If $T\in \mathcal L(V)$ what is $\mathcal T|_U$? Consider for example $\mathcal T:\mathbb R^3\to \mathbb R^3$ s.t. $$(\mathcal T)_{BB}=\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix},$$
where $B=\{u_1,u_2,u_3\}$. Set $W=span\{u_2,u_3\}$ and $B'=\{u_2,u_3\}$. What is $T|_W$ ? Is it  
$$(\mathcal T|_W)_{B'B'}=\begin{pmatrix}2&3\\5&6\\8&9\end{pmatrix} \ \ ?$$

You actually cannot find a matrix $(\mathcal{T}|_W)_{B'B'}$, because ${\rm Im}(\mathcal{T}|_W) \not\subseteq W$, so we can just look for the matrix $(\mathcal{T}|_W)_{B'B}$ (or $_{B B'}$? I never remember which leter corresponds to the "input" basis and which one to the "output" basis; here $B'$ is the input one, $B$ is the output one) and that will be the one you've proposed:
For example, since $\mathcal{T}(u_2) = 2u_1 + 5u_2+8u_3$ (which is described via matrix $\mathcal{T}_{BB}$ as $\mathcal{T}_{BB}\cdot(0, 1, 0)_B^T = (2,5,8)_B^T$, the same must hold for $(\mathcal{T}|_W)_{B'B}$, where however $u_2 = (1,0)_{B'}$, so $(\mathcal{T}|_W)_{B'B}\cdot(1,0)^T_{B'} = (2,5,8)^T_B$. Similarly for $u_3 = (0,1)_{B'}$ and the conclusion follows.
