I have to clarify a bunch of things. My class material is going a little over my head.
Let $V$ be a real inner product space, and $B : V \times V \to \mathbb{R}$ a bilinear form in an orthonormal basis $C$.
I am confused about what makes $B$ symmetric and how this relates to the matrix elements and its basis.
$B$ can be represented with an inner product $B(x,y) = \left< x, Ay \right>$
In this orthonormal basis $B(x,y) = \sum_{j,k=1}^{n,n} x_j y_k A_{jk}$
According to this equation it is obvious that if the matrix elements of $B$ are symmetric then $B(x,y) = B(y,x)$.
However, suppose the matrix elements of $B$ are not symmetric, but that $B$ is diagonalizable (I don't know whether nonsymmetry and diagonalizable is possible). If this is possible, then it seems to me that using the criterion of whether matrix elements are symmetric is a misnomer, because in another basis they might be.
This means I have to check whether $B$ is diagonalizable in order to know whether it is symmetric. I can't merely look at the matrix elements (this is for a homework assignment).
If $A$ is symmetric, then it can be decomposed into $$ A = U^T DU $$ where $U=(u_1,\dots,u_n)$ is your orthonormal basis such that your scalar product B induced by A is diagnoal. Simply being diagonalizeable only requires $$ A = U^{-1} D U $$ with diagonal $D$. But if you want your new basis $U$ to be orthogonal, then you necessariliy have $\langle u_i, u_j\rangle = \delta_{ij}$ and this implies $$ U^T U = \mathbb{I} $$ and therefore $U^{-1}=U$. So if you find an orthonormal basis in which $A$ is diagonal, then you necessarily have $U^{-1}=U^T$ and thus $$ A = U^T DU = (U^T DU)^T = A^T $$ So $A$ must be symmetric. Together with our first statement we find, that this is an "if and only" if condition.