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?
$$
A = \frac{1}{2} A + \frac{1}{2} A^\top
$$
 A: The equality $x^\top\left(\frac12A+\frac12A^\top\right)x=x^\top Ax$ holds indeed, but you cannot deduce from it that $\frac12A+\frac12A^\top=A$. In fact, you also always have $x^\top Ax=x^\top A^\top x$, but it is not true in general that $A=A^\top$.
A: 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.
A: 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.
A: 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.
