The tensor product of the connectivity matrices of two diagrams is the connectivity matrix of their cartesian product, wherein pairs of nodes are linked according to whether both members of the pair are each linked in their own diagram. Sometimes a careful arrangement of the nodes in the tensor product will partially diagonalize the connectivity matrix, rendering some property of the arrangement more evident than it would otherwise have been.
A case in point consists in grouping all the pairs composed of equal
elements together in a subset 
, and the rest into the subset
. It then becomes evident whether links in a tensor square
describe pairs of links in the original diagram which have a node in
common by linking 
to 
or the reverse, both nodes in
common by linking 
to 
, or neither by linking
to 
.
Partition 
accordingly
and consider the eigenvalue equation with a similarly partitioned eigenvector
The following submatrix equations are the result:
If 
is already an eigenvalue belonging to P, it follows that
Qy=0; whether this requires that Q=0 depends somewhat on S, but
the condition is quite sufficient. If 
is not an eigenvalue
of S, 
is invertible whence 
and
would have to be singular unless y were 0.
Consideration of a left eigenvector leads to similar conclusions
regarding 
, while if 
is an eigenvalue of S as well
as P, Rx must vanish. Broadly speaking, if either of the diagonal
submatrices already has the eigenvalue, it had better be decoupled from
the other, but the details of the uncoupling can get messy.
In practice, another line of reasoning is slightly more compact. Consider
If P were a de Bruijn matrix it would be an irreducible nonnegative
matrix and 
would have a maximum eigenvalue strictly larger
than 
unless QR=0; Q=0 or R=0 would be sufficient. In
any event, continuing shows that
with
Gradually the series of requirements
develops, which can be summed to recover
The very slightly stronger result is due to the additional knowledge about P.
To illustrate this result with respect to the three rules of
Sec. 8, rearrange the indices of the matrix N to group paired
indices 
and then complementary indices 
; we obtain
0.45em
The off diagonal submatrices for Rule 14, for which 
, force
an eigenvalue larger than the minimum, whereas the other two matrices
are partially diagonal and maintain the minimum value. Rule 6 is
self-complementary, as evidenced by the equality of its diagonal
submatrices P and S.
