15
$\begingroup$

I would like to know, is there a connection between the rank of a matrix and whether it is positive definite? Specifically, if I can prove that a matrix is not full rank, then can I say that it is not positive definite? If so, why?

Thanks a lot for your help.

$\endgroup$
3
  • 14
    $\begingroup$ If your matrix is a square matrix and has not full rank, then zero is an eigenvalue, therefore it cannot be positive definite. $\endgroup$ Nov 11, 2015 at 17:54
  • 10
    $\begingroup$ If a square matrix $M$ is not full rank then it has a nontrivial kernel, meaning that there exists a nonzero vector $v$ such that $Mv=0$. In particular $v^{\intercal}Mv=0$, so $M$ is not positive definite. $\endgroup$
    – Servaes
    Nov 11, 2015 at 17:54
  • $\begingroup$ Awesome, thanks! $\endgroup$ Nov 11, 2015 at 17:55

1 Answer 1

7
$\begingroup$

Another fact that wasn't pointed out is that: since the determinant is a product of the eigenvalues, and both PD and ND matrices have either all strictly-positive or all strictly-negative eigenvalues, you can deduce the determinant is non-zero, which makes the matrix invertible (i.e. the matrix has full rank).

$\endgroup$
2
  • 2
    $\begingroup$ @Leucippus It definitely answers the question: "Specifically, if I can prove that a matrix is not full rank, then can I say that it is not positive definite? If so, why?". $\endgroup$ Apr 26, 2019 at 6:12
  • 3
    $\begingroup$ Jack, your answer is correct, but claiming that this wasn't pointed out is rather daring: the two comments above state the same, and have been here for $4$ years. $\endgroup$ Apr 26, 2019 at 6:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .