= Eigenspace invariance for commuting linear maps
{title2=$AB=BA\implies B(E_\lambda(A))\subseteq E_\lambda(A)$}
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 $B$.
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