1

Let $\mathscr S$ be a subspace of $\mathbb R^p$, and let G be positive semidefinite such that $\operatorname{span}(G) \subset \mathscr S$. The following statements are equivalent:

  1. there is a basis $v_1,...,v_d$ of $\mathscr S$ such that $V^T GV>0$, where V is the $p\times d$ matrix ($v_1,...,v_d$);

  2. for all $\nu \in \mathscr S, \nu \neq 0,$ we have $\nu^T G \nu >0$;

  3. $\operatorname{span}(G) = \mathscr S$.

Proof. $2 \Rightarrow 1$. If statement 1 is not true, then there is a $w \in \mathbb R^d, w \neq 0$, such that $w^T V^T GVw=0$. So $\nu^T G \nu=0$ for $\nu=Vw$. Because V is a basis matrix of $\mathscr S$, $\nu \in \mathscr S$, and $w \neq 0$, we have $Vw = \nu \neq 0$. Hence statement 2 is not true.

$1 \Rightarrow 2$. Let $\nu \in \mathscr S, \nu \neq 0$. Then $\nu = \sum\limits_{i=1}^d \alpha_i \nu_i$, where $||\alpha|| \neq 0, \alpha$ being the vector $(\alpha_1,...,\alpha_d)^T$. Hence

$$\nu^T G \nu=\alpha^T (V^T GV) \alpha>0.$$

Thus statement 2 holds.

$3 \Rightarrow 2$. If statement 2 is not true then there exists $\nu \in \mathscr S, \nu \neq 0$, such that $\nu^T G \nu=0$. Then $\nu \perp \operatorname{span}(G)$ and statement 3 is not true.

Question: I still lack a proof of either $2 \Rightarrow 3$ or $1 \Rightarrow 3$.

My attempt:

Since $\operatorname{span}(G) \subset \mathscr S$, I only need to prove $ \mathscr S \subset \operatorname{span}(G)$.

Suppose an arbitrary vector $\nu \in \mathscr S$. By statement 2, if $\nu \neq 0$, then we have $\nu^T G \nu >0$. Hence, G is not only positive semidefinite, but G is also positive definite.

I don't know how to prove. Any proof of either $2 \Rightarrow 3$ or $1 \Rightarrow 3$ is helpful.

Jonathen
  • 1,044

2 Answers2

1

We can decompose any $v \in \mathscr{S}$ into $v = v_G + v_{\perp}$, where $v_G \in \mathrm{span}(G)$ and $v_{\perp} \in \mathrm{span}(G)^{\perp} = \mathrm{ker}(G)$. Note that since we assume $\mathrm{span}(G) \subset \mathscr{S}$, $v_{\perp} = v - v_G \in \mathscr{S}$. Now assume that (2) holds, and as you identified, you want to argue that $\mathscr{S} \subset \mathrm{span}(G)$, which is equivalent to proving that $v_{\perp} = 0$ for all $v \in \mathscr{S}$. In this case, what happens if $v_{\perp} \ne 0$ for some $v \in \mathscr{S}$?

JKL
  • 2,139
  • If $\nu_{\perp} \neq 0$, then by statement 2, we have $\nu_{\perp}^T G \nu_{\perp} >0$. Then what to do next? – Jonathen Jan 18 '24 at 19:57
  • 1
    Note that $G v_{\perp} = 0$ by definition of $v_{\perp}$ being in the kernel of $G$. So if $v_{\perp} \ne 0$, we have both $v_{\perp}^T G v_{\perp} > 0$ by assuming 2 is true, and also $v_{\perp}^T G v_{\perp} = 0$, which is a contradiction. – JKL Jan 18 '24 at 21:55
0

I assume that the "span" of $G$ refers to the column space of $G$.

$2 \implies 3$: As I state on your other post, $v^TGv = 0$ if and only if $Gv = 0$, which holds if and only if $v \in \text{span}(G)^\perp$.

The other answer provides one way to fit this into a proof of this implication; here's an alternative.

Suppose that $v^TGv > 0$ for all non-zero $v \in \mathscr S$. Thus, we have $\mathscr S \cap \text{span}(G)^\perp = \{0\}$. It follows that $$ n \geq \dim(S + \text{span}(G)^\perp) = \dim(S) + \dim \text{span}(G)^\perp \implies \\ \dim(S) \leq n - \dim \text{span}(G)^\perp \implies\\ \dim(S) \leq \dim \text{span}(G). $$ Thus, we have $\text{span}(G) \subset \mathscr S$ and $\dim \text{span}(G) \geq \dim(\mathscr S)$. It follows that $\mathscr S = \text{span}(G)$, which is what we wanted.

Ben Grossmann
  • 225,327