1
$\begingroup$

Let $H_n$ be a $(n+1)\times (n+1)$ real symmetric matrix, and let $D_0,D_1,\dots, D_n$ be the leading principal minors of $H_n$.

What I know is:

  1. If $H_n$ is positive definite (resp. positive semi definite), then $D_n> 0$ (resp. $D_n\geq 0$).
  2. If $D_k>0$ for all $0\leq k\leq n$, then $H_n$ is positive definite (by Sylvester's criterion).

What I want to know is, assuming that $H_n$ is positive semi-definite,

$\quad$ Q1. If $D_n>0$, then $H_n$ is positive definite.

$\quad$ Q2. If $H_n$ is not positive definite, then $D_n=0$.

For Q1: I believe it's done by induction over $n$. For $n=0$: If $D_0>0$, then $H_0$ is positive definite, by second point. For $n=1$: If $D_1>0$, how do you know that $D_0\neq 0$, so that we can use second point again?

For Q2: We know that $H_n$ is positive semi-definite by assumption, so $D_n\geq 0$ by first point. But, since $H_n$ is not positive semi-definite, we can't have $D_n>0$, so $D_n=0$. Is that it?

$\endgroup$
5
  • $\begingroup$ The answer to both questions is yes. In other words, a positive semidefinite matrix is positive definite if and only if it is invertible (has non-zero determinant). $\endgroup$ Aug 19 '20 at 18:58
  • $\begingroup$ @BenGrossmann Wow, that's an useful info you gave me there. Could you please inform me where you got this from? I'd love to know the proof of "if"-part. $\endgroup$
    – James2020
    Aug 19 '20 at 19:56
  • $\begingroup$ It's a pretty standard fact, normally taken as a consequence of the following: a symmetric matrix is positive definite if and only if its eigenvalues are real and positive semidefinite if and only if its eigenvalues are non-negative. From there, the determinant of a matrix is the product of its eigenvalues $\endgroup$ Aug 19 '20 at 20:13
  • $\begingroup$ For a more direct proof, it suffices to note that for a (symmetric) positive semidefinite matrix $H$, we have $x^THx = 0 \iff Hx = 0$. In my post here, I prove this in a few different ways. From there, note that a matrix has zero determinant if and only if its nullspace (AKA kernel) is non-trivial) $\endgroup$ Aug 19 '20 at 20:16
  • $\begingroup$ It is settled, thank you. If you can write your two first comments in an answer, I'll accept it. $\endgroup$
    – James2020
    Aug 21 '20 at 14:27
1
$\begingroup$

A positive semidefinite matrix is positive definite if and only if it is invertible (has non-zero determinant).

This is normally taken as a consequence of the following: a symmetric matrix is positive definite if and only if its eigenvalues are real and positive semidefinite if and only if its eigenvalues are non-negative. From there, we note that the determinant of a matrix is the product of its eigenvalues.

For a more direct proof, it suffices to note that for a (symmetric) positive semidefinite matrix $H$, we have $x^THx = 0 \iff Hx = 0$. In my post here, I prove this in a few different ways. From there, note that a matrix has zero determinant if and only if its nullspace (AKA kernel) is non-trivial.

$\endgroup$

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.