Let $f$ be a map from $R^n$ to $R$ so that $f(x) = \frac{1}{2} x \cdot (Ax) - x^T v + \alpha $ where $A$ is an $n by n$ symmetric and positive definite matrix. $v \in \mathbb{R}^n$ and $\alpha $ is a scalar. IS there an easier way to compute the gradient and hessian of $f$?
It seems not too complicated but we still have a ltot of equations involved. Any ideas?
The wikipedia page of matrix calculus is helpful for such stuff.
Note that $A$ is symmetric positive definite.
$$f(x)=\frac12 x^TAx-x^Tv+\alpha$$
$$\nabla f(x)=\frac12 (2Ax)-v=Ax-v$$
If we differentiate by $x$ again, $v$ as a constant vector will vanish and the first part is just linear combination of $x_i$.
$$\nabla^2 f(x)=A.$$
Compare the result with lower dimensions to make sure they agree.