Derivative of quadratic form 3

Question

Could someone provide me with step by step derivation of derivative of quadratic form: $\alpha=\sum_{i=1}^{m}\sum_{j=1}^{m}a_{ij}x_ix_j $ ? Specifically, I am interested in obtaining derivative in such form : $\frac{d\alpha}{dx_k}=\sum_{j=1}^{m}a_{kj}x_j+\sum_{i=1}^{n}a_{ki}x_i$ which is $\frac{d\alpha}{dx}=\textbf{x}^T(A^T+A)$ which is simply $2x^TA$ when A is symmetric matrix.

Answer 1

So another detailed justification is the following. Expand $\alpha = x^TAx$ as you did, namely: $$\alpha = x^TAx = \sum_{i=1}^{m}\sum_{j=1}^{m}a_{ij}x_ix_j $$ Derive w.r.t the $k^{th}$ element of $x$ (say $x_k$), you get $$\frac{\partial \alpha}{\partial x_k} = 2a_{kk}x_k + \sum\limits_{i=1 \\ i\neq k}^m (a_{ik} + a_{ki})x_i = \sum\limits_{i=1}^m (a_{ik} + a_{ki})x_i \qquad k = 1\ldots m$$ So $$\frac{\partial \alpha}{\partial x}= \begin{bmatrix} \frac{\partial \alpha}{\partial x_1} \\ \frac{\partial \alpha}{\partial x_2} \\ \vdots \\ \frac{\partial \alpha}{\partial x_m} \end{bmatrix} = \begin{bmatrix} \sum\limits_{i=1}^m (a_{i1} + a_{1i})x_i \\ \sum\limits_{i=1}^m (a_{i2} + a_{2i})x_i \\ \vdots \\ \sum\limits_{i=1}^m (a_{im} + a_{mi})x_i \end{bmatrix} =x^T(A + A^T) $$

Answer 2

So $\alpha = x^T A x$.

Using $\frac{d x^T A}{d x} = A$, then $\frac{d x^T Ax}{d x} = Ax + A^Tx$