[proofplan]
We expand the dot product of the two incidence vectors in coordinates. Each coordinate product is $1_{\mathbb F_q}$ exactly at the points lying in both blocks, and is $0_{\mathbb F_q}$ otherwise. Thus the coordinate sum is a sum of one copy of $1_{\mathbb F_q}$ for every point of $B \cap C$, which is precisely the integer $|B \cap C|$ interpreted inside $\mathbb F_q$. The self-dot product follows by taking $C=B$.
[/proofplan]
[step:Expand the dot product in coordinates]
Fix blocks $B,C \in \mathcal B$. The standard [dot product](/page/Dot%20Product) is the map
\begin{align*}
\delta: \mathbb F_q^v \times \mathbb F_q^v &\to \mathbb F_q.
\end{align*}
For vectors $a=(a_1,\dots,a_v) \in \mathbb F_q^v$ and $b=(b_1,\dots,b_v) \in \mathbb F_q^v$, this map is defined by
\begin{align*}
\delta(a,b)=\sum_{i=1}^v a_i b_i.
\end{align*}
We write $a \cdot b$ for $\delta(a,b)$. Applying this definition to $a=\mathbf{1}_B$ and $b=\mathbf{1}_C$ gives
\begin{align*}
\mathbf{1}_B \cdot \mathbf{1}_C
=
\sum_{i=1}^v (\mathbf{1}_B)_i(\mathbf{1}_C)_i.
\end{align*}
[/step]
[step:Identify exactly which coordinates contribute to the sum]
For each index $i \in \{1,\dots,v\}$, the definition of the incidence vector gives the following two alternatives. If $x_i \in B \cap C$, then
\begin{align*}
(\mathbf{1}_B)_i(\mathbf{1}_C)_i = 1_{\mathbb F_q}.
\end{align*}
If $x_i \notin B \cap C$, then
\begin{align*}
(\mathbf{1}_B)_i(\mathbf{1}_C)_i = 0_{\mathbb F_q}.
\end{align*}
Indeed, if $x_i \in B \cap C$, then both factors are $1_{\mathbb F_q}$, so their product is $1_{\mathbb F_q}$. If $x_i \notin B \cap C$, then at least one of the two factors is $0_{\mathbb F_q}$, so the product is $0_{\mathbb F_q}$.
[guided]
We now inspect one coordinate at a time. Fix an index $i \in \{1,\dots,v\}$. By definition of the incidence vector, $(\mathbf{1}_B)_i$ records whether the point $x_i$ belongs to $B$, and $(\mathbf{1}_C)_i$ records whether the same point belongs to $C$.
There are two cases. If $x_i \in B \cap C$, then $x_i \in B$ and $x_i \in C$, so
\begin{align*}
(\mathbf{1}_B)_i(\mathbf{1}_C)_i = 1_{\mathbb F_q} \cdot 1_{\mathbb F_q} = 1_{\mathbb F_q}.
\end{align*}
If $x_i \notin B \cap C$, then $x_i \notin B$ or $x_i \notin C$. In the first subcase $(\mathbf{1}_B)_i=0_{\mathbb F_q}$, and in the second subcase $(\mathbf{1}_C)_i=0_{\mathbb F_q}$. Therefore in either subcase,
\begin{align*}
(\mathbf{1}_B)_i(\mathbf{1}_C)_i
=
0_{\mathbb F_q}.
\end{align*}
Thus each coordinate product is exactly the indicator, with values in $\mathbb F_q$, of the condition $x_i \in B \cap C$: it is $1_{\mathbb F_q}$ when $x_i \in B \cap C$, and it is $0_{\mathbb F_q}$ when $x_i \notin B \cap C$.
This is the point where incidence vectors turn set intersection into multiplication of coordinates.
[/guided]
[/step]
[step:Convert the coordinate sum into the size of the intersection]
Let
\begin{align*}
I_{B,C}:=\{i \in \{1,\dots,v\}: x_i \in B \cap C\}
\end{align*}
be the index set of points lying in both $B$ and $C$. Since $X=\{x_1,\dots,x_v\}$ is listed without repetition, the map $\beta_{B,C}: I_{B,C} \to B \cap C$ defined by $\beta_{B,C}(i)=x_i$ is a bijection. Hence $|I_{B,C}|=|B \cap C|$. Using the coordinate computation from the previous step, we first have
\begin{align*}
\mathbf{1}_B \cdot \mathbf{1}_C = \sum_{i=1}^v (\mathbf{1}_B)_i(\mathbf{1}_C)_i.
\end{align*}
Only the indices in $I_{B,C}$ contribute a nonzero summand, so
\begin{align*}
\mathbf{1}_B \cdot \mathbf{1}_C = \sum_{i \in I_{B,C}} 1_{\mathbb F_q}.
\end{align*}
A finite sum of $|I_{B,C}|$ copies of $1_{\mathbb F_q}$ is $|I_{B,C}| \cdot 1_{\mathbb F_q}$, and $|I_{B,C}|=|B \cap C|$. Therefore
\begin{align*}
\mathbf{1}_B \cdot \mathbf{1}_C = |B \cap C| \cdot 1_{\mathbb F_q}.
\end{align*}
The expression $|B \cap C| \cdot 1_{\mathbb F_q}$ means the image of the integer $|B \cap C|$ under the canonical ring homomorphism $\mathbb Z \to \mathbb F_q$, $n \mapsto n \cdot 1_{\mathbb F_q}$. Since $\operatorname{char}(\mathbb F_q)=p$, this image depends only on $|B \cap C|$ modulo $p$ and lies in the prime subfield of $\mathbb F_q$.
[/step]
[step:Specialize to the self-dot product]
Taking $C=B$ in the formula already proved gives $B \cap B=B$. Therefore
\begin{align*}
\mathbf{1}_B \cdot \mathbf{1}_B = |B| \cdot 1_{\mathbb F_q}.
\end{align*}
This proves the stated self-dot product formula and completes the proof.
[/step]
