I'm currently stuck on a problem for my Artificial Intelligence class. The assignment is provided at the following link: http://courses.engr.illinois.edu/cs440/HW1.pdf
The problems that I'm stuck on are #3 parts d-f. I have found the Hessian matrix to be
$$ f(x,y,z) = xln(x) + ylny + zlnz + \alpha(x + y + z - 1) $$
$$ \nabla f(x,y,z) = (lnx + 2)\vec{i} + (lny + 2)\vec{j} + (lnz + 2)\vec{k} $$ $$ \left( \begin{array}{ccc} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{array} \right) $$
In order to prove that the function is convex I need to determine if it is positive definite. I believe that the function as a whole is not positive definite and therefore not convex.
However, it is later asked in part e to tell if the function is convex on the domain $[0,1]$. This will be convex. I'm just not entirely sure how to prove whether or not the function as a whole is indeed not convex.
I'm not looking for an answer just guidance as to how to best approach it.
EDIT
Based on the feedback from jubobs. Thanks a lot by the way.
I've calculated the eigenvalues for the hessian which ended up being:
$$ \lambda_1 = \frac{1}{x} $$$$ \lambda_2 = \frac{1}{y} $$$$ \lambda_3 = \frac{1}{z} $$
What follows are only hints, as I don't want to give you the whole solution.
(c)
You've correctly derived the Hessian, already. Well done.
(d)
Now, ask yourself: over $f$'s domain, is the Hessian positive semidefinite? Remember that the eigenvalues of a diagonal matrix are easy to figure out, and that the signs of those eigenvalues tell you something about whether the matrix is positive semidefinite or not. And you already know that, if the Hessian is positive semidefinite, this unconstrained optimisation problem is convex.
(e)
Be careful there. That question is not about whether the expression of $f$ is a complex function over some set, but whether $f$'s domain is convex (which is a necessary condition for function $f$ to be convex). You should know that an interval of $\mathbb{R}$ is convex, and you should know that the Cartesian product of nonempty sets is convex. Therefore [...]