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?