Confirming an error in a textbook: O'Neill's Semi-Riemannian Geometry I'm working on the following exercise in O'Neill's "Semi-Riemannian Geometry":
(Page 53, 12) Let $b$ be a symmetric bilinear form on $V$. The nullspace of $b$ is $N = \{v : b(v,w) = 0, \, \forall w \in V\}$. The nullcone of $b$ is the set $\Lambda$ of all null vectors in $V$. Let $A = \Lambda \cup 0$, so $A \supset N$. Prove: (a) $N$ is a subspace, but $A$ is not unless $A=0$ or $V$. (b) $b$ is non-degenerate iff $N=0$; $b$ is definite iff $A=0$. (c) $b$ is semidefinite iff $N=A$.
I think two of the claims here are false.

$A$ is not a subspace unless it is $0$ or $V$.

Counterexample:  Take $V = \mathbb{R}^2$, and let $b(e_1,e_1) = b(e_2,e_2) = 1$ but $b(e_1,e_2) = -1$. Then if we write $v = (x,y)$, the condition $b(v,v) = 0$ corresponds to the equation
$$0 = x^2 - 2xy + y^2 = (x-y)^2$$
whose solution space is the diagonal subspace spanned by $(1,1)$, which is a subspace equal to neither $0$ nor $V$.

$N=A$ implies $b$ semidefinite.

Counterexample: Take $V = \mathbb{R}^2$, $b(e_1,e_1) = -1$, $b(e_2,e_2) = b(e_1,e_2) = 0$. Then if $v = (x,y)$, we have $b(v,v) = -x^2$, so $A$ is the set of vectors with zero in the first coordinate. However, if $v$ has zero in its first coordinate, then it is in $N$, so $A=N$. However, $b$ is not semidefinite.

Are these indeed counterexamples, or am I simply confused? Thanks for your time.
 A: For part (a), a much easier counterexample is 
$$ b = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$
in which case we have that $A = N = \mathrm{span}\{e_2\}$. 
In fact, this tells us where the typo in the textbook is. The correct statement is

(a) The set $A$ is a subspace $\iff$ $A = N$ 

Proof: The implication $\Leftarrow$ is obvious. So we show the other way. Suppose $A\setminus N$ is nonempty. Let $v$ be an element of $A\setminus N$. By definition there exists $w$ such that $b(v,w) \neq 0$. 
Consider the vector $z = kv + w$. We have that
$$ b(z,z) = b(w,w) + 2k b(v,w) $$
If $b(w,w) \neq 0$, we can choose $k\neq 0$ such that $b(z,z) = 0$. If $b(w,w) = 0$, then for for any $k\neq 0$ we have $b(z,z) \neq 0$. In any case we have found three linearly dependent vectors $v,w,z$, two of which belonging to $A$ and not the third. This implies that $A$ is not a subspace. q.e.d.
Incidentally, the case $A = V$ is included, since

Lemma: if $b(v,v) = 0$ for every $v\in V$, then $b\equiv 0$. 

Proof: from the polarisation identity 
$$ b(v+w,v+w) = b(v,v) + b(w,w) + 2 b(v,w) $$
so if the first three terms all vanish by assumption, we have that $b(v,w) = 0$ for any $v,w\in V$, and hence $b$ is identically zero. In particular, $V = N$. q.e.d.
A: Daniel Fischer has confirmed the mistake in part (a). For part (c), semidefiniteness is presumed to mean either positive or negative semidefinite. I figure I might as well post my solution to that part, in case somebody is stuck on this matter.
(c) $b$ is semidefinite iff $b(v,v) \geq 0$ or $b(v,v) \leq 0$. We already know that $N \subset A$, so we have to show that $A \subset N$. If $b$ is positive semidefinite, this follows easily from Cauchy-Schwarz inequality. If $b$ is negative semidefinite, then $-b$ is positive semidefinite, so if $b(v,v) = 0$ then $-b(v,v) = 0$ so $-b(v,w) = 0$ for all $w$ (by the prior step), so $b(v,w) = 0$ for all $w$. For the converse, suppose that $N=A$. We claim then that $b$ is either positive or negative semidefinite. Suppose not, for the sake of contradiction, so $b(v,v) < 0$ and $b(w,w) > 0$. Write
$$b(v+\lambda w, v + \lambda w) = b(v,v) + \lambda^{2} b(w,w) + 2\lambda b(v,w)$$
Three cases emerge. If $b(v,w) > 0$, then this is negative for $\lambda = 0$, and positive for $\lambda >> 0$, so we can find some $\lambda$ for which this is zero. If $b(v,w) < 0$, then the same is true (since $\lambda^2$ dominates $2\lambda$), so we can again choose $\lambda$ to make this zero. If $b(v,w) = 0$, it is even easier to find the appropriate $\lambda$. To recap this step, we have found $v + \lambda w$ which is in $A$ for $\lambda > 0$. We claim it is not in $N$. Consider
$$b(v+ \lambda w, v - \lambda w) = b(v,v) - \lambda^{2} b(w,w)$$
If this is zero, then we can add it to the above polarization identity to get
$$2b(v,v) + 2\lambda b(v,w) = 0$$
Since $\lambda > 0$, this impossible unless $b(v,w) > 0$. But then
$$b(v+\lambda w, w) = b(v,w) + \lambda b(w,w) > 0$$
In any case, we must conclude that $v + \lambda w \in A \setminus N$.
