RC Schroeppel: The quaternion matrix "determinant" for 4x4 matrix with all 4 rows = (a,b,c,d) and all letters quaternions, is 0. Fred Lunnon: argues this is true due to row operations enabling reduction to real# times form where a=1, b=i, c=j, d=k. WDS: to complete that argument, first of all, we would need to know that a row-op (e.g. adding a row to another) leaves the "determinant" unaltered. But det( ab 00 ) = 0 while det( ab ab ) = ab-ba is in general NOT zero for quaternions, so Lunnon's row-op-based demonstration is not legitimate. ---- So what is really going on here? Due to the distributive law (valid for quaternions) and the fact reals commute with everything, it would suffice to show the identity is valid when a,b,c,d are selected from the basis set {1,i,j,k} only, which is 4^4=256 cases to check. You could check that by brute force, but there is a better way. Since any two distinct among {i,j,k} anticommute, but everything commutes with itself and 1, it is possible to see the 256 cases work in your head without actually checking any of them. Essentially: in the commuting cases, it works. In the anticommuting cases, the terms T in question yield T = -T, which works exactly when T=0, which happens because of symmetry: with 4 letters abcd, each from {1,i,j,k}, by the pigeonhole principle it is necessary that either at least two must be identical or at least one must be 1. This pigeonhole observation also means you can always perform one well-chosen single interchange (an odd perm) without changing the result, and then all the other terms on the left side of the equation arise from the same orbit as on the right side. QED (that was really 2 distinct proofs, so I should add a second QED). This also shows that you can use any number>=4 of quaternion letters -- Schroeppel was using 4, but 5, 6,... also work. Note 2 and 3 do NOT work. Now let me generalize that: THEOREM: Let N>=1. Let each letter a,b,c,... be an NxN matrix with real (or complex) entries. Then: Sum of all multilinear monomials made of N*N distinct letters using odd-permutations only = same sum using even-perms only. (Each monomial term has degree=N*N. The total number of letters in the alphabet must be at least N*N otherwise we get the trivial 0=0 identity.) PROOF: Suffices to prove where each letter is a basis matrix (e.g. single entry 1 with rest 0... also the identity matrix is a basis matrix and we agree to get rid of 1 extra arbitrary basal). By pigeonhole principle >=two matrices must be same or >=one must be identity matrix which commutes with everything. Either way for them we can do an interchange which converts each term on left side into equal term on right making it obvious left=right. QED Similarly, even though I have not checked this at all, essentially the same argument I just made, should also show that sum of even-perm-based monomials (abcdefgh) = sum of odd-perm-based monomials (abcdefgh) where each letter is an octonion and where all monomials are parenthesized in all possible binary-tree manners. (You need to specify parenthesizations since octonion-mult is non-associative so ab*c and a*bc unequal.)