Extending a linear map In exercise 11), section 3A, of Linear Algebra Done right, we're asked to prove:

V is a finite dimensional space. Prove that every linear map in 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 $S \in
> \mathcal{L}(U,W)$, then there exists $T \in \mathcal{L}(V,W) $ such
  that $Tu=Su \ \forall_{u\in S}$.

In this site, the suggested answer initially made sense, until I looked at exercise 10. It seems as if exercise 10 is stating that the function given in 11) is not a linear map...
When I was trying to solve this exercise, I was thinking of $V=span(u_1,...,u_n,v_1,...,v_m)$, where the $u$'s were a basis for $U$ and the whole set of $u$'s and $v$'s a basis for $V$. Then for any vector in $V$, $Tv=\sum^n c_i S(u_i) + \sum^m d_iT(v_i)$. The problem is what value may I give to $T(v_i)$? Not zero, because then, by 10), it would not be a linear map, right?
Any help would be appreciated.
 A: I have considered it more carefully since I wasn't satisfied with the conclusions being drawn in the comments and this is what I have determined. There is a subtlety between problem $10$ and this problem.
First, extend $u_1,\ldots,u_m$ to a basis $B = (u_1,\ldots,u_m,u_{m+1},\ldots,u_{m+k})$ of $V$. Define $T: B \to W$ (note this function is defined on a basis!) as
$$
Tu_i = \cases{Su_i, & $1 \le i \le m$ \\
0_W, & $m+1\le i\le m+k$.}
$$
Since $B$ is a basis of $V$, $u \in V$ is uniquely determined by the coefficients in its representation $u = \sum_{j=1}^{m+k}c_ju_j$. Define the linear extension $\tilde T: V \to W$ of $T$ by
$$
\tilde Tu = \tilde T(\sum_{j=1}^{m+k}c_ju_j) = \sum_{j=1}^{m+k}c_jTu_j. 
$$
You can check that this $\tilde T$ is well-defined, linear, and satisfies the problem's constraint that $\tilde Tu = Su$ for each $u \in U$.
Problem $10$ defines $T_{10}:V\to W$, the extension of $S:U\to W$ as
$$
T_{10}v = \cases{Sv & if $v \in U$ \\
0_W & if $v \in V$ but $v \notin U$.}
$$
The subtlety arises here. In problem $10$, the extension that I am calling $T_{10}$ of $S$ is defined so that if any vector in $V$ is not in $U$, it is automatically $0_W$. This is what stops $T_{10}$ from satisfying additivity. Observe that if $u \in U$ and $v \in V$ but $v \notin U$ then $(u + v) \notin U$. Hence $T_{10}(u + v) \ne T_{10}u + T_{10}v$.
Notice that this problem is resolved with $\tilde T$. If $u=\sum_{r=1}^mc_ru_r \in U$ and $v=\sum_{r=m+1}^{m+k}c_ru_r \in V$ but $v \notin U$ then $(u + v) \notin U$ as before. Observe that $v = \sum_{r=1}^{m+k}c_ru_r$, but for $1 \le r \le m$, we have that $c_r = 0$. We can now see that
\begin{align}
\tilde T(u + v) &= \tilde T(\sum_{r = 1}^{m}c_ru_r + \sum_{r = m+1}^{m+k} c_ru_r) =\tilde T(\sum_{r = 1}^{m+k}c_ru_r)\\
&= \sum_{r=1}^{m+k}c_rTu_r = \sum_{r=1}^{m}c_rTu_r + \sum_{r=m+1}^{m+k }c_rTu_r = Su + 0_W = \tilde Tu + \tilde Tv,
\end{align}
and this function does not suffer from the same problem that $T_{10}$ did! This analysis reveals that we cannot define the extension to be $0_W$ on arbitrary vectors in $V$ but not $U$. Instead, we have to individually assign each of the the basis vectors $u_i\mapsto 0_W$, and extend that function linearly in the manner I showed above.
A: $0$ is fine, as well as any vector you want.
10) is an example equal to $0$ on $V \setminus U$, in your case $T$ equals to $0$ on the (much smaller) complementary space of $U$.
