I have a question from Numerical linear algebra book by Trefethen & Bau :

Let $\|\cdot\|$ denote any norm on $C^m$. The corresponding dual norm $\|\cdot\|'$ is defined by the formula $\|x\|' = sup_{\|y\|=1}|y^*x|$.

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 $|z^*x|= \|z\|^{'}\|x\|$.

I don't know how to relate the given lemma to the question ( I assumed that $z(s)$ in given lemma and main question are not the same). (2) when it used $\|\cdot\|$ should I assume $\|\cdot\|_2$?


Let $z$ be such that $|z^* x| = \|z\|' \|x\| = \|z\|' = \sup_{v:\|v\|=1} |z^* v|$ as given by the lemma. By re-scaling $z$, we may assume $|z^* x|=1$

Then with $B=yz^*$, we have $\|B\| = \sup_{v : \|v\|=1} \|yz^* v\| = \|y\| \sup_{v : \|v\|=1} |z^* v| = \|y\| |z^* x| = 1$.

Finally, $Bx = y(z^*x)$. It is not clear to me how we can show that $z^* x=1$; perhaps someone else can finish this argument...

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    $\begingroup$ Just change $z$ by $z/z^*x$ $\endgroup$ – Luiz Cordeiro May 27 '16 at 1:15
  • $\begingroup$ Thank you for your answer, I have one question : I assumed according to the formula for computing matrix norm in term of the image of unit vector (which is $\| A \| = Sup_{x \in C^{n}, \| x \|_{(n)} = 1} \| Ax \|_{m}$) you wrote : $\| B \| = Sup_{v:\| v \| = 1} \| yz^{*}v \|$ but I don know how from there I should get to $Sup_{v:\| v \| = 1} \| yz^{*}v \| = \| y \|$$ Sup_{v:\| v \| = 1} \left| z^{*}v \right| $ ? $\endgroup$ – Crimson May 27 '16 at 14:51
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    $\begingroup$ @Zereshki Basically for a scalar $c$, we have $\|cy\| = |c| \|y\|$. Here the scalar is $z^* v$. $\endgroup$ – angryavian May 27 '16 at 17:21

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