What conditions does positive semidefiniteness impose on the matrix elements?

Question

Let $A$ be an Hermitian $2\times 2$ complex matrix. We can always write it as $$A =\begin{pmatrix}a & \alpha \\ \bar\alpha & b\end{pmatrix}$$ for some $a,b\in\mathbb{R}$ and $\alpha\in\mathbb{C}$. One can then see that $A\ge0$ is then equivalent to having $a,b\ge0$ and $|\alpha|\le \sqrt{ab}$. This can be obtained computing the eigenvalues of $A$, which have the form $$\lambda_\pm = \frac{a+b}{2} \pm \sqrt{\left(\frac{a-b}{2}\right)^2 + |\alpha|^2},$$ and thus positivity amounts to $$\frac{a+b}{2} \ge \sqrt{\left(\frac{a-b}{2}\right)^2 + |\alpha|^2} \iff (a+b)^2 \ge (a-b)^2 + 4|\alpha|^2 \iff |\alpha|^2\le ab.$$ Is there any similar "simple" statement to be made for larger matrices? Of course, one can always apply the above reasoning on two-dimensional diagonal blocks, but I'm referring to the "stricter" conditions that probably will involve more than two coefficients.

Answer 1

Yes, a simple condition is that all the principal minors be nonnegative.

In the $2\times2$ case this equivalent to saying that both $a\ge0$, $b\ge0$, and $ab-|\alpha|^2\ge0$.

In the general case where $A\in\mathbb{C}^{n\times n}$ and $A=A^*$, those conditions are that $\det(A^{(I)})\ge0$ for all $I\subset\{1,\ldots,n\}$, where $A^{(I)}$ is the matrix obtained from picking the rows/columns in $I$ from $A$.