Prove that $f_{\beta'} = (f_{\beta})^{-1}$ I'm stuck on this problem for few days and can't find the solution.Hope some one here can help me solve this. I'm so grateful for any any help:


Let $V$ be a finite-dimensional vector space and $f$ a non-degenerate symmetric bilinear form on $V$. With every basis $\beta = \{\alpha_{1}, ..., \alpha_{n}\}$ of $V$ there exists a unique basis $\beta' = \{\alpha_{1}', ..., \alpha_{n}'\}$ of $V$ such that $f(\alpha_{i}, \alpha_{j}') = \delta_{ij}$. Prove that $f_{\beta'} = (f_{\beta})^{-1}$ 


 A: Denote the bilinear $f$ represented in the basis $\beta$ by $U$ and in the basis $\beta'$ by $U'$. Then given any vector $x=\sum c_i \alpha_i=\sum c'_i\alpha'_i$ and $y=\sum d_i \alpha_i=\sum d'_i\alpha'_i$ we have
$$f(x,y)=c^TUd=c'^TU'd'$$
Further we have 
$$f(x,y)=\sum_{ij} c_id'_j f(\alpha_i,\alpha'_j)=\sum_{ij} c_id'_j \delta_{ij}=\sum_i c_i d'_i=c^Td'$$
Since this holds for all vectors $x$ and the bilinear form is non-degenerate it follows that $Ud=d'$ and also $c'^TU'=c^T\Rightarrow U'^Tc'=c\Rightarrow U'c'=c$, since $U'=U'^T$ due to symmetry.
Interchanging the roles of $x$ and $y$ we can conclude
in the same way as above $Uc=c'$ and $U'd'=d$. Now we have
$$c=U'c'=U'(Uc)=(U'U)c$$
Again, since this holds for any $x$, i.e. $c$, and thus $U'U=I$ follows.
I have the feeling, that this proof can be somewhat simplified, but I'll leave that up to you ;-)
A: In reality, I found another solution using 2 ways to calculate $\alpha_{i}'$.
Denote  $f_{\beta} = A$. 
For every $\alpha$ in $V$, we can express:
$\alpha = \sum_{i=1}^{n}b_{i}\alpha_{i}$
Then we have $f(\alpha, \alpha_{j}') = f(\sum_{i=1}^{n}b_{i}\alpha_{i}, \alpha_{j}') = b_{j}$. So we have the formula:
 $$\alpha = \sum_{i} f(\alpha, \alpha_{i}')\alpha_{i}$$
In particular, $\alpha_{i}' = \sum_{j} f(\alpha_{i}', \alpha_{j}')\alpha_{j}$
Denote $\beta_{i} = \sum_{i} (A^{-1})_{ij}\alpha_{j}$. Then we can easily prove that $f(\alpha_{i}, \beta_{j}) = \delta_{ij}$, but $\alpha_{j}'$ is unique, so we have $\beta_{j} = \alpha_{j}'$. Compare 2 formulas, then we have $f_{\beta'} = (f_{\beta})^{-1}$
I know this solution is not natural and complex. But I still want to post it here for everybody as a reference. Thanks :-)
