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For $$A\in C_{n\times n} \text{ define the norm of A to be } ||A||=max{|[a]_{ij}|}.$$

Prove: $$ ||A+B|| \leq ||A||+||B|| $$

I really don't know how to approach this proof.

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up vote 4 down vote accepted

Let $C=A+B$. Then $c_{ij}=a_{ij}+b_{ij}$, so $|c_{ij}|\le|a_{ij}|+|b_{ij}|$ for each pair $i,j$ with $1\le i,j\le n$. Moreover, each $$|a_{ij}|\le\max_{1\le k,\ell\le n}|a_{k\ell}|=\|A\|\;.$$

Can you finish it from here? Use the fact that if $S$ is a set of numbers, and $x$ is a number such that $s\le x$ for every $s\in S$, then $\max S\le x$ as well.

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can I then say $$|b_{ij}|\leq||B||$$ and then say $$|c_{ij}|\leq||C||$$ and since $$|c_{ij}|\leq|a_{ij}|+|b_{ij}|$$ can I conclude that $$ ||A+B||\leq||A||+||B||$$ – sarah jamal Oct 1 '12 at 9:24
@sarah: I’d say it a bit differently, but I think that you probably have the idea: each $|c_{ij}|\le|a_{ij}|+|b_{ij}|\le\|A\|+\|B\|$, so the maximum of all the $|c_{ij}|$ must be $\le\|A\|+\|B\|$. But the maximum of all the $|c_{ij}|$ is $\|A+B\|$, so $\|A+B\|\le\|A\|+\|B\|$. – Brian M. Scott Oct 1 '12 at 9:27

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