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 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? 
 A: 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.
A: The line:

Hence, $T(w)=kw$, for all $w$ in $V$, and $k$ any scalar.

isn't justified (where does this $k$ come from?), and I don't understand why you write "$k$ any scalar". It reads like $k$ could depend on $w$, which isn't what you want. Here's how one could fill the missing details:
For every $u \neq 0$, there is a scalar $\lambda_u$ such that $T(u) = \lambda_u u$. If $V = 0$ there's nothing to prove, otherwise pick a fixed nonzero $u_0 \in V$ and call $\lambda = \lambda_{u_0}$. Now for all nonzero $v \in V$, either:


*

*$u$ and $v$ are linearly independent, and then you have proved that $\lambda_v = \lambda$;

*or they are linearly dependent, therefore $v = au_0$ for some scalar $a$. But then $T(v) = \lambda_v v = a \lambda_v u_0 = T(au_0) = a T(u_0) = a \lambda u_0 \implies \lambda_v = \lambda$ ($a \neq 0$ because $v \neq 0$).

*and obviously $T(0) = \lambda 0$.


Finally $\lambda_v = \lambda$ for all $v$ and $T(v) = \lambda_v v = \lambda v$ for all $v$, where $\lambda$ doesn't depend on $v$.
