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

Observe that
$A = \dfrac{1}{2}(A + A^T) \Longleftrightarrow A = A^T, \tag 1$
that is, 
$A = \dfrac{1}{2}(A + A^T) \tag 2$
if and only if $A$ is a symmetric matrix.  This is easily seen as follows:  if 
$A = A^T, \tag 3$
then
$A = \dfrac{1}{2}(2A) = \dfrac{1}{2}(A + A) = \dfrac{1}{2}(A + A^T); \tag 4$
going the other way, if
$A = \dfrac{1}{2}(A + A^T), \tag 5$
then
$A = \dfrac{1}{2}A + \dfrac{1}{2}A^T, \tag 6$
whence
$\dfrac{1}{2}A = A - \dfrac{1}{2}A = \dfrac{1}{2}A^T, \tag 7$
so
$A = A^T. \tag 8$
We thus see that (2) is false if $A$ is not symmetric.
Note we need not introduce $\mathbf x \in \Bbb R^n$ to establish this.
