Given $\nu^T G \nu=0$, how do we get $\nu \perp \operatorname{span}(G)$?

Question

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:

for all $\nu \in \mathscr S, \nu \neq 0,$ we have $\nu^T G \nu >0$;
$\operatorname{span}(G) = \mathscr S$.

Proof.

$3 \Rightarrow 2$. 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: Given $\nu^T G \nu=0$, how do we get $\nu \perp \operatorname{span}(G)$?

I know by definition, if I want to say $\nu$ is orthogonal of a subspace $W$, then for every vector $w \in W$, we should have $\nu^T w=0$. But here I have span, which confuses me.

$\text{span}(G)$ is not standard usage of the "span" operator. Do you mean the column-space of $G$? — Ben Grossmann, Jan 18 '24 at 21:29
@BenGrossmann Yes, in the textbook, the author clearly defines that he choose to use span(G) to denote the column space of G, which means the space spanned by the columns of G. I guess this is just the usage of the span. — Jonathen, Jan 19 '24 at 02:24

Ben Grossmann · Accepted Answer · 2024-01-19T03:20:39.727

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I assume that the "span" of $G$ is its column-space.

The key fact here is that because $G$ is positive semidefinite, $v^TGv = 0$ if and only if $Gv = 0$; I present several proofs of this fact on my post here.

So, the fact that $v^TGv = 0$ implies that $Gv = 0$, which is to say that $v$ is an element of the null space of $G$, which (because $G$ is a symmetric matrix) is the orthogonal complement to the column space of $G$.

edited Jan 19 '24 at 03:20

answered Jan 18 '24 at 21:34

Ben Grossmann

225,327

Thanks! But why do you emphasize that " (because G is a symmetric matrix)"? – Jonathen Jan 19 '24 at 02:34
The key fact here is that $v^TGv = 0$ if and only if $Gv = 0$. I guess you need to say provided that $G$ is positive definite? – Jonathen Jan 19 '24 at 02:48
Can you give me a proof about "$v^TGv = 0$ if and only if $Gv = 0$" here? Of course, the other way is obvious. – Jonathen Jan 19 '24 at 03:08
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In general, the null space is the orthogonal complement of the column space of $G^T$. So, it’s important that $G^T=G$. – Ben Grossmann Jan 19 '24 at 03:22
Regarding the next two comments, see the link that I had put in my answer – Ben Grossmann Jan 19 '24 at 03:23

Given $\nu^T G \nu=0$, how do we get $\nu \perp \operatorname{span}(G)$?

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