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 $$A\in C_{n\times n}$$ define the norm of A to be $$||A||=max{|[a]_{ij}|}$$ Prove the following:

  1. $$B>>A\Rightarrow ||B||\geq ||A||$$

  2. $$||ABC||\leq n^{2}||A||\centerdot ||B|| \centerdot ||C||$$

  3. $$ \{A_{m}\} \rightarrow 0 \Leftrightarrow \{||A_{m}||\} \rightarrow 0$$

Proofs just stump me, I don't know how to approach them :(

share|improve this question
1  
What are your definitions of $B>>A$ and $|A_m|$? –  Brian M. Scott Oct 1 '12 at 9:35
    
$$|A_{m}|$$ is a matrix sequence $$>> \text{means majorizes}$$ –  sarah jamal Oct 1 '12 at 9:39
    
I understood that you had a sequence; what I don’t understand is the $|\cdot|$ notation, as distinct from the norm $\|A\|$. –  Brian M. Scott Oct 1 '12 at 9:45
    
@BrianM.Scott Im sorry, I didn't realise my curly brackets didn't come up. I'm still learning the commands, I have edited it now. –  sarah jamal Oct 1 '12 at 10:02
add comment

1 Answer

up vote 1 down vote accepted

For (1) the argument depends on which notion of majorization you’re using. I’m guessing that $B>>A$ means that $b_{ij}\ge|a_{ij}|$ for each pair $i,j$, or possibly that $|b_{ij}|\ge|a_{ij}|$; in either case this one really is completely straightforward. $\|A\|$ is the largest of the $n^2$ $|a_{ij}|$’s, and it’s at most as big as the corresponding $|b_{ij}|$, so ... ?

Here’s a hint for (2). Start with something simpler: prove that $\|AB\|\le n\|A\|\cdot\|B\|$. Not only is this simpler, but the result in (2) follows very easily from it with no extra calculation.

To prove this, note that the $(i,j)$-entry in $AB$ is $$a_{i1}b_{1j}+a_{i2}b_{2j}+\ldots+a_{in}b_{nj}=\sum_{k=1}^na_{ik}b_{kj}\;.$$

Thus, if I let $D=AB$ so that I can conveniently call this entry $d_{ij}$, I have

$$|d_{ij}|=\left|\sum_{k=1}^na_{ik}b_{kj}\right|\le\sum_{k=1}^n|a_{ik}||b_{kj}|\;.$$

You know that each $|a_{ik}|\le\|A\|$ and each $|b_{kj}|\le\|B\|$ by the definition of the norm, so

$$|d_{ij}|\le\sum_{k=1}^n|a_{ik}||b_{kj}|\le\sum_{k=1}^n\|A\|\|B\|\;.$$

And from here it’s very much like the argument that we just did for the other problem.

(3) should follow almost immediately from your definition of $\langle A_m:m\in\Bbb N\rangle\to\mathbf{0}$. I’m guessing that you’ve defined matrix convergence something like this:

$\langle A_m:m\in\Bbb N\rangle\to B$ iff for each $\epsilon>0$ there is an $n_0\in\Bbb N$ such that for all $m\ge n_0$, $|a^m_{ij}-b_{ij}|<\epsilon$ for all pairs $i,j$. (Here $a^m_{ij}$ is the $(i,j)$-entry in $A_m$.)

That just says that for each $\epsilon>0$ there is an $n_0\in\Bbb N$ such that $\|A_m-B\|<\epsilon$ for each $n\ge n_0$, which means that the sequence of real numbers $\langle \|A_m-B\|:m\in\Bbb N\rangle$ converges to $0$.

share|improve this answer
add comment

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.