# Matrix Norm set

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

-
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
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

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 $(6)$.
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
@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