Is symmetry a necessary condition for positive (or negative) definiteness? Is symmetry a necessary condition for positive (or negative) definiteness?
If not:
It can be proved that if $\mathbf{A} \in \mathbb{R}^{m\times m}$ is a square (non-symmetric) matrix, then 
$$ \mathbf{z'Az=z'Bz},~~\mathbf{B=B'= \frac{A+A'}{2}} $$
On the other hand, a positive definite matrix is a symmetric matrix for which:
$$\mathbf{z'Bz}>0,~~ \mathbf{z\ne 0}$$
Can we imply that $\mathbf{A}$ which is a non-symmetric matrix, is positive definite?
 A: It looks like you are working with real matrices.
Most often the definition of positive definite includes symmetric, but sometimes this is not required. 
In any case, if $z'Az\geq0$ then $z'A'z=(z'Az)'=z'Az\geq0$. So
$$
z'Az=\frac12(z'Az+z'A'z)=z'\left(\frac{A+A'}2\right)z.
$$
A: Not sure if that was your question, but if it is

does there exist a nonsymmetric matrix s.t. $v^T A v >0$ for all $v\neq 0$

then the answer is yes. Take for example the matrix
$$
A = \begin{bmatrix}
1 & 1 \\
-1 & 1
\end{bmatrix}.
$$
You can check that for any $v=(x,y)$ you get $v^T A v = x^2+y^2 > 0$, if $x$ or $y$ is not $0$, or equivalently $v$ is not the zero vector.
A: Yes, it is possible. In fact, it follows easily from the properties of taking the transpose:
$$0 < z^{\mathrm{T}}\left(A+A^{\mathrm{T}}\right) z = z^{\mathrm{T}}Az + z^{\mathrm{T}}A^{\mathrm{T}}z = z^{\mathrm{T}}Az + (z^{\mathrm{T}}Az)^{\mathrm{T}} = 2z^{\mathrm{T}}Az$$
Taking the transpose of a real number doesn't change anything. Therefore $z^{\mathrm{T}}Az> 0$ if and only if $z^{\mathrm{T}}Bz>0$.
