Exercise 1.11.11

• Equation (21), it satisfies $|A{|}_{\infty }>0$, $|cA{|}_{\infty }=|c\left|\right|A{|}_{\infty }$. Also, $|A+B{|}_{\infty }\le |A{|}_{\infty }+|B{|}_{\infty }$.

For $|AB{|}_{\infty }=\text{largest L1 norm of the rows of }AB$, we have

${(AB)}_{i,j}={\sum <!--FUNCTION\; APPLICATION-->}_k{a}_{ik}{b}_{kj}$, so for the $l^1$ norm of row $i$, we have $|(AB){i,}| = $∑ ⁡ j| ∑ ⁡ k a{ik}b{kj}|≤∑ j∑ k |a{ik}b{kj}|≤∑ j ∑ k |a{ik}||b{kj}| = ∑ k |a{ik}|∑ j |b{kj}| = ∑ k |a{ik}||B{k,}| = |A_{i,}||B_{k,}|≤|A|{∞}|B|{∞}$So |AB|∞ = max|(AB)i,|≤ max|A|∞|B|∞ = |A|∞|B|∞   niuers 2020-03-20 00:00 Comments Comment Add answer Add answer Error: Text (in LaTeX) Cancel Submit Creating... Edit Error Chapter Problem name Text (in LaTeX) Cancel Submit Saving... Sign In × Email Password Close Sign In Forgot your password? Reset Not a member? Sign up $