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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

Rodrigo de Azevedo
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FASCH
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1 Answers1

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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 $(6)$.

23rd
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    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. –  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. –  Apr 20 '13 at 19:19
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    @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
  • @ShanaKugimiya: You are welcome. I hope you have solved the questions by yourself. Otherwise, I still suggest that you should start a new post. By the way, I don't have the book you mentioned. – 23rd Apr 20 '13 at 20:21
  • The idea was not to use Duality Theory, but the way, I like your answer so much!! I really enjoyed it, Thanks @Landscape. – FASCH Apr 20 '13 at 21:07
  • @FASCH: You are welcome and thank you for your bounty and appreciation. – 23rd Apr 20 '13 at 21:18
  • @FASCH: I just found a typo in the second last displayed equation(i.e. the one between $(5)$ and $(6)$) and now it is fixed. – 23rd Apr 20 '13 at 21:20
  • @Landscape I did what you said. Thanks (http://math.stackexchange.com/questions/367752/matrix-norm-set-2) –  Apr 20 '13 at 23:12