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?
$$
A = \frac{1}{2} A + \frac{1}{2} A^\top
$$

It is true that if
$$x^T A y = x^T B y$$
for all vectors $x,y$, then $A=B$. The proof is by plugging in basis vectors $e_i$ (with a $1$ in the $i$th entry, and zeroes elsewhere):
$$A_{ij} = e_i^T A e_j = e_i^T B e_j = B_{ij}.$$
Now if you are forced to multiply by the same vector on both sides, instead of two arbitrary vectors, you can no longer conclude
$$x^TAx = x^TBx \quad \not\Rightarrow \quad A=B,$$
and you have already found a counterexample. You can try to extend the above argument, and easily prove that $A_{ii} = B_{ii}$, but you have no way to isolate the off-diagonal terms.
