Okey here i am stuck in a problem in matrix norm proof, i don't need to the proof of Matrix norm , i just need to know how the calculations are done to part (1) and (2) in figure.
Here is the complete problem sol. https://www.slader.com/textbook/9780538733519-numerical-analysis-9th-edition/442/exercises/8/
Here is where i am stuck.
On
At (1) notice that on the left side of the equation you're dealing with a finite sum so that you can change the order of summation without altering the result. Thus we can first take the sum over $k$ and then over $j$: \begin{equation} \sum^n_{i=1}\sum^n_{j=1}\sum^n_{k=1}|a_{ik}||b_{kj}| = \sum^n_{i=1}\sum^n_{k=1}\sum^n_{j=1}|a_{ik}||b_{kj}| = \sum^n_{i=1}\sum^n_{k=1}|a_{ik}|\sum^n_{j=1}|b_{kj}| \end{equation} where in the last equation we just use distributivity.
As for (2), everyithing is already there: We observe that independently of $k$ we get $|b_{kj}|\leq \sum^n_{l=1}|b_{lj}|$, and thus \begin{equation} \sum^n_{i=1}\sum^n_{k=1}|a_{ik}|\sum^n_{j=1}|b_{kj}|\leq \sum^n_{i=1}\sum^n_{k=1}|a_{ik}|\sum^n_{j=1}\sum^n_{l=1}|b_{lj}|. \end{equation}
In going from line 4 to line 5, they pull out $|a_{ik}|$ from the inner-most sum since it is constant with respect to $j$.
In line 5, they note in the underbrace that $|b_{kj}|$ is less than the sum $|b_{1j}|+|b_{2j}|+\cdots+|b_{nj}|$, i.e. the absolute value of one entry is less than the sum of the absolute values of all the entries in that column. Hence, they can replace the single term $|b_{kj}|$ with the sum $\sum_{l = 1}^{n}|b_{lj}|$ since $|b_{kj}| \le \sum_{l = 1}^{n}|b_{lj}|$ .