Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.