<s>[INST] If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$why $A = \frac{1}{2} A + \frac{1}{2} A^\top $ is not always true? If $A \in \mathbb{R}^{n\times n}$  and $\textbf{x}\in \mathbb{R}^{n}$, 
It is possible to prove that
$$
\textbf{x}^\top A \textbf{x} = \textbf{x}^\top(\frac{1}{2} A + \frac{1}{2} A^\top)\textbf{x}
$$
That could let us think that 
$$
A = \frac{1}{2} A + \frac{1}{2} A^\top ?
$$
However if we take
$$A = \begin{bmatrix}1&2\\ 3&4 \end{bmatrix}$$
and compute
$$B = \frac{1}{2} A + \frac{1}{2} A^\top$$
we get $$B = \begin{bmatrix}1&2.5\\ 2.5&4 \end{bmatrix}\ne A$$
Although
$$
\textbf{x}^\top A \textbf{x} = \textbf{x}^\top B \textbf{x}
$$
Here is my question:
How can we explain that the following is not always true?
$$
A = \frac{1}{2} A + \frac{1}{2} A^\top
$$
 [/INST] In the product $\mathbf{x}^\top A \mathbf{x}$ we get terms like $x_1 A_{12} x_2 + x_2 A_{21} x_1 = (A_{12}+A_{21}) x_1 x_2$ so that for example $\begin{bmatrix}1&2.5\\ 2.5&4 \end{bmatrix}$, $\begin{bmatrix}1&5\\ 0&4 \end{bmatrix}$ and $\begin{bmatrix}1&3\\ 2&4 \end{bmatrix}$ give the same result since $2.5+2.5 = 5+0 = 3+2$. Thus only the symmetric part $\frac12(A+A^\top)$ matters; the antisymmetric part $\frac12(A-A^\top)$ does not matter.
</s>