linear mapping question If $f$ is a linear mapping from $R^3$ $\rightarrow$ $R^3$, and $fof$=$2$$f$.
Suppose that $v$ $\in$ $Imf$
Can i say that $fof$($v$)=$2$$f$($v$) $\Rightarrow$ $fof$($v$)=$f$($2$$v$) $\Rightarrow$ $f^{-1}$($f$($f$($v$))=$f^{-1}$($f$($2$$v$)) $\Rightarrow$ $f$($v$)=$2$$v$ ? i know the other way by considering an element $u$ $\in$ $R^3$ s.t $f$($u$)=$v$ and its done. but am i able to use the method i wrote before? 
($N.B$: since $v$ $\in$ $Imf$ $\Rightarrow$ $v$ $\in$ $R^3$ so that $f$($v$) exists.)
 A: Take any basis in $\Bbb R^3$ and write the matrix $A$ of the map $f$ in the basis. YOu initial condition essentially says that $$A^2=2A.$$This implies that the eigenvalues of $A$ lie in the set $\{0,2\}$.
Without losing generality we can suppose that $A$ is already in its Jordan normal form. There are several possibilities for the JNF of this matrix:
$$\begin{pmatrix}
2&1&0\\0&2&1\\0&0&2
\end{pmatrix}, \begin{pmatrix}
2&1&0\\0&2&0\\0&0&2
\end{pmatrix}\\
\begin{pmatrix}
2&1&0\\0&2&0\\0&0&0
\end{pmatrix},\begin{pmatrix}
2&0&0\\0&2&0\\0&0&2
\end{pmatrix}\\
\begin{pmatrix}
0&1&0\\0&0&0\\0&0&2
\end{pmatrix},\begin{pmatrix}
0&1&0\\0&0&0\\0&0&0
\end{pmatrix}\\
\begin{pmatrix}
2&0&0\\0&0&0\\0&0&0
\end{pmatrix},\begin{pmatrix}
0&0&0\\0&0&0\\0&0&0
\end{pmatrix},\begin{pmatrix}
2&0&0\\0&2&0\\0&0&0
\end{pmatrix}
$$
It is easy to see that any JNF with a Jordan cell of order $2$ or $3$ does not fit the description $A^2=2A$. One can argue that this is the consequence of the fact that all roots of the polynomial $x^2-2x$ are simple.
Therefore, the only options we have are 
$$ 
 \begin{pmatrix}
2&0&0\\0&2&0\\0&0&2
\end{pmatrix},
\begin{pmatrix}
2&0&0\\0&0&0\\0&0&0
\end{pmatrix},\begin{pmatrix}
0&0&0\\0&0&0\\0&0&0
\end{pmatrix},\begin{pmatrix}
2&0&0\\0&2&0\\0&0&0
\end{pmatrix}
$$
Finally, in the general case the matrix $A$ is described as $P^{-1}JP$ where $P$ is an arbitrary invertible matrix and $J$ is one of the four options above.
In terms of linear operators it means that there exists a basis $v_1,v_2,v_3$ such that $f(v_i)= l_iv_i$ with $l_i\in\{0,2\}$. Essentilly, we have only four such operators. The operator $\forall w\,f(w)=2w$ is only one of them.
