda Bv$. Hence $B$ preserves each eigenspace of $A$. On a one-dimensional eigenspace, $Bv$ is a scalar multiple of $v$, so each eigenvector for a simple eigenvalue of $A$ is also an eigenvector of $B$. For repeated eigenvalues the invariant eigenspace can have higher dimension; arbitrary vectors in it need not individually be eigenvectors of $B$.