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Suppose V is a vector space and T is a linear transformation. If every subspace of V is T-invariant, prove that T is a scalar multiple of the identity map.

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Let $0\ne v\in V$. Then $kv$ is $T$-invariant, hence $Tv=a_v v$ for some $a_v\in k$. Then for any multiple $w\in kv$, clearly $Tw=a_vv$ as well. Likewise, if $w\in V\setminus kv$ we obtain $Tw = c_ww$ and $T(v+w)=c_{v+w}(v+w)$ with $c_w,c_{v+w}\in k$. From $T(v+w)=Tv+Tw$ we infer $$c_{v+w}(v+w)=c_vv+c_ww $$ i.e. $$(c_{v+w}-c_v)v + (c_{v+w}-c_w)w=0.$$ Now $v,w$ are linearly independant, so that we conclude $c_{v+w}=c_v=c_w$. In other words, $Tx=c_vx$ for all $x\in V$.

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  • $\begingroup$ Shouldn't it be $Tw = a_v w$? And given that you use $c_v$ further down, maybe you want to change $a_v$ into $c_v$ at the beginning? Or am misunderstanding your answer? $\endgroup$ Oct 20, 2014 at 11:58
  • $\begingroup$ Hello? I think your answer contains a few typos. $\endgroup$ Oct 26, 2014 at 7:15

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