I'm seeing different definitions of a positive definite matrix:
Definition (i): A square matrix $A$ is positive definite if there is a positive scalar $\alpha$ such that:
\begin{gather*} x^T A x \ge \alpha x^T x, \quad \forall x \in \mathbb{R}^n \\ \end{gather*}
Definition (ii): A square matrix $A$ is positive definite if there is a positive scalar $\alpha$ such that:
\begin{gather*} x^T A x > 0, \quad \text{when} \, x \neq 0, x \in \mathbb{R}^n \\ \end{gather*}
Clearly definition (i) implies definition (ii).
My first question: Is the converse true? Is the converse true if we assume finite dimensions?
Some definition, such as wikipedia, also say that a matrix must be symmetric to be positive definite. Some texts I'm studying offer definitions that don't include this property in the definition.
My second question: Can a non-symmetric matrix satisfy the above definitions? Or can we deduce from above definitions that the matrix is symmetric? If not, it would seem that definitions that require the symmetric property and those that don't are conflicting definitions?
Let $f(x) = x^TAx$. It is obvious that $f$ is continuous. Also notice that $f(x) = \|x\|^2 u^TAu$, where $\|u\|=1$. Therefore, the positive condition can be rewritten as $$u^T Au>0, \|u\|=1.$$ However, the set $\|u\|=1$ is compact in finite dimensions. Since $f$ is continuous over a compact set, it attains its minimum value and this value must be greater than zero. Let $\alpha$ be the minimum and we are done.
For the second question, the answer is no; consider $A = \begin{pmatrix} 1&1\\-1&1\end{pmatrix}$. There is a good discussion on these types of matrices here.