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Let $V$ be a vector space over a field $F$ and let $T : V \to V$ be a linear operator. Show $T(W) ⊂ W$ for any subspace $W$ of $V$ iff there is a $\lambda \in F$ such that $Tv = \lambda v$ for all $v \in V$.


My progress in it is:

Let $v \in V$, $v \ne 0$. Then $[v]$ is a one dimensional subspace of $V$ and $T[v] ⊂ [v]$, so, $Tv = \lambda v$ for some $\lambda \in F$.

I'm able to prove that if $v_1, v_2 \in V$ and $Tv_1 = \lambda_1v_1$ and $Tv_2 = \lambda_2v_2$, then $\lambda_1=\lambda_2$. But how can I prove the converse

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  • $\begingroup$ I think your question is fine, but you should edit your title so that it better describes your specific question. This will make it easier for others to find and probably draw more attention to the question. $\endgroup$ – Tyberius Apr 6 '17 at 5:15
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Suppose that $T(v) = \lambda v$ for all $v$. Then given a subspace $W$, for any $w \in W$, $Tw = \lambda w \in W$, so $T(W) \subset W$. This is the easy part.

The other way, suppose that $T(W) \subset W$ for all $W$. Pick a basis of $W$, say $w_1, w_2 , .., w_n$. You have concluded that $Tw_i = \lambda_i w_i$ for all $i$, for some $\lambda_i$, by considering the subspaces generated by each basis element separately.

Now, note that $T([w_1 + w_2]) \subset [w_1 + w_2]$, so this means that $T(w_1 +w _2) = \lambda(w_1+w_2)$. From this, you see that $\lambda_1 = \lambda_2 = \lambda$. Doing this for other indices, we get that $Tv = \lambda v$ for all $v$.

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