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(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?

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    $\begingroup$ In general no, but this is a Gramian matrix, for which a special argument holds, see for instance the Wikipedia article. $\endgroup$ – Andreas Caranti Apr 15 at 7:08
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    $\begingroup$ Of course not. But a Gram matrix is a matrix of the form $M^H M$, so it is always positive-semidefinite, and if it is non-singular, it is positive-definite. $\endgroup$ – Abhimanyu Pallavi Sudhir Apr 15 at 7:17
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    $\begingroup$ How about $$\begin{pmatrix} -1 &\ 0 \\ 0 &\ -1 \end{pmatrix}$$ is this a positive definite? $\endgroup$ – Fareed AF Apr 15 at 7:19
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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.

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