2x2 Matrix Inequality Is it true that if I have a positive definite matrix $m = \left( \begin{smallmatrix} m_{11} & m_{12}\\ m_{21} & m_{22} \\\end{smallmatrix} \right)$ in $\mathrm{M}(\mathbb{C};2)$ the following inequality holds
$\textrm{Tr }m = m_{11} + m_{22} > 2\textrm{ Im }(m_{12})$ ?
 A: Yes. $z^* m z$ is real and positive for any non-zero complex vector $z$.
In particular,
$$ \begin{bmatrix}1 & -i\end{bmatrix} \begin{pmatrix} m_{11} & m_{12} \\ m_{21} & m_{22}\end{pmatrix} \begin{bmatrix}1 \\ i\end{bmatrix}  = m_{11} + m_{22} + i(m_{12}-m_{21}) > 0. $$
Now if $m_{12} = a + ib$, then $m_{21}$ is its complex conjugate $a - ib$,
and $m_{12}-m_{21} = 2ib$,
so we have $m_{11} + m_{22} - 2b > 0$
which is the inequality you're looking for.
A: Let $M=(m_{ij})$ be a $2\times2$ complex symmetric matrix. Since $m_{21}=\overline{m}_{12}$, for every non-zero $z=(z_1,z_2) \in \mathbb{C}^2$ we have
\begin{eqnarray}
0< \langle Mz,z\rangle&=&(m_{11}z_1+m_{12}z_2)\bar{z}_1+(m_{21}z_1+m_{22}z_2)\bar{z}_2\\
&=&m_{11}|z_1|^2+m_{12}\bar{z}_1z_2+m_{21}z_1\bar{z}_2+m_{22}|z_2|^2\\
&=&m_{11}|z_1|^2+m_{22}|z_2|^2+m_{12}\bar{z}_1z_2+\overline{m}_{12}z_1\bar{z}_2\\
&=&m_{11}|z_1|^2+m_{22}|z_2|^2+2\Re(m_{12}\bar{z}_1z_2).
\end{eqnarray}
Choosing $z=(1,i)\in \mathbb{C}^2\setminus\{(0,0)\}$ i.e. $z_1=1,z_2=i$, we get
$$
0<m_{11}+m_{22}+2\Re(im_{12})=\operatorname{tr}(M)-2\Im m_{12},
$$
i.e.
$$
\operatorname{tr}(M)> 2\Im m_{12}.
$$
