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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 $ac \geq |b|^2$ ?

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    $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Feb 10, 2016 at 8:57
  • $\begingroup$ "semi-definite" positive imply "selfadjoint" hence $a,c$ are reals $\endgroup$
    – Zouba
    Feb 10, 2016 at 13:18

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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 $|b|^2 \le ac$.

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  • $\begingroup$ We can also use the determinant(=product of eigenvalues) $\endgroup$
    – Zouba
    Feb 11, 2016 at 15:55

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