Linear Maps Proof

[spice mantle]

Since $U$ is a subspace of $V$, we have $\dim U \le \dim V$. Let $\dim U = m$ and $\dim V = n$, and suppose ${v_1, \ldots, v_m}$ is a basis of $U$. This basis is linearly independent in V and can thus be extended to a basis of V: ${v_1, \ldots, v_n}$. Now suppose $S \in \mathscr{L} (U,W)$. S maps the basis vectors ${v_1, \ldots, v_m}$ to vectors in W. Suppose $Sv_i = w_i$ for i \in {1, \ldots, m}$
For $i \in {m+1, \ldots, n}$, let $w_i$ be arbitrary vectors in $W$.
By the Linear Map Lemma, there exists a unique $T \in \mathscr{L} (V,W)$ such that $Tv_i = w_i$ for $i \in {1, \ldots, n}$. For $i \in {1, \ldots, m}, Tv_i = Sv_i = w_i$
For each basis vector in $v_1, \ldots, v_m$ of $U$, since $T$ and $S$ map it to the same element in $W$, it follows that $Tu = Su$ for all $u \in U$. Thus, such a $T \in \mathscr{L} (V,W)$ exists.