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.

I need help with this problem:

Let $\|\cdot\|$ and $\|\cdot\|^{\prime}$ two matrix norms, and consider the relation $$\|\cdot\| \leq \|\cdot\|^{\prime}\ \Leftrightarrow\ \|A\| \leq \|A\|^{\prime},$$ which provides a partial ordering of the set $\mathcal{N}$ of matrix norms defined over the ring $M_n$.

  1. If $\|\cdot\|$ and $\|\cdot\|^{\prime}$ are matrix norms subordinate to the vector norms $|\cdot|$ and $|\cdot|^{\prime}$, respectively, and if $\|A\| \leq \|A\|^{\prime}$ for all matrices $A\in M_n$ of rank 1, show that there exists a constant $c$ such that $$|v|\ =\ c|v|^{\prime},\;\; \mbox{for every vector }v.$$
  2. Show that if a matrix norm is subordinate to two vector norms $|\cdot|$ and $|\cdot|^{\prime}$, then $|v|\ =\ c|v|^{\prime},\;\; \mbox{for every vector }v.$

Somebody knows how solve it? Thanks in advance

share|improve this question
1  
Sorry, I may have made a mistake. I have deleted my answer and see if it can be fixed. –  user1551 Apr 15 '13 at 16:48
    
Thank you @user1551 –  FASCH Apr 16 '13 at 12:55
1  
I want to know that answer too. Thanks –  user70195 Apr 17 '13 at 11:22
    
Question 2 follows from question 1, since $||A||\leq||A||$. –  Brian Rushton Apr 19 '13 at 13:47

1 Answer 1

up vote 2 down vote accepted
+50

Since assertion 2 is an immediate corollary of assertion 1, it suffices to prove assertion 1.

Denote the linear space $\mathbb{R}^n$ by $V$, and denote its dual by $V^*$, i.e. $$V^*:=\{f:V\to\mathbb{R}\mid f\text{ is linear }\}.$$ Given any norm $|\cdot|$ on $V$, it induces an norm on $V$, still denoted by $|\cdot|$, in the following way: $$|f|:=\sup_{v\in V\setminus\{0\}}\frac{|f(v)|}{|v|}=\sup_{v \in V,|v|=1}|f(v)|,\quad\forall f\in V^*.\tag{1}$$ Now given a nonzero vector $v\in V$ and a nonzero linear function $f\in V^*$, we can define $$A: V\to V,\quad w\mapsto f(w)\cdot v.\tag{2}$$ Then for the norm $\|A\|$ induced by $|\cdot|$, we have: $$\|A\|=\sup_{w \in V,|w|=1}|A(w)|=|v|\cdot\sup_{w \in V,|w|=1}|f(w)|=|v|\cdot |f|.\tag{3}$$ Similarly, for the norm $|\cdot|'$, we can define $|f|'$ as in $(1)$ and hence for $\|A\|'$ induced by $|\cdot|'$, we have: $$\|A\|'=|v|'\cdot |f|'.\tag{4}$$ Note that by $(2)$, the rank of $A$ is $1$, so from the assumption in assertion 1 we know that $\|A\|\le \|A\|'$. From $(3)$ and (4) it follows that $$|v|\cdot |f|\le |v|'\cdot |f|',\quad \forall v\in V\setminus\{0\}, \forall f\in V^*\setminus\{0\},$$ i.e. $$c:=\sup_{v\in V\setminus\{0\}}\frac{|v|}{|v|'}\le\inf_{f\in V^*\setminus\{0\}}\frac{|f|'}{|f|}:=c'.\tag{5}$$ However, since $V$ is finite dimensional, for every $v\in V\setminus\{0\}$, there exists $f_v\in V^*\setminus\{0\}$, such that $$|v|'\cdot|f_v|'=|f_v(v)|\le |v|\cdot |f_v|,$$

i.e. $$c\ge \frac{|v|}{|v|'}\ge\frac{|f_v|'}{|f_v|}\ge c'.\tag{6}$$ The conclusion follows from $(5)$ and $(6)$.

share|improve this answer
1  
Form this, how you can get that: 3. if $\|\cdot\|^{\prime}$ be any matrix norm, then there exists (at least) one subordinate matrix norm $\|\cdot\|$ satisfying $\|\cdot\|\leq \|\cdot\|^{\prime}$? 4. tha matrix norm $\|\cdot\|$ is subordinate if and only if it is minimal element of the set $\mathcal{N}$? 5. there exists matrix norms $\|\cdot\|$ satisfying $\|I\| = 1$, which are yet not subrodinate? Thanx for the time. –  user70195 Apr 20 '13 at 16:05
    
@ShanaKugimiya: Your questions are beyond this post, and I don't think it is suitable to discuss them in comments, because they cannot be explained clearly in a few words. So my suggestion is: could you please post another separate question? By the way, I may not be able to give all the answers promptly. –  23rd Apr 20 '13 at 16:16
    
I see. I just tried to complete the problem 1.4-5 from the book: Numerical Linear Algebra and Optimisaton by Ciarlet. But thank you. –  user70195 Apr 20 '13 at 19:19
1  
@Landscape thanks for the answer and ShanaKugimiya for complete the problem. I have a question, why it is clear that $A$ have rank 1? Thanks again –  FASCH Apr 20 '13 at 19:57
    
@FASCH: It follows from the definition of $A$ in $(2)$. Note that the image of $A$ is spanned by a single vector $v$. –  23rd Apr 20 '13 at 20:17

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.