Positive Definite and Hermitian Matrices If we know that $C$ is positive definite and Hermitian, how can we prove that there exists a matrix $Q$ such that $Q^∗CQ=I$. Where, $Q^∗$ is complex conjugate.
The definition of positive definiteness for a Hermitian Matrix I am using is if all principal minors are positive.
I am also looking for a link between this definition and another equivalent definition i.e. $x^*Ax>0$ for all $x$.
 A: Sylvester's Law of Inertia tells you that two hermitian matrices are congruent if, and only if, they have the same inertia. (Wikipedia only deals with the reals, but everything works out the same over $\mathbb C$).
Since $C$ is positive definite, its eigenvalues are all positive, thus it has the same inertia as the identity matrix and what you want follows.
The link you're looking for between your definition of positive definite matrix $A$ and the property $\forall x\in \mathbb C^{n\times 1}\setminus \{0_{\mathbb C^{n\times 1}}\}\left(x^*Ax>0\right)$ is the following:
$$A \text{ is positive definite} \iff \forall x\in \mathbb C^{n\times 1}\setminus \{0_{\mathbb C^{n\times 1}}\}\left(x^*Ax>0\right).$$
The proof of this isn't particularly short so if you want it I suggest you open a book or ask a new question.
A: Hint: if $C$ is positive definite, then there is a unitary matrix $U$ for which $C = UDU^*$ where
$$
D = \pmatrix{\lambda_1 &&\\&\ddots&\\ &&\lambda_n}
$$
is the diagonal matrix of eigenvalues (note that $\lambda_i > 0$).  Find a matrix $V$ for which 
$$
VDV^* = \pmatrix{1 &&\\&\ddots&\\ &&1}
$$
(hint: there is a diagonal $V$ that works here). Now we have $D = UCU^*$, and $I = V^*DV$.  It follows that $I = V^*UCU^*V = (U^*V)^*C(U^*V)$
