A Bilinear form on $(\mathcal{P}(X),\Delta)$, as a vector space over $\mathbb{Z}/2\mathbb{Z}$

Question

This is exercise 3.6 from Elstrodt's Maß- und Integrationstheorie.

In English: Let $X$ be a set and consider the symmetric difference operator $\triangle$ on its subsets. Then $(\mathcal{P}(X),\triangle)$ becomes an abelian group with $0=\varnothing$, and $A\triangle A=0$ holds for all $A\in\mathcal{P}(X)$. Hence, it can be thought as a vector space over $\mathbb{Z}/2\mathbb{Z}$ in a natural way. For example, when $X$ has $n$ elements, one can readily see that the dimension of the vector space becomes $n$.

Now Suppose $X$ is a finite set and consider the map $$\langle-,-\rangle:\mathcal{P}(X)\times\mathcal{P}(X)\rightarrow\mathbb{Z}/2\mathbb{Z}$$ $$\langle A,B\rangle:=|A\cap B|+2\mathbb{Z}$$ where $|A\cap B|$ denotes the number of elements $A\cap B$ has. It is easy to check that this is a symmetric bilinear form on $\mathcal{P}(X)$, and when given a subspace $V$ of $\mathcal{P}(X)$ we may restrict $\langle-,-\rangle$ to $V\times V$ and consider it as a symmetric bilinear form on $V$.

The question is: For which $V$, is the bilinear form $\langle-,-\rangle$ non-degenerate?

I checked several individual cases such as when $V=\mathcal{P}(X)$ or when the dimension of $V$ is $\leq 3$, but could not recognize any satisfying patterns. Could someone enlighten me?

Answer 1

Not sure if this helps, but I picture the space like this. If $X$ has $n$ elements and $A\subseteq X$, then $A$ can be represented as a vector of $0$'s and $1$'s, where an entry is $0$ if the corresponding element is not in $A$ and $1$ if it is. It is not too hard to check that $A\Delta B$ is then just the sum of their vectors modulo two.

Example: $X=\{1,2,3,4,5\}$, $A=\{1,2,3\}$ and $B=\{2,3,5\}$, then $A\Delta B=\{1,5\}$. In our notation, $A=(1,1,1,0,0)$ and $B=(0,1,1,0,1)$, adding these modulo 2 gives $(1,0,0,0,1)$.

It's not too difficult to check that $A\cap B$ is given by taking the coordinate-wise products of the vectors of $A$ and $B$. So in our example $A\cap B=\{2,3\}$ and you can check that this works.

Maybe using this to see the bilinear form can help you...