# inequality with a positive matrix

Let $$A=\left[ \begin{array}{cc} a & b\\ \overline{b} & c\\ \end{array} \right]$$ be a positive semi-definite positive of $M_2(\mathbb{C})$. How prove the inequality $ac \geq |b|^2$ ?

• The inequality makes no sense unless those numbers are real, and even then the inequality would have to be the other way around for positiveness. Feb 10, 2016 at 8:57
• "semi-definite" positive imply "selfadjoint" hence $a,c$ are reals Feb 10, 2016 at 13:18

For positive semi-definiteness, we need $z^*Az$ to be real and non-negative for all $z = [x, \;y]^T$ where $x, y \in \mathbb C$. Thus we need $$a|x|^2+c|y|^2+bx^*y+b^*xy^*$$ to be real and non-negative for every choice of $x, y \in \mathbb C$.
As we may have either of $x, y = 0$ it is immediate that $a, c$ must both be real and non-negative.
Further, selecting $x = b, y=t \in \mathbb R$ we have $a|b|^2+ct^2+2|b|^2t \ge 0$ which means we need by the discriminant condition, $|b|^4 \le ac|b|^2 \implies |b|^2 \le ac$ or $|b| = 0$.
In either case, $A$ being semi-definite implies $|b|^2 \le ac$.