# Linear Algebra munkres analysis on manifolds question.

If $A$ is an $n$ by $m$ matrix and $B$ is an $m$ by $p$ matrix, then

$$|AB| \leq m|A||B|$$ where $|A| = \max\{|a_{ij}| : i = 1,\ldots,n \text{ and} j = 1,\ldots,m\}$

Attempt: $|AB| = \max\{| \sum_{j=1}^{m}a_{ij}b_{jk} |: i = 1,\ldots,n \text{ and } k = 1,\ldots,p\} \leq \max\{ \sum_{j=1}^{m}|a_{ij}b_{jk}| : i = 1,\ldots,n \text{ and } k = 1,\ldots,p\} \leq m\max\{|a_ij|\}\max\{|b_{jk}|\} = m|A||B|$

Is this correct?

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We have $\vert a_{ij} \vert \leq \vert A \vert$ forall $i,j$ and $\vert b_{jk} \vert \leq \vert B \vert$ forall $j,k$. (Follows from definition of $\vert A \vert$ and $\vert B \vert$).
Hence, we have $$\left \vert (AB)_{ik} \right \vert = \underbrace{\left \vert \sum_{j=1}^m a_{ij} b_{jk} \right \vert \leq \sum_{j=1}^m \vert a_{ij} \vert \vert b_{jk} \vert}_{\text{Triangle inequality}} = \overbrace{\sum_{j=1}^m \vert a_{ij} \vert \vert b_{jk} \vert \leq \sum_{j=1}^m \vert A \vert \vert B \vert}^{\text{Definition of \vert A \vert  and \vert B \vert}} = m\vert A \vert \vert B \vert$$ This is true forall $i,k$. Hence, we have $$\max_{i,k}\left \vert (AB)_{ik} \right \vert \leq m\vert A \vert \vert B \vert$$
Yes. $|AB| = \max\{| \sum_{j=1}^{m}a_{ij}b_{jk} |: i = 1,...,n \text{ and} k = 1,...p\} \leq \max\{ \sum_{j=1}^{m}|a_{ij}b_{jk}| : i = 1,...,n \text{ and} k = 1,...p\} \leq \max\{ \sum_{j=1}^{m}|a_{ij}||b_{jk}| : i = 1,...,n \text{ and} k = 1,...p\}\leq \max\{ \sum_{j=1}^{m}max\{|a_{ij}|\}max\{|b_{jk}|\}\}=max\{ \sum_{j=1}^{m}|A||B|\}=m|A||B|$