# Proving that a Hessian Matrix is positive definite

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:

$$\lambda_1 = \frac{1}{x}$$$$\lambda_2 = \frac{1}{y}$$$$\lambda_3 = \frac{1}{z}$$

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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 [...]

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Hey, thanks for your help I appreciate it. I think I'm starting to understand a bit more. Now that I have the eigenvalues I know that for the matrix to be positive semidefinite they must be greater than zero. The part I'm getting hung up on is because they are functions of x,y and z. Therefore, the way I've seen it is that it is only positive semidefinite for values of x,y,z > 0. Would the function therefore be neither convex nor concave since it would be convex for values greater than 0 and concave for values less than 0? – dmcqu314 Jan 31 '14 at 5:48
@dmcqu314 To answer your question, have a closer look at the domain of function $f$... Can $x$, $y$, or $z$ be negative? – Jubobs Jan 31 '14 at 5:49
It cannot when the domain is specified as [0,1]. Although, I thought that part d was to tell if the entirety of the function was convex, concave or neither? – dmcqu314 Jan 31 '14 at 5:53
$f$ is only defined over its domain... er, tautology. Anyway, if the Hessian is positive semidefinite over the interior of $f$'s domain, i.e. over the open set $(0,1)^3$, and if $f$' domain is convex, then $f$ is convex. – Jubobs Jan 31 '14 at 5:56
Ok, that makes a ton more sense now. My last confusion has to do with the minimums and maximums of the function over that domain. I've never done multivariate optimas so i'm a bit lost. – dmcqu314 Jan 31 '14 at 6:02