Change in basis leads to change in bilinear matrix 
Let $V,W$ be finite dimensional vector spaces over $K$.Let $g:V\times W\to K$ be a bilinear map, written $g(v,w)=\langle v,w\rangle$. Show that for each pair of bases $\mathscr{B}$ of $V$ and $\mathscr{B}´$ of $W$ we can associate a matrix $C$, such that, if $v\in V$ and $w\in W$, and $X=M_{\mathscr{B}}(v)\:,\:Y=M_{\mathscr{B}}(v)$, then
$g(v,w)=X^{t}CY$

I do not understand this question.
I know that $\langle\mathscr{B}\mathscr{B}´\rangle$ gives rise to the matrix $C$. Therefore changes in the basis will change the matrix $C$.
Questions:
1) How can I prove that change in basis is reflected in the matrix $C$? Which procedures?
2) How can $X=M_{\mathscr{B}}(v)\:,\:Y=M_{\mathscr{B}}(v)$?
Thanks in advance!
 A: Let $\mathscr{B}=(v_1,\dots, v_n)$ and $\mathscr{B}’=(w_1,\dots, w_n)$. Let 
$v=\sum_{i=1}^n \lambda_i v_i$ and $w=\sum_{j=1}^m \mu_j w_j$. Then I guess that $X=M_{\mathscr{B}}(v)=(\lambda_1,\dots, \lambda_n)^t$ and 
$Y=M_{\mathscr{B}’}(w)=(\mu_1,\dots, \mu_m)^t$. Using bilinearity of $g$, we obtain $$g(v,w)=\sum_{i=1}^n\sum_{j=1}^m \lambda_i\mu_j g(v_i,w_j).$$
Thus if we put $C=\|c_{ij}\|=\| g(v_i,w_j)\|$, we obtain $g(u,v)=X^tCY$ .
That is for each pair of bases we can associate such a matrix $C$.
Now assume that we changed the bases $\mathscr{B}$ and $\mathscr{B}’$ to bases $\mathscr{B_1}$ and $\mathscr{B_1}’$, respectively. Let 
$$A=[M_{\mathscr{B_1}}(v_1)\dots M_{\mathscr{B_1}}(v_n)]$$ and 
$$B=[M_{\mathscr{B_1}’}(w_1)\dots M_{\mathscr{B_1}’}(w_n)].$$
Then $X'=M_{\mathscr{B_1}}(v)=AM_{\mathscr{B}}(v)=AX$ and 
$Y'=M_{\mathscr{B_1}’}(w)=AM_{\mathscr{B}’}(w)=BY$. 
Thus if we put $C’=\|c’_{ij}\|=\| g(v’_i,w’_j)\|$, we obtain 
$g(u,v)=X'^tC'Y'=X^tA^tC’BY$ for all $X^t\in K^n$ and $Y^t\in K^m$, which implies $A^tC’B=C$.
