# Proving that $\|A\|_{\infty}$ the largest row sum of absolute value of matrix $A$

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 $\|x\|_{\infty}$.

• You are talking about definition, why you want to prove it? Jul 12 '13 at 9:37
• 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. Jul 12 '13 at 9:37
• @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. Jul 12 '13 at 9:40
• This doesn't answer my question. Jul 12 '13 at 9:40
• @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. Jul 12 '13 at 9:49

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.

• By showing there exists such an x, how do we know that it is the tightest possible upper bound? – Aug 24 '18 at 19:10
• @SridharThiagarajan because for one of such $x$ the chain of inequalities will become a chain of equalities. Aug 24 '18 at 19:13

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.

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