Positive definite implies positive semi-definite

Question

I have read in the post, but there is something missing I'd like to know. I do believe it's not a duplicate post. I have a specified question, probably a trivial one.

First, let $(s_n)_{n\geq 0}$ be a sequence of real numbers, and define the $n$th Hankel matrix $H_n=(s_{i+j})_{0\leq i,j\leq n}$.

If it's assumed that $\det H_n>0$ for all $n\geq 0$, then by Sylvester's Criterion [note: $H_n$ is Hermitian as it's symmetric], we get that $H_n$ is positive definite. Does it then imply that $H_n$ is positive semi-definite as well?

TL;DR: Is a real positive definite matrix always a positive semi-definite matrix?

I would say yes, but I want someone to confirm it for me.

Answer 1

Yes.

If $x^TAx > 0$ for all $x \in \mathbb{R}^n \setminus \{\mathbf{0}\}$, then $x^TAx \geq 0$ for all $x \in \mathbb{R}^n \setminus \{\mathbf{0}\}.$

As we always have $\mathbf{0}^TA\mathbf{0} = 0$, it also follows that $x^TAx \geq 0$ for all $x \in \mathbb{R}^n.$

Answer 2

Yes, if $\forall x \ne 0, x^TAx > 0$, then we have $\forall x \ne 0, x^TAx \ge 0$.

All positive numbers are nonnegative.