Luke Wiljanen

I'm a math graduate student studying number theory.

Some of My Favorite Theorems/Topics

Fun Proofs

Recreational Math

Commutativity Relations

The goal is to classify which relations of the form ab = bwa, where w is a word in a and b, imply commutativity for semigroups. If such a relation implies commutativity, then we call it a commutativity relation.

Example: Let S be a semigroup such that ab = baba for all a and b in S. Then, S is commutative.

Proof: Let a and b be elements of S. Then,

ab = baba
   = (abab)(abab)
   = (ab)(ab)(ab)(ab)
   = abab
   = ba.
Therefore, S is commutative.

I do not know of any example of noncommutative semigroups where a relation of the form ab = bwa is satisfied.

Some Known Commutativity Relations:

  • ab = (ba)n for each positive integer n
  • ab = bnam for each pair of positive integers (n,m)
  • ab = ba3ba3

Delayed Multiplying Matrices

The goal is to try to compute the characteristic polynomial of certain matrices which exhibit nice factorization properties for their characteristic polynomials.

6 by 6 matrix with 9 ones and the rest of the entries 0, on the top right 3 by 3 identity matrix, on the left hand side there is a 3 by 3 identity matrix with each row is repeated twice in a row

The above matrix is the third delayed doubling matrix. It is constructed by taking the three by three identity matrix, repeating each row twice, and then putting the doubled identity matrix on the left side of a six by six matrix, putting the three by three identity matrix on the top right, and putting zeros on the bottom right.

For a positive integer n, we can construct a 2n by 2n matrix in a completely analogous way. We denote that matrix Mn. We write pn for the characteristic polynomial of Mn.

Theorem: Let n be a power of 2, and let m be a positive integer. Then, each entry of the matrix (Mn)m is a Fibonacci number.

After computing a few of the characteristic polynomials, one finds that they seem to all factor in a very nice way.

Conjecture: For each positive integer n, there exists a non-negative integer t, and non-negative integers n0, n1, ..., nt such that pn(x) = xn0(x2 - x - 1)(xn1 - 1)···(xnt - 1).

After further computations, one is seemingly led to a formula for t.

Conjecture: For n = 2rm with m odd, we have 1+t = ∑d|m | (ℤ/dℤ)×/⟨ 2 ⟩ |.

For a positive integer N, we write [N] = {0,1,...,N-1}. For a positive integer n, the delayed doubling map Tn : [2n] → [2n] is the map defined by Tn(m) = 2m if m is in [n] and by Tn(m) = m - n if m is in [2n] \ [n].

Using the maps Tn we also have a way to seemingly predict the exponents n1, ..., nt. We show case this when n = 15 below. Note that p15(x) = x7(x2 - x - 1)(x7 - 1)(x6 - 1)(x5 - 1)(x3 - 1).

1 to 2 to 4 to 8 to 16 to 1, 3 to 6 to 12 to 24 to 9 to 18 to 3, 5 to 10 to 20 to 5, 7 to 14 to 28 to 13 to 26 to 11 to 22 to 7, (15 to 0 to 0)

In the example, we see that the non-degenerate orbit lengths coincide with the exponents in the xni - 1 factors of the characteristic polynomial.

Conjecture: The exponents n1, ..., nt are exactly the orbit lengths of the non-degenerate orbits of the map Tn.

Ingredients for a Potential Proof: The Coefficient Theorem for Digraphs allows you to relate the coefficients of the characteristic polynomial in terms of linear subdigraphs. This should allow us to connect the orbits of Tn to the characteristic polynomial of Mn. I haven't attempted to formalize this outline due to other projects; please let me know if you carry out such a proof.

(For a reference for the Coefficient Theorem for Diagraphs, one can see Cvetković, Doob, and Sachs's book Spectra of Graphs, or Peña and Rada's paper Energy of digraphs (2008).)