Show that $\quad \vec{x}^TA\vec{x}=\frac{1}{2}\vec{x}^T(A^T+A)\vec{x}$ Let $A\in\mathbb{R}^{n\times n}$ and $\vec{x}\in\mathbb{R}^{n\times 1}$
Show that $\quad \vec{x}^TA\vec{x}=\frac{1}{2}\vec{x}^T(A^T+A)\vec{x}$
My try:
We know that $$A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)$$
Then $$\vec{x}^TA\vec{x}=\vec{x}^T(\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T))\vec{x}$$
$$=\frac{1}{2}\vec{x}^T((A+A^T)+(A-A^T))\vec{x}$$
$$=\frac{1}{2}\vec{x}^T((A+A)\vec{x}$$
Which is not quite what I want. Any suggestions of how to keep going would be great!
 A: It's trivial once you show that $x^\top Ax = x^\top A^\top x \qquad$ :)
Note that $(x^\top A x)^\top = x^\top (x^\top A)^\top = x^\top A^\top x$
But $(x^\top A x)$ is a scalar and the transpose of a scalar equals itself! So we must have that $$(x^\top A x) = (x^\top A x)^\top = x^\top A^\top x$$
And the rest of the proof follows immediately
A: Since $\vec{x}^\top \!A\, \vec{x}$ is a scalar (a $1 \times 1$ matrix), it is always equal to its transpose, thus
$$
\vec{x}^\top \!A\, \vec{x} 
= \bigl( \vec{x}^\top \!A\, \vec{x} \bigr)^\top 
= \vec{x}^\top \!A^\top\, \vec{x}, 
$$
and so
$$
\vec{x}^\top \bigl( A - A^\top \bigr)\, \vec{x} 
= \vec{x}^\top \!A\, \vec{x} - \vec{x}^\top \!A^\top\, \vec{x} 
= 0.
$$
You can probably finish it from here.
A: Note that
$$
\frac{1}{2}x^T (A+A^T) x 
= \frac{1}{2}(x^T (A x + A^T x))
= \frac{1}{2}(x^T A x + x^T A^T x)
= \frac{1}{2}(x^T A x + (x^T A x)^T).
$$
Since $x^T A x$ is a scalar, it follows that $x^T A x = (x^T A x)^T$, so
$$
\frac{1}{2}(x^T A x + (x^T A x)^T)
= \frac{1}{2}(x^T A x + x^T A x)
= \frac{1}{2}(2x^T A x)
= x^T A x.
$$
