Let $T$ be a linear operator on vector space $V$.Suppose W is an invariant subspace of $V$ under $T$. Prove that it $\vec{v_1},\cdots,\vec{v_k}\in V$ are eigenvectors of $T$ corresponding to distinct eigenvalues $\lambda_1,\cdots,\lambda_k\in \mathbb{F}$ such that $\vec{v_1}+\cdots+\vec{v_k}\in W$,then $\vec{v_i}\in W$ for all $i = 1,\cdots,k$.How do I use induction on k to prove that?


1 Answer 1


$W$ is also invariant for $T-\lambda_k \operatorname{id}$, hence $W$ contains $$(T-\lambda_k \operatorname{id})(v_1 + \dotsb v_k) = \mu_1v_1 + \dotsb \mu_{k-1}v_{k-1}$$ where all $\mu_i = \lambda_i-\lambda_k$ are non-zero by assumption for $i=1, \dotsc, k-1$.

Note that all $\mu_i v_i$ are still eigenvectors of $T$ for distinct eigenvalues, hence you can use induction and get $\mu_iv_i \in W$ for $i = 1, \dotsc, k-1$.

This of course shows $v_1, \dotsc, v_{k-1} \in W$ and $v_k \in W$ is now a trivial consequence.

  • $\begingroup$ how to get $\mu_iv_i \in W$ by induction? $\endgroup$
    – user227158
    Oct 25, 2017 at 10:09
  • $\begingroup$ It is just the statement for $k-1$ instead of $k$...That is the whole point of induction... $\endgroup$
    – MooS
    Oct 25, 2017 at 10:44

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