Abstract
Given two noncommuting matrices, A and B, it is well-known that AB and BA have the same trace. This extends to cyclic permutations of products of A's and B's. Thus if A and B are fixed matrices, then products of two A's and four B's can have three possible traces. For 2 × 2 matrices A and B, we show that there are restrictions on the relative sizes of these traces. For example, if M1 = AB2AB2, M2 = ABAB3, and M3 = A2B4, then it is never the case that Tr(M2) > Tr(M3) > Tr(M1), but the other five orderings of the traces can occur. By utilizing the connection between Lucas sequences and powers of a 2×2 matrix, a formula is given for the number of orderings of the traces that can occur in products of two A's and n B's.
| Original language | English (US) |
|---|---|
| Pages (from-to) | 200-211 |
| Number of pages | 12 |
| Journal | Fibonacci Quarterly |
| Volume | 56 |
| Issue number | 3 |
| State | Published - Aug 2018 |