Suppose $V$ is finite dimensional. Prove that every linear map on a subspace of $V$ can be extended to a linear map on $V$. In other words, show that if $U$ is a subspace of $V$ and $S \in \mathcal L(U,V)$, then there exists $T \in \mathcal L(V,V)$ such that $Tu = Su$ for all $u \in U$.

Let $u_1,\dots,u_m$ be a basis for $U$ then extend it to a basis $u_1,\dots,u_m,v_1,\dots,v_n$ of $V$ using 2.33. Define $Tu_j = Su_j$ and $Tv_j = 0$ for all $j$.

$T$ is clearly linear because we defined it in terms of how it acts on the basis vectors. And clearly $Su = Tu$.

This is fun to contrast with 3a/10 where we proved the naive extension doesn't work