How is it shown that a Hermitian matrix will be positive definite? 

The following Hermitian matrix is given for all $m \in \mathbb{Z}$:

$$A_m = \left( \begin{array}{cc} m & i
 \\ -i & m\\ \end{array} \right)$$
(i) Show that $A_1$ is not positive definite.
(ii) Show that $A_2$ is positive definite.
(iii) Determine all $m \in \mathbb{Z}$
   such that $A_m$ is positive definite.

So my understanding is: when given a matrix, to check if it is positive definite, one calculates: $\overline{z}^{t}Az$. Would this always be the same as $z^{t}A\overline{z}$? (When I calculated it out, the answer seemed to be "no"). I got the former off wikipedia, but I think I've seen the later as well...
Using the first version, I get $\overline{z}^{t} A_{m}z = \left( \begin{array}{cc} m \overline{z_{1}}-i\overline{z_{2}}, & i\overline{z_{1}}+m\overline{z_{2}} \end{array} \right) \left(\begin{array}{c}z_{1} \\ z_{2} \end{array} \right)  = mz_{1}\overline{z_{1}}-iz_{1}\overline{z_{2}}+i\overline{z_{1}}z_{2}+mz_{2}\overline{z_{2}}$ and since I'm examining whether they are positive definite, I think I should try to express as many terms as squared so I think I can write $=m|z_{1}|^{2}-iz_{1}\overline{z_{2}}+i\overline{z_{1}}z_{2}+m|z_{2}|^{2}$. Here is where I don't know what to do with regard to parts (ii) and (iii)... Is there some sort of way to factor or rearrange the terms to make a case for which $m\in \mathbb{Z}$ this expression must be $\geq 0$? I think I've managed part (i) since $m=1, 0 \neq z = (-1,-i) \Rightarrow 1-i(-1)(i)+i(-1)(-i)+1 = 1+i^{2}+i^{2}+1 = 0$. The problem is that I just got that through playing around and don't really know why it should be so...
Also, just to make sure, the same approach will work for a Hermitian matrix of higher order as well, right?
 A: In this case, you can compute the eigen values and find out for what $m$'s the eigen values are positive. (Caution: In general, if the matrix is positive definite the eigen values are positive. However it is not always the case that if all the eigen values are positive the matrix is positive definite. You need the geometric multiplicity of the eigen values should be equal to the algebraic multiplicity to conclude the matrix is positive definite (which happens to be true this case for $m >1$))
A: You could try to prove (and use) the following criterion: $$ \begin{pmatrix} a & b \\ \bar{b} & c \end{pmatrix} $$ is positive definite iff $a > 0$ and $ac - |b|^2 > 0$ (here $a,c$ are real and $b$ is complex).
Hint: apply to the vector $\begin{pmatrix} b \\ -a \end{pmatrix}$
This criterion can be generalized to higher dimensions: if $(a_{i,j})_{1 \leq i,j \leq n}$ is hermitian, it is positive definite iff for any $1 \leq k \leq n$, $\det \left( (a_{i,j})_{1 \leq i,j \leq k} \right) > 0$.
To prove it, you need to show that the maximal subspaces on which a hermitian form is positive definite have the same dimension, and proceed by induction on $n$.
A: The trace of you matrix is 2m and its determinant is m^2-1 thus you just need you m to be 1)strictly positive 2)m^2>1 thus Am is definite positive for any m integer greater than or equal 2
