Let $V$ be a finite dimensional inner product space with basis $B$ = {$v_1, v_2,..., v_n$} Then there exists a symmetric $n\times n$ matrix $A$ such that for all vectors $x,y∈V$.
$\langle x,y \rangle = [x]_B^TA[y]_B$
Prove this for the case $n=2$.
I believe I know what the question is asking, and I know what all of the terms mean in the problem. But I'm not sure how to start the proof at all.
I'll do it for $n=2$ as requested, but it generalizes quite easily. I'm also going to assume that your vector space is real, or at least that your inner product is symmetric and not conjugate symmetric.
Let $x=\alpha_1v_1+\alpha_2v_2$ and $y=\beta_1v_1+\beta_2v_2$. Then $\langle x,y\rangle=\langle \alpha_1v_1+\alpha_2v_2,\beta_1v_1+\beta_2v_2\rangle=\alpha_1\beta_1\langle v_1,v_1\rangle +\alpha_1\beta_2\langle v_1,v_2\rangle+\alpha_2\beta_1\langle v_2,v_1\rangle+\alpha_2\beta_2\langle v_2,v_2\rangle$.
What this shows is us that the value of the inner product is a function of the inner products of the basis vectors with each other.
Now if we let $a_{ij}=\langle v_i,v_j\rangle$ and call the matrix composed of these values $A$, it's not hard to verify via matrix multiplication that $[x]_b^TA[y]_b=\langle x,y\rangle$.