TITLE: Let V be a vector space. If every subspace of V is T-invariant, prove that there exist a scalar multiple c such that T=c1v
QUESTION [6 upvotes]: I wrote a proof for the above question, but I am not sure whether it is right or not since I assumed linear independence. 
Here's the proof: 
Let $u$,$v$ be linearly independent vectors in $V$.
$span(u)$, $span(v)$, $span(u+v)$ are all $T-invariant$. 
$T(v)$ is an element in $span(v) \Longrightarrow T(v) = av$
$T(u)$ is an element in $span(u) \Longrightarrow T(u) = bu$
$T(u+v)$ is an element in $span(u+v) \Longrightarrow T(u+v) = T(u)+T(v)= bu+av = c(u+v)$
Hence, $bu+av-c(u+v) = 0 \Longrightarrow (b-c)u + (a-c)v = 0$
Since u and v are linearly independent, $(b-c)=(a-c)=0 \Longrightarrow b=a$ 
Hence, $T(w) = kw$, for all $w$ in $V$, and $k$ any scalar. 
Is it right to assume linear independence? And is there any problem with my proof?

REPLY [4 votes]: First note that, by your initial work, one knows that for each $v\in V$, there exists $a_v\in \mathbb F$ such that $$Tv=a_v v$$
Because $T0=0$, we can choose $a_0$ to be any number in $\mathbb F$, but for $v\in V-\{0\}$ the value $a_v$ is determined by the equation above.
We must show that $a_v$ is independent of $v\in V-\{0\}$. To do this, suppose $v,w\in V-\{0\}$. We want to show that $a_v=a_w$. Consider the two cases:

If $(v,w)$ is linearly independent.

We have $$a_{v+w}(v+w)=T(v+w)$$
$$=Tv+Tw$$ $$=a_v v + a_w w$$ which implies that $$(a_{v+w}-a_v)v+(a_{v+w}-a_w)w=0$$
Then, because $(v,w)$ is linearly independent, $$a_w=a_{v+w}=a_v$$ as desired.

If $(v,w)$ is linearly dependent

Then there exists $b\in \mathbb F$ such that $w=bv$. Hence we have
$$a_w w=Tw$$
$$=T(bv)$$
$$=b(Tv)$$
$$=b(a_v v)$$
$$=a_v w$$ as desired.