A matrix $A\in\mathbb C^{n\times n}$ is positive semidefinite (All eigenvalues have nonnegative real part) iff
- $x^H A x \ge 0 \qquad \forall x\in\mathbb C^n$
- If all principal minor determinants are $\ge 0$. Principal minors are the Matrices $M_I = (a_{ij})_{i,j\in I}, \ I\subset \{1,\ldots,n\}$ where some rows and columns with the same index are removed.
- If the ODE $\dot x= -Ax, x(0) = x_0$ satisfies $\lim\limits_{t\to\infty} x(t) = 0$ for all $x_0 \in \mathbb C^n$
The system is called stable in this case.
If $A\in\mathbb R^{n\times n}$ is symmetric ($A=A^T$), it only has real eigenvalues and we can weaken condition 1 to $x^T A x \ge 0 \ \forall x\in\mathbb R^n$
Note that $A\succeq 0$ (positive semidefinite) is not equivalent to all eigenvalues being nonnegative! A real matrix can have complex eigenvalues if it's not symmetric. In this case you need to require $\Re (\lambda_i) \ge 0$ (the real part). For symmetric matrices, this coincides.