Given $x \in \mathbb{R}^n$ and $A \in \mathbb{R}^{n \times n}$ show that there exists a symmetric matrix $B \in \mathbb{R}^{n \times n}$ (which means $B^T = B$) for all A such that: $$x^TBx=x^TAx$$
This statement totally makes sense to me when I try it with different examples but not sure how to formally prove it.
You just have to balance the coefficients of the quadratic form given by $A$. Just take $b_{ij}=(a_{ij}+a_{ji})/2$.
Note that $A = \dfrac 12(A + A^T) + \dfrac 12(A - A^T)$.
Since $x^TAx$ is a scalar, then $x^TAx = (x^TAx)^T = x^TA^Tx$
Let $B = \dfrac 12(A + A^T)$. Then
\begin{align} x^TAx &= \dfrac 12x^T(A + A^T)x + \dfrac 12x^T(A - A^T)x \\ &= x^TBx + \dfrac 12x^TAx - \dfrac 12x^TA^Tx \\ &= x^TBx + \dfrac 12x^TA^Tx - \dfrac 12x^TA^Tx \\ &= x^TBx \end{align}
