Hessian matrix of $x^T Q x$ where $Q=Q^T$ and $Q$ is positive-definite

Question

I'm interested in the Hessian matrix $\nabla^2f(x)$ of $f(x) = x^T Q x$ where $Q$ is symmetric and positive definite. I understand that $\nabla f(x) = x^T(Q+Q^T) = 2Qx$, but am not sure how to calculate the Hessian matrix without considering each partial directly.

Thank you!

Answer 1

In the denominator layout (more commonly used), with $\nabla f(x)=2Qx$ $$\begin{split}\nabla^2 f(x) &=2Q^T\\ &=2Q \text { because Q is positive definite}\end{split}$$

That is, $\nabla_x (Ax)=A^T$ in general. The gradient of a matrix vector product wrt the vector is the transpose of the matrix.