Follow-up to this question: Gradient and Hessian of this function
Trying to figure out if the objective function from that problem is concave, so I need to check if the Hessian is negative semidefinite.
@john316 derived:
\begin{align*} H(x) &= fpp^T - pg^T - gp^T - fM^TM \cr && \text{Hessian} \\ g(x) &= M^Ty-fp \cr && \text{Gradient} \end{align*}
Where \begin{align*} M &= \alpha A \\ p &= M^Tz \\ \beta^2 &= b^Tb \\ b &= Ax \\ z &= \alpha b \\ 1 &= \alpha \beta \\ f &= y^Tz \cr\cr \end{align*} And $A \in \mathbb{R}^{m\times n}$, $y \in \mathbb{R}^m$ are given.
So, is $H(x)$ negative semi-definite?
In other words, is this true?
$$w^{\top} H(x) w \leq 0 \, \forall \, x \in \mathbb{R}^n, w \in \mathbb{R}^n$$
The first-order conditions are satisfied at any point where $g=0$
At those points, the Hessian reduces to $$\eqalign{ H &= f\,(pp^T-M^TM) \cr &= f\,M^T\,(zz^T-I)\,M }$$ Note that $z$ is a unit vector. To simplify things, let's use a coordinate system where $z$ is the first basis vector.
Let $v = Mw$, then $$\eqalign{ w^THw &= f\,v^T(zz^T-I)v \cr &= f\,v_1^2 - f\,(v_1^2 + v_2^2 + \ldots + v_n^2) \cr &= f\,v_1^2 - f\,(v_1^2 + v_2^2 + \ldots + v_n^2) \cr &= -f\,(v_2^2 + ... + v_n^2) \cr }$$ Whether this is positive or negative depends on the sign of $f$.