We start with a new scalar but where now the vector \( \boldsymbol{y} \) is replaced by a vector \( \boldsymbol{x} \) and the matrix \( \boldsymbol{A} \) is a square matrix with dimension \( n\times n \).
$$ \alpha = \boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x}, $$with \( \boldsymbol{x} \) a vector of length \( n \).
We write out the specific sums involved in the calculation of \( \alpha \)
$$ \alpha = \sum_{i=0}^{n-1}\sum_{j=0}^{n-1}x_i a_{ij}x_j, $$taking the derivative of \( \alpha \) with respect to a given component \( x_k \) we get the two sums
$$ \frac{\partial \alpha}{\partial x_k} = \sum_{i=0}^{n-1}a_{ik}x_i+\sum_{j=0}^{n-1}a_{kj}x_j, $$for \( \forall k =0,1,2,\dots,n-1 \). We identify these sums as
$$ \frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{x}^T\left(\boldsymbol{A}^T+\boldsymbol{A}\right). $$If the matrix \( \boldsymbol{A} \) is symmetric, that is \( \boldsymbol{A}=\boldsymbol{A}^T \), we have
$$ \frac{\partial \alpha}{\partial \boldsymbol{x}} = 2\boldsymbol{x}^T\boldsymbol{A}. $$