Matrix Norm set #2 As a complement of the question
Matrix Norm set
and in order to complete the Problem 1.4-5 from the book: Numerical Linear Algebra and Optimisaton by Ciarlet. I have this additional conditions:
(3) if $\|\cdot\|^{\prime}$ be any matrix norm, then there exists (at least) one subordinate matrix norm $\|\cdot\|$ satisfying $\|\cdot\| \leq \|\cdot\|^{\prime}$.
(4) the matrix norm $\|\cdot\|$ is subordinate if and only if it is minimal element of the set $\mathcal{N}$.

Then, show that there exists matrix norms $\|\cdot\|$ satisfying $\|I\| = 1$, which are yet not subrodinate. Where my definition of matrix norm is:
\begin{eqnarray*}
\|A\| & = & 0 \Leftrightarrow A=0, \mbox{ and } \|A\|\geq 0\\
\|\alpha A\| & = & |\alpha| \|A\|\\
\|A + B\| & \leq & \|A\| + \|B\|\\
\|AB\| & \leq & \|A\|\cdot\|B\|
\end{eqnarray*}
for all $A,B\in M_n$ and $\alpha\in\mathbb{R}$. Besides, my definition of subordinate matrix norm is: there exists a vector norm $\|\cdot\|$, such that
$$\|A\|\ =\ \sup_{v\neq 0}\frac{\|Av\|}{\|v\|}.$$
Please someone knows how prove it? Thanks in advance.
 A: Part (3)
If a matrix $\|\cdot\|'$ is given on $M_n(\mathbb{R})$, we can define a norm $|\cdot|$on $\mathbb{R}^n$ by embedding $\mathbb{R}^n$ into $M_n(\mathbb{R})$. For example, 
consider the following map 
$$L:\mathbb{R}^n \to M_n(\mathbb{R}),\quad v\mapsto (v,0,\dots,0).\tag{1}$$
That is to say, $L(v)$ is the $n\times n$ matrix whose first column is $v$ and other columns are zero. Then we can define 
$$|\cdot|:\mathbb{R}^n\mapsto [0,\infty),\quad v\mapsto \|L(v)\|'. \tag{2}$$
It is easy to verify that $|\cdot|$ defined in $(2)$ is a norm on $\mathbb{R}^n$.
Note that for any $A\in M_n(\mathbb{R})$ and $v\in \mathbb{R}^n$, 
$$L(Av)=(Av,0,\dots,0)=A\cdot L(v),$$
and hence 
$$|Av|=\|L(Av)\|'=\|A\cdot L(v)\|'\le\|A\|'\cdot\|L(v)\|'=\|A\|'\cdot|v|.\tag{3} $$
Now let $\|\cdot\|$ be the matrix norm subordinate to  $|\cdot|$. Then for every $A\in M_n(\mathbb{R})$, from $(3)$ we know that
$$\|A\|=\sup_{v\ne 0}\frac{|Av|}{|v|}\le \sup_{v\ne 0}\frac{\|A\|'\cdot|v|}{|v|}=\|A\|'.$$

Part (4)
"$\Rightarrow$" direction. Let $\|\cdot\|$ be a subordinate norm. To show $\|\cdot\|$ is minimal, let $\|\cdot\|'\le \|\cdot\|$, and it suffices to show that $\|\cdot\|'=\|\cdot\|$. According to the conclusion in part (3), there exists a subordinate norm $\|\cdot\|''$, such that  $\|\cdot\|''\le \|\cdot\|'$. Therefore, $\|\cdot\|''\le \|\cdot\|$ are both subordinate norms, then as an immediate corollary of the conclusion in part (1) here, $\|\cdot\|''=\|\cdot\|$, i.e. $\|\cdot\|$ is minimal.
"$\Leftarrow$" direction. Let $\|\cdot\|$ be a minimal norm. Then by the the conclusion in part (3), there exists a subordinate norm $\|\cdot\|'$, such that  $\|\cdot\|'\le \|\cdot\|$. Since $\|\cdot\|$ is minimal, these two norms must coincide, i.e. $\|\cdot\|$ is subordinate.

The Last Part
Let $\|\cdot\|'$ be an arbitrary matrix norm with $\|I\|'=1$, and fix an arbitrary matrix $P\in M_n(\mathbb{R})$ which does not commute with some other matrix in $M_n(\mathbb{R})$. Then define
$$\|\cdot\|:M_n(\mathbb{R})\mapsto [0,\infty),\quad A\mapsto \|A\|'+\|AP-PA\|'. \tag{4}$$
Claim: $\|\cdot\|$ is a matrix norm which is not subordinate to any vector norm, and $\|I\|=1$.
Proof: By definition, $\|I\|=1$, and only the condition $\|AB\|\le\|A\|\cdot\|B\|$ is nontrivial in verifying that $\|\cdot\|$ is matrix norm. Note that
$$ABP-PAB=A(BP-PB)+(AP-PA)B,$$
so by $(4)$,
\begin{eqnarray*}
\|AB\|&= &\|AB\|'+\|ABP-PAB\|'\\
&\le &\|A\|'\cdot\|B\|'+\|A\|'\cdot\|BP-PB\|'+\|AP-PA\|'\cdot\|B\|'\\
&\le &(\|A\|'+\|AP-PA\|')\cdot(\|B\|'+\|(BP-PB)\|')\\
& =  &\|A\|\cdot\|B\|,
\end{eqnarray*}
i.e. $\|\cdot\|$ is a matrix norm. On the one hand, by $(4)$, $\|\cdot\|\ge\|\cdot\|'$. On the other hand, there exists $A\in M_n(\mathbb{R})$, such that $A$ does not commute with $P$, so by $(4)$, $\|A\|>\|A\|'$, i.e. $\|\cdot\|\ne\|\cdot\|'$. According to part (4), subordinate matrix norms are minimal, so $\|\cdot\|$ is not subordinate.
