Homepage › Solution manuals › Gilbert Strang › Linear Algebra and Learning from Data › Exercise 1.11.12
(AB)ij = ∑ kaikbkj, so
$|||AB|||{∞} = max|(AB){ij}| = max|∑ k a{ik}b_{kj}|≤ max∑ k |a{ik}b_{kj}|≤ max∑ k |a{ik}||b_{kj}|≤ max∑ k |a{ik}|≤|||B|||{∞}max∑ k |||A|||{∞} = n|||A|||{∞}|||B|||_{∞}$This is equivalent to (mp|||AB|||∞) ≤ (mn|||A|||∞)(np|||B|||∞)
