$T(u)=2u$ for $u\in Im(T)$ implies $V=Im(T)\oplus Ker(T)$ Let $V$ be a finite dimensional space. Let $T:V\to V$ a linear operator. If  $T$ satisfies that
$$T(u)=2u \hspace{5mm} \text{for} \hspace{2mm} u\in Im(T)$$
does this implies that $V=Im(T)\oplus Ker(T)$?
In particular, I am having difficulties proving that if $v\in V$, then $v\in Im(T) +Ker(T)$.
I tried doing something similar than the case when $T$ is idempotent, that is, writing $v$ as $v=T(v)+(v-T(v))$ and seeing that the term $T(v)$ is in the image and the term $(v-T(v))$ is in the kernel, but I get an extra $-T(v)$ term. Any help will be appreciated.
 A: If $T$ is the zero operator or if $T$ is invertible $\text{Im}(T)\oplus\text{Ker}(T) = V$ is immediate.
Otherwise let $r$ be the rank of $T$ and let $n$ be the dimension of $V$. Let $y_{r+1},\dots,y_n$ be a  basis of the kernel of $T$ and it can be extended to a basis of $V$ say $y_1,\dots,y_r,y_{r+1},\dots,y_n.$
It is sufficient to show $y_1,y_2,\dots,y_r$ span $\text{Im}(T).$
We show that $Ty_1,Ty_2,\dots,Ty_r$ forms a basis of $\text{Im}(T)$. 
To see that note that any $x \in V$ is of the form $x = a_1 y_1 + \dots + a_n y_n.$ 
So, $T(x) = a_1 T(y_1) + a_r T(y_r)$, since $T(y_{r+1})=\dots=T(y_n)=0.$
So $T(y_1),\dots,T(y_r)$ span $\text{Im}(T)$.
Next note if $a_1T(y_1)+ \dots + a_rT(y_r) = T(a_1y_1+\dots+a_ry_r)=0$ then $a_1y_1+a_2y_2+\dots+a_ry_r \in \text{Ker}(T) \in \text{span}\{y_{r+1},\dots,y_n\}$, but by the construction of $y_1,\dots,y_r$ this is possible only if  $a_1y_1+a_2y_2+\dots+a_ry_r=0$ and independence leads us to: $a_1=\dots=a_r=0$. 
So $T(y_1),T(y_2), \dots, T(y_r)$ froms a basis of $\text{Im}(T)$
or $2y_1,\dots,2y_n$, i.e., $y_1,y_2,\dots,y_r$ forms a basis of $\text{Im}(T).$ 
