(Skagestad, 2005) states the following conclusion in page 128.
"The system $(A, B)$ is state controllable if and only if the Gramian matrix $W(t)$ has full rank (and thus is positive definite) for any $t > 0$."
Is actually a full rank square matrix necessarily a positive definite matrix?
Is a full rank square matrix necessarily a positive definite matrix?
No, as pointed out in the comments.
Is a full rank controllability Grammian necessarily positive definite?
Yes. To see this note that the controllability Grammian is defined to be
$$ W(t)\triangleq\int_{0}^{t}e^{A\tau}BB^{T}e^{A^{T}\tau}d\tau. $$
Now suppose that $W(t)$ is full rank. It follows that $\ker(W) = \{0\}$, and thus, for all $x\in \mathbb{R}^n\setminus\{0\}$, $x^TW(t)x \neq 0$. Furthermore
\begin{align} x^TW(t)x &= \int_{0}^{t}x^Te^{A\tau}BB^{T}e^{A^{T}\tau}xd\tau\\ % &= \int_{0}^{t}(B^{T}e^{A^{T}\tau}x)^T(B^{T}e^{A^{T}\tau}x)d\tau\\ % &= \int_{0}^{t}\|B^{T}e^{A^{T}\tau}x\|^2d\tau.\\ \end{align}
Since the norm of a vector is always positive or zero it follows that the integral is positive or zero. Since the quadratic form cannot be zero, by virtue of the fact that $W(t)$ is full rank, it follows that the integral must be positive, and thus $x^TW(t)x > 0$, which is the condition that $W(t)$ be positive definite.