Prove $B^TC = \mathbf 0$, given that $U = Col(B)$ , $W = Col(C)$, $U \bot W$. 
Prove $$B^TC = \mathbf 0$$ given that if $U = Col(B), W = Col(C)$, then $U \bot W$.

Well, I wrote that $$U \bot W \iff u^T w = 0,\;\;\;\;\forall u \in U,\; w \in W$$
But I'm not sure that I can deduce from that the $B^TC = \mathbf 0$. I'd like to know what do you guys think.
 A: For the multiplication to be possible $B$ must be a $n\times k$ and $C$ a $n\times l$ matrix, with $n,k,l\in \Bbb N$. Then, for $1\le i\le k$ and $1\le j\le l$ you have that the $(i,j)$-th element of the product matrix $B^TC$ is given by
$$(B^TC)_{ij}=B_{\cdot i}C_{\cdot j}=\sum_{k=1}^nB_{k i}C_{k j}$$ where $B_{\cdot i}$ denotes the $i$-th column of matrix $B$ and similarly $C_{\cdot j}$ the $j$-th column of matrix $C$. To see that the previous equation is indeed true, observe that the rows of $B^T$ are just the columns of $B$. Now the result is immediate by your observation since $B_{\cdot i} \in U$ and $C_{\cdot j}\in W$.

To visualize this sum (if you are not familiar with it) just try a simple $2\times 2$ example. Choose a simple $B$ for example $$B=\begin{pmatrix}1&-1\\0&0\end{pmatrix}$$ and pick $C$ to have columns perpendicular to the columns of $B$, for example $$C=\begin{pmatrix}0&0\\-1&1\end{pmatrix}$$ Now the element $(B^TC)_{(11)}$ of the product matrix $B^TC$ is given by $$B^TC_{(11)}=B_{\cdot 1}C_{\cdot 1}=\sum_{k=1}^2B_{k1}C_{k1}=1\cdot0+0\cdot(-1)=0$$
