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I am studying matrix norms. I have read that $\|A\|_{\infty}$ is the largest row sum of absolute value and $\|A\|_{1}$ is the highest column sum of absolute values of the matrix $A$. However, I am not able to prove this. Are there any proof of these statements? Please help and thanks for your time.

Edit: Where $\|A\|_{\infty}$ is the matrix norm induced by the vector norm $\|x\|_{\infty}$.

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  • $\begingroup$ You are talking about definition, why you want to prove it? $\endgroup$
    – CosimoDS
    Jul 12 '13 at 9:37
  • $\begingroup$ This sounds like a good exercise. You might want to first establish that the suspected value is an upper bound, then show that the value can be achieved. $\endgroup$
    – Tunococ
    Jul 12 '13 at 9:37
  • $\begingroup$ @Tunococ I have studied operator norm. I want to apply same definition to evaluate matrix norms but I am not able to do so properly. Thanks for your reply. $\endgroup$
    – mathscrazy
    Jul 12 '13 at 9:40
  • $\begingroup$ This doesn't answer my question. $\endgroup$
    – mathscrazy
    Jul 12 '13 at 9:40
  • $\begingroup$ @mathscrazy CosimoDS does answer your question, as $\|A\|_\infty=\max_i\sum_j|a_{ij}|$ is indeed the definition of the maximum row sum norm in some books. If you define $\|A\|_\infty$ as the matrix norm induced by the vector norm $\|x\|_\infty$, it is your responsibility to make it clear in your question. $\endgroup$
    – user1551
    Jul 12 '13 at 9:49
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I assume that the author tries to derive the matrix norms $\|A\|_1$ and $\|A\|_\infty$ induced by vector norms $\|x\|_1$ and $\|x\|_\infty$.

Let's take, for example, $\|\cdot\|_\infty$. We write

$$\|Ax\|_\infty=\sup_i \left|\sum_j A_{ij}x_j\right|\le \sup_i \sum_j |A_{ij}||x_j|\le \sup_j |x_j| \sup_i \sum_j |A_{ij}|.$$ Hence, a good candidate for $\|A\|_\infty$ is $\sup_i \sum_j |A_{ij}|$. We need to prove that this boundary is indeed achieved; it is true, since we can take $i$ where that supremum is achieved and impose $x_j=\mathrm{sign}(A_{ij})$. With such $x$ all our inequalities degenerate to equalities and we conclude that $\|A\|_\infty = \sup_i \sum_j |A_{ij}|$ is a matrix norm induced by $\|x\|_\infty = \sup_j |x_j|$.

The case $\|\cdot\|_1$ is done likewise.

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  • $\begingroup$ By showing there exists such an x, how do we know that it is the tightest possible upper bound? – $\endgroup$ Aug 24 '18 at 19:10
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    $\begingroup$ @SridharThiagarajan because for one of such $x$ the chain of inequalities will become a chain of equalities. $\endgroup$ Aug 24 '18 at 19:13
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I will give you some guidelines then. Suppose an $m$-by-$n$ matrix $A = (a_{ij})$ is given. For any $n$-dimensional vector $x$,

\begin{align*} \|Ax\|_1 & = \sum_{i=1}^m \left|\sum_{j=1}^n a_{ij}x_j\right| \\ & \le \sum_{i=1}^m \sum_{j=1}^n \left|a_{ij}x_j\right| \\ & = \sum_{j=1}^n \left|x_j\right| \sum_{i=1}^m \left|a_{ij}\right| \\ & = \sum_{j=1}^n \left|x_j\right| A_j \end{align*} where I define $A_j = \sum_{i=1}^m \left|a_{ij}\right|$. If $J = \operatorname{argmax}_j A_j$, i.e., $A_J$ is a maximum among all $A_j$'s, then $$ \|Ax\|_1 \le A_J \sum_{j=1}^n |x_j| = A_J \|x\|_1. $$ This shows that $\|A\|_1 \le A_J$. Next, you show that there exists $x$ such that $\|Ax\|_1 = A_J\|x\|_1$. This is quite simple as you can choose $x_i = 0$ for all $i \ne J$, and $x_J = 1$ or $-1$.

For the case of $\|\cdot\|_\infty$, the situation is quite similar. (Choosing $x$ in the last step might be a bit tricky as you need to pick the right signs.) I will leave that to you.

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    $\begingroup$ Funny enough, you decided to take $\|\cdot\|_1$ and I took $\|\cdot\|_\infty$. Combined together these two posts give a complete answer) $\endgroup$ Jul 12 '13 at 10:08

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