Quadratic forms of two real matrices are equal

Question

I have the expression $$x^TAx=x^TBx$$ where $x\in \mathbb{R}^{n\times1}$ and $A,B\in \mathbb{R}^{n\times n}$. In addition, $A$ is known to be symmetric.

What can be argued for matrix $B$?

Answer 1

$$ \frac{1}{2} \left( B + B^T \right) = A $$

first, in general, square matrices. $$ x^T (E+F) x = x^T Ex + x^T F x $$

Next, if $G^T = - G.$ Remember that the transpose of a scalar ( a one by one matrix) is itself. $$ x^T G x = ( x^TGx)^T = x^T G^T x = - x^T G x, $$ so $$ 2 x^T G x = 0. $$ The skew symmetric part of a square matrix contributes nothing to the quadratic form, only the symmetric part contributes.

Answer 2

$x^TAx=x^TA^Tx$ $[$ $A$ is symmetric $]$
$=(x^TAx)^T=(x^TBx)^T=x^TB^Tx.$

$\Rightarrow x^TBx=x^TB^Tx$

Therefore, $B$ is also a real symmetric matrix.

Edit:

Combining the above argument with @Wills answer,

We Get $A=B.$