I am having trouble with the following problem:

Show that the vector norm $||x||_1$ gives the subordinate matrix norm: \begin{equation} ||A||_1 = \max_{1\leq j\leq n}\sum_{i=1}^n|a_{ij}| \end{equation}

I really do not have any starting point for this question. I though maybe we could use $||x||_1$ norm for the rows or columns of $A$ but I did not get anywhere with that.

Note: \begin{equation} ||A|| = \sup{||Au||: u\in \mathbb{R}^n, ||u||=1} \end{equation}

All help is greatly appreciated!

EDIT: This is what I have so far: \begin{align} ||A||_1 =& \sup_{||u||_1=1}||Au|| \\ =& \sup_{||u||_1=1}||\sum\limits_{i=1}^n|(Au)_i|\,|| \end{align}

Here is where I am having trouble. The maximum value depends on the entries in $A$. It just seems to make sense that we would pick the largest values in each row.

1.How do I finish the proof from here?

  • $\begingroup$ Do you know what you're supposed to do? $\endgroup$
    – Git Gud
    Jul 26 '13 at 13:18
  • 4
    $\begingroup$ Looks like a typical numerical analysis homework problem to me. $\endgroup$ Jul 26 '13 at 13:22
  • 2
    $\begingroup$ There are two inequalities to prove. For $\leq$, use the triangular inequality on $\sum_{i} |(Ax)_i|=\sum_i |\sum_j a_{ij}x_j|$. For $\geq$, pick the good vector $x$ with only one nonzero coordinate. $\endgroup$
    – Julien
    Jul 26 '13 at 14:23
  • $\begingroup$ This certainly must have been asked before. Here's one instance: math.stackexchange.com/questions/441912/… $\endgroup$
    – Tunococ
    Jul 29 '13 at 7:37

Let $A$ be a $m\times n$ matrix. Let us denote $$\gamma:=\max_{1\le j\le n}\sum_{i=1}^m |a_{ij}|$$ Then \begin{align} ||A||_1 =& \sup_{||u||_1=1}||Au||_{1} \\ =&\sup_{||u||_1=1}||v||_{1} \end{align} where $$v=\begin{bmatrix}A_{1*}u & A_{2*}u & \cdots & A_{m*}u \end{bmatrix}$$ where $A_{i*}$ is the $i$th row of $A$. Hence
\begin{align} \|v\|_{1}=&\sum_{k=1}^m |A_{k*}u|\\ \ =&\sum_{k=1}^m |\sum_{j=1}^n a_{kj}u_j|\\ \ \le& \sum_{k=1}^m \sum_{j=1}^n |a_{kj}||u_j|\ (\mbox{By triangle inequality})\\ \ \le& \left(\max_{1\le j\le n}\sum_{k=1}^m|a_{kj}|\right)\sum_{j=1}^{n}|u_j|=\gamma \|u\|_{1} \end{align}

So, $$\|A\|_{1}\le \gamma\tag{1}$$

Now, let $$k=\arg{\max_{1\le i\le m}}\sum_{j=1}^n |a_{ij}|$$ and let $e_k$ be the $n\times 1$ unit vector with $$(e_k)_{i}=\left\{\begin{array}{rl} 1 & \mbox{if}\ i=k\\ 0 & \mbox{else} \end{array} \right. $$ for $1\le i\le n$. Then $\|e_k\|_{1}=1$ and \begin{align} \|Ae_k\|_{1}=&\sum_{i=1}^m|a_{ik}|=\gamma\le \sup_{\|u\|_{1}=1}\|Au\|_{1}=\|A\|_{1}\tag{2}\\ \end{align} Hence, from $(1)$ and $(2)$, $$\|A\|_{1}=\gamma\hspace{0.6cm}\Box$$


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