Proving an injective map exists from $V$ to $W \iff$ dim $V$≤ dim$W$ I'm reading Axler's Linear Alg Done Right and working through problems. For this one, I proved the forward direction already using Fundamental Theorem of Linear Maps.  
And for the backwards direction I assumed dim $V$≤ dim $W$. Let $v_1,\ldots, v_n$ be a basis for $V$ and  $w_1,\ldots, w_n$ be a basis for $W$.  Then $T(a_1v_1 \ldots a_nv_n)=(a_1w_1\ldots a_mw_m)$. 
I know I first I have to prove this is indeed a linear map, which I can do. Then I prove this is injective. I know an injective map null($T$)=$0$. 
I'm stuck on  proving this map is injective.  Any help/guidance is appreciated. 
 A: The map $T$ is defined by
$$
T(a_1v_1+a_2v_2+\dots+a_nv_n)=a_1w_1+a_2w_2+\dots+a_nw_n
$$
(you forgot the + signs). It is injective by the fundamental theorem, because the image is spanned by $\{w_1,\dots,w_n\}$, which is linearly independent. Thus the image has dimension $n$.
A: I'm not sure how your map is defined.
Take $v_1, \dots,v_n$ a basis for $V$ and $w_1, \dots, w_n, w_{n+1}, \dots w_m$ a basis for $W$.
Try your idea again by sending $T(v_i)= w_i$ in other words
$T(a_1v_1+ \dots +a_n v_n)=a_1w_1+\dots +a_nw_n$
or in your notation,
$T(a_1, \dots,a_n)=(a_1, \dots, a_n,0, \dots ,0)$
My hint would be take a vector $a_1v_1+ \dots +a_n v_n=v \in V$
and suppose that $T(v)=a_1w_1+\dots +a_nw_n=0$.
Now what do we know about sums likes this when we have basis vectors?
A: 1) The statement is wrong because there are no restriction on map exept injectivity. So we can find injective map from $\mathbb R^2$ to  $\mathbb R$ because they are have the same cardinality.
2) The statement will be true if a linear injective mapping is implied
