Let $A\in \mathbb{R}^{n\times n}$ be diagonalizable, then there is and symmetric possitive definite matrix $C\in \mathbb{R}^{n\times n}$ such that $A$ is symmetric w.r.t $\langle\cdot,\cdot\rangle_C$, where $\langle x,y\rangle_C = \langle Cx,y\rangle$.
I have no idea to find $C$. Can anyone give me a hint? Thank you in advance !
Let $\mathcal{B} = (v_1, \ldots, v_n)$ be a basis of $\mathbb{R}^n$ consisting of eigenvectors of $A$. Define an inner product $(\cdot, \cdot)$ on $\mathbb{R}^n$ by letting $(v_i, v_j) := \delta_{ij}$ and extending bilinearly. With respect to the inner product $(\cdot, \cdot)$, we have
$$ (Av_i, v_j) = (\lambda_i v_i, v_j) = \lambda_i (v_i, v_j) = \lambda_i \delta_{ij} = \lambda_j \delta_{ij} = (v_i, \lambda_j v_j) = (v_i, Av_j) $$
so $A$ (or more precisely, the linear map $T_A \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ associated to $A$, given by $T_A(v) = Av$) is symmetric.
Finally, show that given any inner product $(\cdot, \cdot)$ on $\mathbb{R}^n$, there exists a positive definite matrix $C$ such that
$$ (v, w) = \left< Cv, w \right> \,\,\,\,\,\, \forall v,w \in \mathbb{R}^n $$
where $\left< \cdot, \cdot \right>$ is the standard inner product on $\mathbb{R}^n$.