= Solution
Let $Av=\lambda v$ with $v\ne0$. A vanishing <commutator> gives $A(Bv)=B(Av)=\lambda Bv$, so $Bv$ belongs to the <eigenspace> $E_\lambda$. This is <eigenspace invariance for commuting linear maps>. A <nondegenerate eigenvalue> has a one-dimensional <eigenspace>, so $Bv=\beta v$ for a <scalar> $\beta$; $\beta=0$ is allowed. Therefore \b[$v$ is also an <eigenvector> of $B$]. Without the simple-<eigenvalue> condition, $B$ can mix <vectors> inside a larger <eigenspace>, and an arbitrarily chosen <eigenvector> of $A$ need not be an <eigenvector> of $B$.
Back to article page
