Let $\|.\|$ denote any norm on $C^m$. The corresponding dual norm $\|.\|'$ is defined by the formula $\|x\|' = sup_{\|y\|=1}|y^*x|$.
(a)Prove that $\|.\|'$ is a norm?
(b) Let $x, y \in C^m $ with $\|x\|=\|y\|=1$ be given. Show that there exists a rank-one matrix $B=yz^*$ such that $Bx=y$ and $\|B\| =1$, where $\|B\|$ is the matrix norm of B induced by the vector norm $\|.\| $. You may use the following lemma, without proof: given $x\in C^m $ there exists a nonzero $z \in C^{m}$ such that $|z^*x|= \|z\|^{'}\|x\|$.
(Form Numerical linear algebra by Trefethen & Bau , exercise 3.6.b)
I already figured out part a, but for part b I don't know where to start.
My Thinking:
$B=yz^*$ (we want $Bx=y$ )so I mutliplied both side by x = > $Bx=yz^{*}x$. I need to show $z^{*}x =e_{1}$
Do I approach the problem correctly?
Thank you for any help.
Best Answer
Sorry to resurrect, but figured this could use an answer.
Don't make any assumptions about $||\cdot||$, beyond those inherent in the definition of a norm. This proof depends on the axiom $||\alpha z||=|\alpha|\,||z||$ for all $\alpha\in\mathbb{C}$ and $z\in\mathbb{C}^m$, for both norms. The key is to find $z$ such that $z^*x=1$. (Not $e_1$, the dimensions are wrong, and it wouldn't contract with $y$!)
Using the given lemma, take $w\neq0$ such that $|w^*x|=||w||'\,||x||$. Then take $z=w/x^*w$, so that $$z^*x=w^*x/w^*x=1.$$ This also gives $|z^*x|=||z||'=1$.
Then immediately, $yz^*x=y$. Furthermore, $$||yz^*||=\sup_{||x'||=1}||yz^*x'||=||y||\sup_{||x'||=1}|z^*x'|=||y||\,||z||'=1.$$
If someone has a nice proof of the lemma, that would be cool to see. (The reason I came looking.)