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Let $S,T:V \rightarrow V$ be linear operators such that $ST=TS$. Show that if $\lambda$ is an eigenvalue of T, then Ker$(T-\lambda{I})^{k}$, $k\in{\mathbb{Z}^{+}}$, is invariant under $S$.

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$S$ commutes with $T-\lambda I$, hence with $(T-\lambda I)^k$. Therefore if $v\in \ker((T-\lambda I)^k)$, then $$ (T-\lambda I)^kSv=S(T-\lambda I)^kv=0 $$ so $Sv\in \ker((T-\lambda I)^k)$ as well.

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