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 $M_n$.


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*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
$$|v|\ =\ c|v|^{\prime},\;\; \mbox{for every vector }v.$$

*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
 A: 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 $(6)$.
