# Positive and positive semi-definite matrices

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?

• 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). Aug 19 '20 at 18:58
• @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. Aug 19 '20 at 19:56
• 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 Aug 19 '20 at 20:13
• 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) Aug 19 '20 at 20:16
• It is settled, thank you. If you can write your two first comments in an answer, I'll accept it. Aug 21 '20 at 14:27

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.