Prove a set is not a vector space

[kostas230]

Homework Statement

Let $b$ be a symetric bilinear form on $V$ and $A = \{ v\in V : b\left(v,v\right)=0\}$. Prove that $A$ is not a vector space, unless $A = 0$ or $A = V$.

2. The attempt at a solution

If we suppose that $A$ is a vector space then for every $v,w\in A$ we must have: $b\left(v+w,v+w\right) = b\left(v,w\right)$. This try didn't go anywhere. I think I should construct a counter example, but I wouldn't know from where to start.

 

[Orodruin]

Assume that $V = \mathbb R^2$ and that
$$
b(v,w) =
v^T
\begin{pmatrix}
1 & 0 \\ 0 & 0
\end{pmatrix}
w.
$$
For $v = \begin{pmatrix} x \\ y\end{pmatrix}$, we now have
$$
b(v,v) = x^2
$$
so $A$ is the set of vectors with $x = 0$, i.e., vectors of the form
$$
v = \begin{pmatrix} 0 \\ y \end{pmatrix},
$$
which is basically the vector space $\mathbb R$. Thus $A \neq 0$ and $A \neq V$ while still being a vector space.

Are you sure the problem formulation is correct?

 

[kostas230]

I think so. It's problem 12, page 53 from O' Niell's book "Semi-Riemannian Geometry with Applications to General Relativity".

 

[Orodruin]

I do not have that book but "Semi-Riemannian" suggests that eventually there will be a (pseudo-Riemannian) metric involved. There were no additional constraints on the bilinear form apart from just being symmetric? For example, the light-cone extending from the origin of Minkowski space is obviously not a vector space as addition of vectors on the light-cone can end up being time-like or space-like.

 

[kostas230]

No. V is an finitely dimensional real vector space over the real numbers and b is just a bilinear form. It doesn't mention semi-Riemannian or even Minkowski spaces.

Maybe the author made a mistake...

 

[vela]

I ran across a thread on stackexchange which suggests that the problem should have said that $A$ is not a vector space unless $A=0$ or $A=N$, where $N$ is the nullspace of $b$.

 

[pasmith]

No. V is an finitely dimensional real vector space over the real numbers and b is just a bilinear form.

In that case, you can write $b(v,w) = v^TBw$ for some real symmetric matrix $B$.

It is then easy to show that $\ker B \subset A$. But $A \subset \ker B$ doesn't necessarily hold.

Consider the case where $V = \mathbb{R}^4$ and $B$ is diagonal with eigenvalues 1, 1, -1 and 0 so that $$v^TBw = v_1w_1 + v_2w_2 - v_3w_3.$$ Now $\ker B = \{(0,0,0,t) : t \in \mathbb{R} \}$ and $A = \{(x,y,z,t) \in \mathbb{R}^4 : z^2 = x^2 + y^2\}$ which is the product of a double cone and the real line. It's not a subspace of $\mathbb{R}^4$.

If it happens that $A = \ker B$ then $A$ is indeed a subspace. That $A = \ker B$ need not be $\{0\}$ or $V$ is shown by Orodruin's example.

 

[pasmith]

In that case, you can write $b(v,w) = v^TBw$ for some real symmetric matrix $B$.

And since $B$ is symmetric it is diagonalisable, so there exists a basis of eigenvectors. There is no reason not to use that basis, so you can take $$b(v,w) = \sum_{i=1}^n \lambda_i v_i w_i$$ where $v = \sum v_i e_i$, $w = \sum w_i e_i$ and $e_i$ is an eigenvector of $B$ with eigenvalue $\lambda_i$.

Your aim is to show that if $v \in A \setminus \ker B$ then there exists a $w \in A \setminus \ker B$ such that at least one of $v + w$ and $v - w$ is not in $A$.