Considerthe canonical basis $(e_1,e_2,e_3)$ and its dual basis $e^1 , e^2 , e^3$. Let $A = (a_{ij})$ a matrix in $M_{3 \times 3}(\mathbb{R})$ given by: $$A = \begin{bmatrix} 1 & -3 & 1\\ -3 & -3 & 3\\ 1 & 3 & 1\\ \end{bmatrix}$$ Now consider the following basis of eigen vectors $f_1,f_2,f_3$ from $A$ each of them given, explicitly by: $$B = \{(1,0,1),(1,-1,-1),(1,2,-1)\}$$ and its dual basis $(f^1,f^2,f^3)$.
I have the following $(0,2)$-tensor: $$s(x^i e_i,y^i e_i) = a_{ij}x^{i}y^{j}$$ How can I write the same tensor but as a linear combination of tesonsotrial products between the elements of $(f^1,f^2,f^3)$ basis? For instance I know that $s$ in terms of canonical basis could be expreed as: $$s(x^i e_i,y^i e_i) = a_{ij}(e^i \otimes e^j)(x^i e_i,y^i e_i)$$ My question is what should I put as $\star$ such that the following becomes true: $$s(x^i f_i,y^i f_i) = \star(f^i \otimes f^j)(x^i f_i,y^i f_i)$$