Is the following Proof Correct?
Theorem. Given that both $V$ and $W$ are finite-dimensional. There exists a injective linear map from $V$ to $W$ if and only if $\dim V\leq \dim W$
Proof. $(\Rightarrow).$ Assume for the purpose of contradiction that $T:V\to W$ is an injective linear map and that $\dim V>\dim W$ thus by making use of the Fundamental Theorem of Linear Maps we see that $$\dim \operatorname{null}\ T = \dim V-\dim \operatorname{range}\ T$$ $$\dim \operatorname{null}\ T \ge \dim V-\dim W$$ $$\dim \operatorname{null}\ T > 0$$ but $T$ is injective and thus $\operatorname{null}T = \{0\}$ implying that $\dim \operatorname{null}\ T=0$ resulting in a contradiction.
$(\Leftarrow).$ Assume now that $\dim V\leq\dim W$ since both $V$ and $W$ are finite dimensional we may invoke the existence of the vectors $v_1,v_2,...,v_m\in V$ and $w_1,w_2,...,w_n\in W$ such that they act as basis for $V$ and $W$ respectively, where $m\leq n$.
Consider now the linear map $T\colon V\to W$ defined as follows. $$\forall j\in\{1,2,...,m\}(Tv_j = w_j)$$ Let $u_1$ and $u_2$ be arbitrary members of $V$ and assume that $Tu_1=Tu_2$, in light of $(1)$ we may express $Tu_1$ and $Tu_2$ as follows $$Tv_1 = T\left(\sum_{j=1}^{m}a_jv_j\right)=\sum_{j=1}^{m}a_jTv_j=\sum_{j=1}^{m}a_jw_j\tag{2}$$ $$Tv_2 = T\left(\sum_{j=1}^{m}b_jv_j\right)=\sum_{j=1}^{m}b_jTv_j=\sum_{j=1}^{m}b_jw_j\tag{3}$$ subtracting $(2)$ from $(3)$ we see that $$0=\sum_{j=1}^{m}(b_j-a_j)w_j\tag{4}$$ we know that $w_1,w_2,...,w_n$ is a linearly independent list in $W$ consequently $w_1,w_2,...,w_m$ is also linearly independent, it then follows that $$\forall j\in\{1,2,...,m\}(a_j=b_j)\tag{5}$$ consequently $u_1=u_2$ as every vector in $V$ is uniquely expressed as a linear combination of $v_1,v_2,...,v_m$.
$\blacksquare$
