I know that the quadratic form $x'Ax$ is a concave in vector $x$ if matrix $A$ is negative semi definite. What happens if $A$ depends on $x$ (so that I have $x'A(x)x$), but I still know that $A(x)$ is NSD? Does it still hold? Maybe that's not even a quad.form in that case$\dots$ anyways how do I check concavity in that case?
More generally if let $m$ be a vector valued function of vector $x$, so $m=m(x)$, and let $A$ be a NSD matrix dependent on $x$, so $A=A(x)$. What is then the derivative of $m'Am$ w.r.t. $x$?
Thanks a lot!
Rahul is right, but his function $A$ is not differentiable. Unfortunately, when $A$ is differentiable, then the second derivative is complicated. Let $A=[a_{i,j}]$ and if $U$ is a matrix, then $[U]_i$ denotes its $i^{th}$ column. One has
$\dfrac{\partial^2f}{\partial x_kx_l}:x\rightarrow 2a_{k,l}+x^T\dfrac{\partial^2A}{\partial x_kx_l}x+2x^T([\dfrac{\partial A}{\partial x_k}]_l+[\dfrac{\partial A}{\partial x_l}]_k)$.
Finally, is the following symmetric matrix: $Hess(f)=[\dfrac{\partial^2f}{\partial x_kx_l}]_{kl}$ non-positive ?