Linear Transformation, if $T(x) \neq 0$ for some $x \in V_1$, prove that $Tx \neq 0 \iff x \neq 0$ I don't know if it's true, but it seems to be, altought I cannot prove it yet.
I'm trying to find a way of write any other vector as a sum of $\textbf{x}$ with another vector (I don't know if it will work because I don't know if $T\textbf{y} = 0, \textbf{y} \neq \textbf{x}$) and multiplication by scallar, but don't know how.
I think it's pretty simple but I can't figure out.

This is false.  All you need for a counterexample is a nonzero linear transformation which is not injective.  For instance, the one whose matrix rel the standard basis for $\Bbb R^3$ is $\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}$.
