When is the matrix $\begin{pmatrix} a & b \\ \overline{b} & c\end{pmatrix} \in M_2(\mathbb{C})$ positive?
I showed that it is necessary that $a,c \ge 0$ and that the determinant is positive. I now want to show the converse. So my question boils down to:
If $a,c \ge 0$ and $ac \ge |b|^2$, are the eigenvalues of $\begin{pmatrix} a & b \\ \overline{b} & c\end{pmatrix} \in M_2(\mathbb{C})$ all positive?
Best Answer
The characteristic equation is $$(\lambda - a)(\lambda - c) - |b|^2 = 0$$ or equivalently $$\lambda^2 - (a+c)\lambda + ac - |b|^2 = 0$$ The solutions are given by the quadratic formula: $$\lambda = \frac{1}{2}\left((a+c) \pm \sqrt{(a+c)^2 + 4|b|^2 - 4ac}\right)$$ which simplifies to $$\lambda = \frac{1}{2}\left((a+c) \pm \sqrt{(a-c)^2 + 4|b|^2}\right)$$ from which it's clear that the expression inside the square root is nonnegative. This is expected since the eigenvalues of a self-adjoint matrix are real.
Both roots will be nonnegative if and only if $$(a+c) \geq \sqrt{(a-c)^2 + 4|b|^2}$$ Since the RHS is nonnegative, the LHS $a+c$ must also be nonnegative, and in that case we can square both sides to get the equivalent inequality $$(a+c)^2 \geq (a-c)^2 + 4|b|^2$$ which simplifies to $$ac \geq |b|^2$$ So to summarize, the conditions are
which are equivalent your stated conditions $a,c \geq 0$ and $ac \geq |b|^2$.