I know that a Kähler form $\omega$ looks like: $$\omega=\frac{i}{2}\sum_{j,k}h_{jk}dz_j\wedge d\overline{z}_k$$ such that $\overline{h_{jk}}=h_{jk}$ ($\omega$ is real), $\partial\omega=0$, $\overline{\partial}\omega=0$ ($\omega$ is closed), $\det (h_{jk})\neq 0$ ($\omega$ is non-degenerate).
One other condition is that $g(\cdot,\cdot)=\omega(\cdot,J(\cdot))$ is positive-definite, which should be equivalent to $(h_{jk})$ being positive-definite.
I'm having trouble to see that. I've tried to consider the real $2n\times 2n$ matrix associated with the real bilinear form $g$, which must be positive-definite, but I can't see the relation to that and the complex $n\times n$ matrix $(h_{jk})$.
As written, the matrix $(h_{jk})$ is not positive-definite. And just like Ted said in the comments, one usually writes $$\omega=\frac{i}{2}\sum_{j,k} h_{jk} dz^j\wedge d\overline{z}^k. $$ Now take a vector $$v = \sum_j a^j\frac{\partial}{\partial x^j}+b^j\frac{\partial}{\partial y^j} =\sum_j (a^j+ib^j)\frac{\partial}{\partial z^j}+(a^j-ib^j)\frac{\partial}{\partial \bar z^j} . $$ Check that $$Jv = \sum_j -b^j\frac{\partial}{\partial x^j}+a^j\frac{\partial}{\partial y^j} =\sum_j i(a^j+ib^j)\frac{\partial}{\partial z^j}-i(a^j-ib^j)\frac{\partial}{\partial \bar z^j} . $$
Now you can compute that $$g(v,v) = \omega(v,Jv) = \sum_{jk} h_{jk}(a^j+ib^j)\overline{(a^k+ib^k)}, $$and the conclusion follows. It works out this nicely exactly because of the $i/2$ factor.