I was reading this pdf and on page 6 proposition 8 states:
I don't really understand the steps that bring from
$$\alpha = \sum_{j=1}^n\sum_{i=1}^n a_{ij} x_{i} x_{j}$$
to its derivative
$$\frac{\partial \alpha}{\partial \bf{x}} = \sum_{j=1}^n a_{kj} x_J + \sum_{i=1}^n a_{ik}x_i$$
and then back to the final result:
$$\frac{\partial \alpha}{\partial \bf{x}} = \bf{x}^T A^T + \bf{x}^T A$$
Can someone please help me?
Best Answer
Another way to approach this formula is to use the definition of derivatives in multivariable calculus. The function is $\alpha: \mathbb R^n \to \mathbb R$ and the Jocabian matrix $D\alpha = \frac{\partial \alpha}{\partial x}$ is thus an $n \times n$ matrix and by definition satisfies the following equation \begin{align*} \lim_{\|h\| \to 0} \frac{\alpha(x+h) - \alpha(x) - D\alpha(h)} {h} = 0. \end{align*} But note $\alpha(x+h) - \alpha(h) = (x+h)^T A (x+h)- x^T A x = h^T A x + x^T A h$. Since $h^T A x = x^T A^T h$, we have $\alpha(x+h) - \alpha(h) = x^T(A^T + A)h$. Then it follows $\frac{\partial \alpha}{\partial x} = x^T(A^T+A)$.