Matrix Norm set #2

Question

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 $\mathcal{N}$.

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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.

Answer 1

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 $$\|A\|=\sup_{v\ne 0}\frac{|Av|}{|v|}\le \sup_{v\ne 0}\frac{\|A\|'\cdot|v|}{|v|}=\|A\|'.$$

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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.

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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.