# proving matrix norm properties

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 :(

-
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

## 1 Answer

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 $0$.

-