Literally three different definitions of "The kernel of a quadratic form" in the same proof I'm confused on a couple of propositions in Bourbaki, Lie Groups and Lie Algebras, Chapter 5.  We are given a quadratic form $q$ on $V = \mathbb{R}^n$ which is positive, meaning $q(v) \geq 0$ for all $v$.
To $q$ we can associate a symmetric, positive bilinear form $$B(v,w) = \frac{1}{2}[q(v+w) - q(v) - q(w)]$$
We recover $q$ as $q(v) = B(v,v)$.
By the kernel of $q$, I would normally assume we are talking about one of two things:

1 .  The set of $v \in V$ such that $q(v) = 0$.
2 .  The set of $v \in V$ such that $B(v,w) = 0$ for all $w \in W$.

It is straightforward that the second set is contained in the first.  In the beginning of Lemma 4, they cite a result in Algebra, Chapter IX which says that since $q$ is positive, they are actually equal.  I looked up this result, I still can't figure out if it actually implies what is claimed.  Anyway.
On the other hand, in the beginning of the proof of Lemma 4, they seem to be using yet another definition.  They are saying that if $v = |c_1|a_1 + \cdots +|c_n|a_n$ is in the kernel of $q$, then
$$\sum\limits_i q_{ij}|c_i| = 0$$
for all $j$.  That is...not how you compute $q(v)$.  You don't treat $q$ as a linear transformation and multiply the column vector.  Right?  I thought $q(v)$ would be
$$\sum\limits_{i,j} q_{ij} |c_ic_j| $$


 A: Added: it appears that their positive is positive semidefinite. Not hard to adjust the stuff below to semidefinite case
your early question, positivity of the quadratic form says that both versions of the kernel are the single vector $0.$
Think of a symmetric positive definite real matrix $M.$ If we have a column vector $X$ with $X^T M X = 0,$ then $X = 0.$
The bilinear form is
$$ B(X,Y) = X^T M Y = Y^T M X. $$
If, for fixed $X,$ we always have
$$  ( X^T M) Y = 0, $$ 
we are allowed to take $Y = MX,$ which proves that
$$X^T M = 0, M X = 0.$$ Since $M$ is definite,
$$ X = 0 $$
A: We have $M$  real, symmetric, and positive semidefinite.  We may take an orthogonal matrix $P$ such that $ P^T D P = M,$ where $D$ is real diagonal. Let dimension be $n.$ Let the first $n-r$ diagonal elements of $D$ be zero, that is $D_{jj} = 0$ for $1 \leq j \leq n-r.$ Then  $D_{jj} > 0$ for $n-r+1 \leq j \leq n.$ 
Your condition (i) reads $X^T M X = 0.$ This becomes $X^T P^T D P X = 0.$ Make a new vector
$$  Y = PX. $$ We have
$$  Y^T D Y = 0. $$
That is $$ \sum_{j = n-r+1}^n D_{jj} \; \; y_j^2 = 0. $$
Positivity says  $$y_j = 0 \; \; \; \; \mbox{for} \; \; \; \; n-r+1 \leq j \leq n.$$
Alright, for $i \leq n -r+1,$ we have $D_{ij} = 0.$ So 
$$ (DY)_i = \sum_j D_{ij} y_j = 0 \; \; \; \; \mbox{for} \; \; \; \;i \leq n_r+1.  $$
Different for $i \geq n-r.$
$$ (DY)_i = \sum_j D_{ij} y_j = D_{ii} y_i = 0 \; \; \; \; \mbox{for} \; \; \; \;i \geq n_r.  $$
Condition (i) implies
$$ DPX = 0.  $$ Therefore
$$ P^T D P X = 0, $$
$$  MX = 0. $$
This was your condition (ii).
This material comes under the general heading of Witt's Theorem; it is easier here because we have the real numbers and semidefiniteness, the "null cone" is just a hyperplane.
https://en.wikipedia.org/wiki/Witt's_theorem
I like the book by Cassels for this material. I have a copy and used it for something a week ago, now I can't find it. Found it...
A: Okay, here's what I'm trying to prove.  Let $V = \mathbb{R}^n$, let $M$ be an $n$ by $n$ real symmetric matrix such that $m_{ij} \leq 0$ for all $i \neq j$, and such that $X^TMX \geq 0$ for all column vectors $X$.  The following are equivalent for $X$:
(i): $X^TMX = 0$;
(ii): $MX = 0$;
(ii): $X^TMY = 0$ for all column vectors $Y$.
(ii) $\Rightarrow$ (iii) is clear, since $(MX)^T = X^TM^T = X^TM$, and one can then multiply on the right by $Y$.  (iii) $\Rightarrow$ (i) is also clear.  The difficulty for me is showing that (i) $\Rightarrow$ (ii).
Proof of (i) $\Rightarrow$ (ii) is the case $n = 2$:
Let $M = \begin{pmatrix} a & b \\ b & d \end{pmatrix}$ where $b \leq 0$.  If $X = \begin{pmatrix} x \\ y \end{pmatrix}$, then 
$$X^TMX = ax^2 + 2bxy + dy^2$$
If $M$ is invertible, we already know that (i) and (ii) are equivalent to $X$ being $0$.  So we may assume $ad = b^2$.  Under the assumption $ax^2 + 2bxy + dy^2 = 0$, multiply by $b$ and use the fact that $2b^2 = ad + b^2$ to get
$$0 = abx^2 + 2b^2xy + dy^2 = abx^2 + adxy + b^2xy + dy^2 = (ax + by)(bx + dy)$$
If one of the rows of $M$ is zero, then being symmetric, $M$ must be of the form $\begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix}$ or $\begin{pmatrix} 0 & 0 \\ 0 & d \end{pmatrix}$, and these are easy to work out.  Otherwise, the determinant of $M$ being zero, we can conclude that $(a,b)$ is a nonzero scalar multiple of $(d,b)$, and so $(ax+by)(bx + dy) = 0$ is equivalent to $ax + by$ and $bx + dy$ both being zero, or in other words, $MX = 0$, which was what I wanted to show.
