Let $f(x) = x^\top Q \, x$, where $Q \in \mathbb R^{n×n}$ is NOT symmetric. Show that the Hessian is $H_f (x) = Q + Q^\top$ 
Let $$f(x) = x^\top Q \, x$$ be a quadratic form, where $Q \in \mathbb R^{n×n}$ is NOT symmetric. Show that the Hessian matrix is $$H_f (x) = Q + Q^\top$$ Hint: $x^\top Q \, x = x^\top Q^\top x.$

If $Q$ is symmetric I know that $\nabla f(x) = 2 Q x$ and $H_f(x) = 2 Q$. However, I am not sure as to what I should do when Q is not symmetric. Also, the hint is somewhat misleading to me as it makes it appear that $Q$ is symmetric. Help would be really appreciated! 

HINT:
$$ Q = \frac{ 1}{2} \underbrace{( Q + Q^T)}_{sym} + \frac{1}{2} ( Q - Q^T)$$
If you use the hint provided then
$$\begin{align*} (x,Qx) =& \frac{1}{2} (x,(Q+Q^T)x) + \frac{1}{2}(x,(Q-Q^T)x) \\
=& \frac{1}{2} (x,(Q+Q^T)x) + \frac{1}{2} \underbrace{(x,Qx)-(x,Q^Tx)}_{=0} \\
= & \frac{1}{2} (x,(Q+Q^T)x)
\end{align*}$$
