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Let $T$ be a linear operator on a vector space $V$, and let $c\in F$. Let $W$ be the set of eigenvectors of $T$ with eigenvalue $c$, together with $0$. Prove that $W$ is a $T$-invariant subspace.

So, I need to show $T(W)\subset W$. If I show that $\ker (T-cI)$ is $T-cI$-invariant. But $(T-cI)(v)=0\in \ker (T-cI)$, when $v\in \ker (T-cI)$. Is my argument ok? Thanks.

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In my opinion, you need to flush out your argument a bit more to make sure it is clear. Alternatively, you can do this by just directly applying $T$ to a element of $W$ and showing that the element stays in $W$. That is, take $w \in W$ so that $w$ is an eigenvalue of $T$ corresponding to eigenvalue $c$. By definition, this means $Tw = cw$. But then $T(Tw) = T(cw) = c(Tw)$. This shows that $Tw$ is also an eigenvalue of $T$ corresponding to eigenvalue $c$ and so $Tw \in W$. Thus $W$ is $T$ invariant.

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  • $\begingroup$ Yeah! that was easy, I made it over-complicated. Thanks a lot. $\endgroup$ – algebra_001 Oct 22 '17 at 7:47

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