I have a quadratic function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$,
$f(\mathbf{x}) = (\mathbf{x}-\mathbf{p})^\top \mathbf{Q} (\mathbf{x} - \mathbf{p})$
where $\mathbf{Q}$ is positive definite and $\mathbf{p} \in \mathbb{R}^2$.
I want to find $\mathbf{x}$ satisfying $\| \mathbf{x} \|_2 = 1$ that (locally) minimizes $f$. The condition for a point $\mathbf{x}$ to be a critical point should be:
$$ \nabla f(\mathbf{x}) = \lambda\mathbf{x} $$ $$ 2\mathbf{Q}(\mathbf{x}-\mathbf{p}) = \lambda\mathbf{x} $$ for some $\lambda$.
My questions is:
Does the condition that a critical point $\mathbf{x}$ (locally) minimizes $f$ is as follows?
$$ \left<\nabla \left<\nabla f(\mathbf{x}), \mathbf{x}^\perp \right>, \mathbf{x}^\perp \right> > 0 $$
where $\left< , \right>$ is the dot product, and $\left< \mathbf{x}^\perp, \mathbf{x}\right> = 0$. (the second-order directional derivative of f at $\mathbf{x}$, direction: $\mathbf{x}^\perp$, is greater than zero.)
If not, what is it?
Thanks in advance.
I got this answer from mathoverflow: https://mathoverflow.net/questions/85640/condition-for-a-convex-quadratic-function-restricted-on-a-unit-circle-to-attain-a
The condition for being a local minimal point is stated here: http://www.math.ufl.edu/~hager/papers/Regular/sphere.pdf
Thanks all.