Linear transformation: $T(T(x)) = T(x) + x$ for any $x$ in $\mathbb{R}^3$. Prove that $T$ is injective and surjective. I'm having trouble with this particular question in linear transformation. Much thanks if anyone would help me out on this question!

$T(T(x)) = T(x) + x$ for any $x$ in $\mathbb{R}^3$. Prove that $T$ is injective and surjective.

 A: Given any $v,w\in\mathbb R^3$, let $x=T(v), y=T(w)$ then
\begin{array}{}
T(x)=x+v\\
T(y)=y+w
\end{array}
Now assume that $T(v)=T(w),$ then $x=y,$ so
\begin{align}
T(x)=x+v=y+w=T(y)\implies v=w.
\end{align}
So $T$ is injective, and since $T$ linear and $\dim\mathbb R^3\lt\infty$, one-one implies onto.
A: If 
$T(T(x)) = T(x) + x = x + T(x), \tag 1$
and
$Tx = 0, \tag 2$
then
$x = x + 0 =  x + T(x) = T(T(x)) = T(0) = 0; \tag 3$
that is
$\ker T = \{0\}, \tag 4$
and so $T$ is injective.  We further note that for any $y \in \Bbb R^3$, (1) yields
$y = T(T(y)) - T(y) = T(T(y)) - y), \tag 5$
which shows that every $y$ is the image under $T$ of $T(y) - y$, so $T$ is surjective as well, which may also be concluded directly from (4), since $T$ operates on the finite-dimensional vector space $\Bbb R^3$; in finite dimensions we have "injective $\Longleftrightarrow$ surjective", as is well-known.
N.B:  It is perhaps worth observing that there is nothing intrinsically $3$-dimensional about this result; it apparently holds for $T:V \to V$, where $V$ is any vector space over any field $\Bbb F$, whether or not $\dim V < \infty$; of course, in that case we can't us our little "injective $\Longleftrightarrow$ surjective" trick; we may then instead rely on (5). End of Note.
