1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ There is also Silvester's criterion. $\endgroup$
    – Pp..
    Feb 4, 2015 at 14:47
  • $\begingroup$ @AlexR That is what the link is pointing to. $\endgroup$
    – Pp..
    Feb 4, 2015 at 14:49
  • $\begingroup$ @Pp.. Sorry, I overlooked the last sentence in the article. $\endgroup$
    – AlexR
    Feb 4, 2015 at 14:50

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
12
  • $\begingroup$ This definition of positive semidefinite does not imply positive eigenvalues. We must either impose symmetry or allow $x \in \Bbb C^n$. $\endgroup$ Feb 4, 2015 at 14:48
  • 1
    $\begingroup$ @Omnomnomnom I'm unsure about this... $\endgroup$
    – AlexR
    Feb 4, 2015 at 14:49
  • $\begingroup$ 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" $a$. $\endgroup$
    – Nick Alger
    Feb 4, 2015 at 14:57
  • $\begingroup$ @NickAlger Yes indeed, thanks for pointing out. $\endgroup$
    – AlexR
    Feb 4, 2015 at 14:58
  • 1
    $\begingroup$ @Omnomnomnom $A$ has eigenvalues $1 \pm i$... Where's the problem? $\endgroup$
    – AlexR
    Feb 4, 2015 at 15:03
0
$\begingroup$

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: $x^T x > 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.