determine if the linear transformation is injective ,or surjective $T: \Bbb R^3 \to \Bbb R^4$ given $T(1,-1,3)=(2,0,-2,4)$ $T(2,1,4)=(3,0,-3,6)$ 
determine if the linear transformation is injective ,or surjective $T: \Bbb R^3 \to \Bbb R^4$ given $T(1,-1,3)=(2,0,-2,4)$ $T(2,1,4)=(3,0,-3,6)$ the answer in the textbook says that it is none

My try:
in order for it to be surjective then we need $dimImT=dim \Bbb R^4 =4$ and according to $dimkerT+dimImt=n$ we get $dimkerT+4=3$ which is not possible
and in order to prove injective we need $kerT=\{0\}$ so $ImT \subseteq \Bbb R^4 $ so $dimImT \leq dim \Bbb R^4=4$ but $dimkerT+dimImT=3$ so $dimImT \leq3$ but how can I get from here to $kerT\not=\{0\}$ can $dimkerT=0$? or is it necessarily  $dimkerT \geq 1$ if it has to be greater than one then I got the answer
is there a way I could check this with the given information $T(1,-1,3)=(2,0,-2,4)$ $T(2,1,4)=(3,0,-3,6)$?
thank you!
 A: As @Mark has suggested in the comments, a linear mapping between finite dimensional vector spaces is injective if and only if it maps LI sets into LI sets.
Indeed, if $T:V\to W$ is a linear transformation between finite dimensional vector spaces and we assume that $\dim V = n$ and $\mathcal{B} = \{v_{1},v_{2},\ldots,v_{n}\}\subset V$ is linear independent, we conclude based on the given assumption that
\begin{align*}
\alpha_{1}T(v_{1}) + \alpha_{2}T(v_{2}) + \ldots + \alpha_{n}T(v_{n}) = 0 & \Rightarrow T(\alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{n}v_{n}) = 0\\\\
& \Rightarrow \alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{n}v_{n} = 0\\\\
& \Rightarrow \alpha_{1} = \alpha_{2} = \ldots = \alpha_{n} = 0
\end{align*}
Hence the set $\mathcal{B}' = \{T(v_{1}), T(v_{2}),\ldots, T(v_{n})\}$ is LI, and we are done.
Conversely, if $T$ maps LI vectors on LI vectors, then it is injective.
In order to conclude so, we can proceed as follows:
\begin{align*}
T(\alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{n}v_{n}) = 0 & \Rightarrow \alpha_{1}T(v_{1}) + \alpha_{2}T(v_{2}) + \ldots + \alpha_{n}T(v_{n}) = 0\\\\
& \Rightarrow \alpha_{1} = \alpha_{2} = \ldots = \alpha_{n} = 0\\\\
& \Rightarrow v = 0
\end{align*}
Now it comes the application of such claim to the specific case of your exercise.
Observe that $(1,-1,3)$ and $(2,1,4)$ are LI, but their corresponding images under $T$ are not (one is proportional to the another). Hence there is no possibility that $T$ is injective.
Hopefully this helps !
