Let \sigma_1 and \sigma_3 be Pauli matrices (see https://en.wikipedia.org/wiki/Pauli_matrices) and let I be the identity matrix (all of these matrices are square matrices of size 2).

Now let

B_1 = \sigma_3 \otimes I \otimes \cdots \otimes I,

B_2= \sigma_1 \otimes \sigma_3 \otimes I \otimes \cdots \otimes I,

B_3= \sigma_1 \otimes \sigma_1 \otimes \sigma_3 \otimes I \otimes \cdots \otimes I,

…

B_n= \sigma_1 \otimes \cdots \otimes \sigma_1 \otimes \sigma_3;

(B_i contains exactly one factor \sigma_3 on i-th position, all factors before it are equal to \sigma_3 and all factors after it are equal to the identity matrix I).

Note that the matrices B_i described above appear naturally in the context of quantum mechanics for the following reason: since Pauli matrices anticommute: \sigma_1 \sigma_3=- \sigma_3 \sigma_1, it follows that the matrices B_i anticommute as well: B_i B_j = – B_j B_i whenever i\neq j. Also, (B_i)^2=I\otimes \cdots \otimes I is the identity matrix.

Luckily, we have that A=B_1+\cdots+B_n is a (signed) adjacency matrix for the graph we are interested in.

From the anticommutation of the matrices B_i it follows immediately that

A^2=(B_1+\cdots+B_n)^2= n.

(but I’m not sure if God will vote for it) ]]>