§ Linear algebraic proof of the handshaking lemma

We wish to show that the number odd vertices is even. Let AA be the adjacency matrix of the undirected graph GG. Since GG is undirected, A=ATA = A^T. Now move everything to F2F_2, including AA. This means that AA has entries {0,1}\{0, 1\}. Now, denote the vector of all ones by o(1,1,1)o \equiv (1, 1, \dots 1). See that AoAo counts the partities of the degrees of each vertex, and oT(Ao)o^T(Ao) counts the sum of parities of the degrees of each vertex. Note that the vertices of even degree with add 00 to the sum oTAoo^TAo, while odd vertices will add a 11. Thus, oTAoo^TAo will equal the parity of the number of odd vertices. As we wish to show that the number of odd vertices is even, we want to prove that oTAo=0o^TAo = 0. We will now algebraically simplify oTAoo^TAo (does anyone have a cleaner proof?) giving us:
oTAo=ijoiAijoj=i=joiAijoj+i<joiAijoj+ojAjioi(A is symmetric; Aji=Aij)=i=joiAijoj+i<joiAijoj+ojAijoi=i=joiAijoj+i<j2oiAijoj(F2 has characteristic zero, so 2=0)=i=joiAijoj+0(replace i=j with k)=kokAkkok(Akk=0 since graph has no self loops)=k0ok2=0 \begin{aligned} &o^TAo = \sum_{ij} o_i A_{ij} o_j \\ &= \sum_{i=j} o_i A_{ij} o_j + \sum_{i < j} o_i A_{ij} o_j + o_j A_{ji} o_i \\ &\text{($A$ is symmetric; $A_{ji} = A_{ij}$)} \\ &= \sum_{i=j} o_i A_{ij} o_j + \sum_{i < j} o_i A_{ij} o_j + o_j A_{ij} o_i \\ &= \sum_{i=j} o_i A_{ij} o_j + \sum_{i < j} 2 \cdot o_i A_{ij} o_j \\ &\text{($F_2$ has characteristic zero, so $2 = 0$)} \\ &= \sum_{i=j} o_i A_{ij} o_j + 0 \\ &\text{(replace $i = j$ with $k$)} \\ &= \sum_{k} o_k A_{kk} o_k \\ &\text{($A_{kk} = 0$ since graph has no self loops)} \\ &= \sum_{k} 0 \cdot o_k^2 = 0 \end{aligned}
So, the number of vertices of odd degree is even. I want to avoid this computation with respect to the basis, but I'm not sure how to do that.

§ A simplification from arjun

Since Akk=0A_{kk} = 0, we have that A=B+BTA = B + B^T for BB lower triangular. This allows us to simplify: \begin{aligned} & o^T A o = o^T (B + B^T) o = \\ & =o^T B o + o^T B^T o = \langle o, Bo \rangle + \langle Bo, o \rangle \\ & = 2 \cdot \langle o, Bo \rangle = 0 \end{aligned}