When will eigenvalues of a matrix be positive?

This may be very general question, Are there any set of conditions that if some arbitrary matrix say A fulfils we can say that A will have eigenvalues such that all of them are non negative. In other words what are conditions for a matrix to be positive semi definite.

Regards Ahsan

• There is also Silvester's criterion.
– Pp..
Feb 4, 2015 at 14:47
• @AlexR That is what the link is pointing to.
– Pp..
Feb 4, 2015 at 14:49
• @Pp.. Sorry, I overlooked the last sentence in the article. Feb 4, 2015 at 14:50

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.

• This definition of positive semidefinite does not imply positive eigenvalues. We must either impose symmetry or allow $x \in \Bbb C^n$. Feb 4, 2015 at 14:48
• Err, should there be a minus sign on the ODE example? Consider the 1D example $\dot{f}(t) = a f(t)$ with 1-by-1 "matrix" $a$. Feb 4, 2015 at 14:57
• @Omnomnomnom $A$ has eigenvalues $1 \pm i$... Where's the problem? Feb 4, 2015 at 15:03
Start with the equation $Ax = ax$ for eigenvalue $a$ and eigenvector $x$ of a matrix $A$. Then multiply by the left with $x^T$. Other hint: $x^T x > 0$.