Homepage › Solution manuals › Gilbert Strang › Linear Algebra and Learning from Data › Exercise 3.3.2
(AH)ij = ∑ kAikHkj = AiiHij = 2i−1 2 1 i+j−1 = 2i−1 2(i+j−1) (HA)ij = ∑ kHikAkj = HijAjj = 1 i+j−1 2j−1 2 = 2j−1 2(i+j−1)
So we have (AH)ij + (HA)ij = 2i−1 2(i+j−1) + 2j−1 2(i+j−1) = 2(i+j−1) 2(i+j−1) = 1
Whereas C = AH − HB = AH + HA, so C are all ones.
