# If $P$ is not positive semidefinite, show that the problem is unbounded below

I'm currently taking a convex optimization class and we're using the textbook by Boyd & Vandenberghe. I was solving an exercise problem when I came across a question (for anybody curious the problem is Exercise 9.1). I have two questions in total regarding the same problem. One about a detail within the problem and one regarding the solving itself. Here's the problem:

Consider the problem of minimizing a quadratic function:

$$\text{minimize}\quad f(x) = \frac{1}{2}x^T P x + q^T x + r$$

where $$P \in S^n$$ (i.e. an $$n \times n$$ symmetric matrix) but we do not assume $$P$$ is positive semidefinite.

Show that if $$P$$ is not positive semidefinite (and therefore objective function $$f$$ is not convex) then the problem is unbounded below.

My approach

By definition if a matrix is not positive semidefinite then that means there exists some $$z$$ such that

$$z^TPz \lt 0$$

Following this logic we could say that $$x = \alpha z$$ with $$x, z \in \Bbb{R}^n$$ and $$\alpha \in \Bbb{R}$$. The objective function changes to

$$f(x) = \frac{1}{2}\alpha^2 (z^TPz) + q^T(\alpha z) + r$$

This is where I get stuck. According to the solution manual, it states that if $$\alpha$$ becomes large then $$f \rightarrow -\infty$$ and we can therefore conclude that the function is unbounded below.

My two questions are as follows:

1. Why does the matrix not being positive semidefinite imply the function is not convex?

2. How was the conclusion drawn that if $$\alpha$$ becomes large then the function diverges to $$-\infty$$?

Thank you.

The definition of a convex function (in the context of twice continuously differentiable functions) is that the Hessian is positive semidefinite, and the Hessian of $$f$$ in this case is $$P$$.
Since you've now fixed $$z$$, the expression $$\frac{1}{2}z^TPz\alpha^2 + q^Tz\alpha+r$$ is just a quadratic equation with variable $$\alpha$$, and if the second order term of a quadratic equation is negative then we know its parabola faces downwards.