# Why do these conditions ensure symmetric positive-definiteness?

Forgive me, if I have made a blunder or missed something obvious, I'm not a mathematician!

I'm trying to understand 2 seemingly simple lines of maths - and understand how the conclusions are drawn from these. So here are the three lines (or rather two lines of maths, one statement at the end):

Let:

$$L(n) := \{ A \in M(n) : A^T = A \}$$

$$P(n) := \{ A \in L(n) : A > 0 \}$$

"The boundary of $$P(n)$$ is the set of singular positive semidefinite matrices"

So from what I understand here, the set L contains symmetric ($$A^T = A$$) n dimensional square matrices with real entries ($$A \epsilon M(n)$$). Then I have presumed that the part, $$A > 0$$, refers to the determinant of A, and not A itself.

I am also inferring (possibly incorrectly) that P(n) is therefore the space of SPD matrices..?

So why does ensuring we have square, real, symmetric matrices whose determinant is above 0 (this means they're invertible..?) ensure that the eigenvalues of said matrices are above 0 (my understanding of what makes a matrix positive definite)?

• I think you are misreading $A>0$ as "$A$ has a positive determinant" when you should read it as "$A$ is strictly positive definite". The author might have written $A\succ 0$ to make this somewhat clearer. – kimchi lover Dec 6 at 15:30
• Unfortunately the author used >, but you may be right! If so then my question is rather moot. Would it not make any sense to draw the SPD conclusions from my original interpretation? – bidby Dec 6 at 15:34
• It's all nonsense under the determinant interpretation (the 2 by 2 diagonal matrix with -1, -1 on the diagonal has positive determinant and is not PSD or PD; the diagonal matrix with -1 and 0 on the diagonal is not in the closure of the PSD matrices). Under the other interpretation it all makes sense, in a trite way. – kimchi lover Dec 6 at 15:42

As others have stated, $$P(n)$$ is intended to represent the set of all positive definite $$n \times n$$ matrices (not the set of all matrices with positive determinant, as bidby initially thought). We let $$A \succ 0$$ denote a positive definite matrix and $$A \succeq 0$$ denote a positive semidefinite matrix.
Claim: $$\partial P(n) \overset{\textrm{Def}}{=} \overline{P(n)} \setminus \mathring{P(n)} = \{A \in \mathbb{R}^{n \times n} | A \succeq 0, \nexists A^{-1}\}$$.
Proof: Let $$A \in \partial P(n)$$, $$(A_k)_{k \in \mathbb{N}} \subseteq P(n)$$, and $$z \in \mathbb{R}^n$$. Then $$z^T A z = z^T \left( \lim_{k \to \infty} A_k \right) z = \lim_{k\to\infty} \underbrace{\left( z^T A_k z \right)}_{\geq 0} \geq 0,$$ since the functions $$f : \mathbb{R}^n \to \mathbb{R}, \; x \mapsto z^T x$$ and $$g : \mathbb{R}^{n \times n} \to \mathbb{R}^n, \; B \mapsto B z^T$$ are linear (and linear transformations between finite dimensional vector spaces are continuous). Therefore, $$A$$ is positive semidefinite.
We now show that $$A$$ is singular. Since $$A$$ is positive semidefinite, all its eigenvalues are non-negative. We know that $$A$$ is not positive definite, since $$\mathring{P(n)} = P(n)$$ by this post. But if the eigenvalues were all positive, then $$A$$ would be positive definite. Hence, $$A$$ has $$0$$ as an eigenvalue -- i.e. $$\exists$$ eigenvector $$v \in \mathbb{R}^n \setminus \{0\}$$ with $$Av = 0 \cdot v = 0$$. Therefore, $$\textrm{Ker}(A) \neq \{0\}$$, so $$A$$ is singular (see 3. here).
(A trivial example of such an $$A$$ and such a sequence $$(A_n)$$ is $$A := 0 \in \mathbb{R}^{1 \times 1}$$ and $$A_n := 1/n$$.)