$f(x) = \frac{1}{2}x^TQx+q^Tx$ has a unique minimum if and only if Q is positive definite.

Question

I am trying to prove the above result. I can show one direction but am having trouble with the direction "unique minimum implies positive definite".

Edit: I have seen related questions such as Quadratic Function must be positive definite to have a unique minimum

Unfortunately, I cannot use any results about convex functions as I have not covered that content yet. The tools I have are the gradient, hessian, sufficient/necessary conditions for optimality, and Taylor's theorem.

I know that if we have a unique minimum, $x^*$, we necessarily have $\nabla f(x^*) = Qx^*+p = 0$, $\nabla^2 f(x) = Q \geq 0$ and $f(x) > f(x^*)$ for all $x \not= x^*$. Past this, I have not made progress.

Answer 1

You can go for the contraposition. Assume that $Q$ is not positive definite. Then we find some $x\in \mathbb{R}^n$ such that $x^T Q x\leq 0$.

Let's start with the case $c(x):=x^T Q x<0$ and consider $$ g:\mathbb{R}\rightarrow \mathbb{R}, g(t)=f(tx).$$ We can explicitely compute $g$. Namely, we get $$g(t)=\frac{c(x)}{2} t^2+d(x)t,$$ where $d(x)=q^T x$. Then $\lim_{t\rightarrow\infty} g(t)=-\infty$ and so $f$ will not admit a minimum.

Next we consider the case $c(x)=0, d(x)\neq 0$. Then we get $g(t)=d(x)t$, which again does not admit a minimum.

Finally one would need to consider the case $c(x)=0, d(x) = 0$, which I would not know how to deal with in this way.

In case you know a bit of linear algebra, you can use the fact that we can unitarily diagonalize self-adjoint operators. Thus, after applying such a coordinate transformation, we may assume wlog that $$ f(x)= \frac{1}{2}\sum_{j=1}^n \lambda_j x_j^2 + \sum_{k=1}^n a_k x_k,$$ where $\lambda_j$ are the eigenvalues of $Q$. Then, we see that if some $\lambda_j$ vanishes, we have $f(te_j)= t a_j$, which never has a unique minimum.