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Let $A \colon = [\alpha_{ij}]_{m\times n}$ be a given $m \times n$ matrix of real numbers.

Let $\mathbb{R}^n$ be the normed space of all ordered $n$-tuples of real numbers with the norm defined as follows: $$\Vert x \Vert_{\mathbb{R}^n} \colon= \max_{j=1, \ldots, n} \left( \vert \xi_j \vert \right) \ \ \ \forall x \colon= (\xi_1, \ldots, \xi_n) \in \mathbb{R}^n.$$

Let $\mathbb{R}^m$ be the normed space of all ordered $m$-tuples of real numbers with the norm defined as follows: $$\Vert y \Vert_{\mathbb{R}^m} \colon= \max_{i=1, \ldots, m} \left( \vert \eta_i \vert \right) \ \ \ \forall y \colon= (\eta_1, \ldots, \eta_m) \in \mathbb{R}^m.$$

Let the operator $T \colon \mathbb{R}^n \to \mathbb{R}^m$ be defined as $$T(x) \colon= Ax \ \ \ \forall x \in \mathbb{R}^n;$$ where $x$ and $y$ are to be written as column vectors and $Ax$ denotes the usual matrix product. Of course, $T$ is linear.

What is $\Vert T \Vert$?

Here we are using the following definition for $\Vert T \Vert$: $$\Vert T \Vert \colon= \sup \left\{ \ \frac{\Vert T(x)\vert_{\mathbb{R}^m}}{\Vert x \Vert_{\mathbb{R}^n}} \ \colon \ x \in \mathbb{R}^n, \ x \neq \theta_{\mathbb{R}^n} \ \right\}. $$ Here $\theta_{\mathbb{R}^n} $ denotes the zero vector in $\mathbb{R}^n$.

My effort:

For any $x \colon= (\xi_1, \ldots, \xi_n ) \in \mathbb{R}^n$, we have \begin{eqnarray*} \Vert T(x) \Vert_{\mathbb{R}^m} &=& \max_{i=1, \ldots, m} \left( \left\vert \sum_{j=1}^n \alpha_{ij} \xi_j \right\vert \right) \\ & \leq & \max_{i=1, \ldots, m} \left( \sum_{j=1}^n \vert \alpha_{ij} \xi_j \vert \right) \\ &=& \max_{i=1, \ldots, m} \left( \sum_{j=1}^n \left( \vert \alpha_{ij} \vert \cdot \vert \xi_j \vert \right) \right) \\ &\leq & \Vert x \Vert_{\mathbb{R}^n} \cdot \max_{i=1, \ldots, m} \left( \sum_{j=1}^n \vert \alpha_{ij} \vert \right). \end{eqnarray*} If $x$ is not the zero vector, then upon deviding by the norm of $x$ and then taking the supremum of the quantity on the left hand side, we obtain $$\Vert T \Vert \leq \max_{i=1, \ldots, m} \left( \sum_{j=1}^n \vert \alpha_{ij} \vert \right). $$

How to show that $$\Vert T \Vert = \max_{i=1, \ldots, m} \left( \sum_{j=1}^n \vert \alpha_{ij} \vert \right)? $$

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It suffices to show that at least one $x$ with $\|x\|_{\mathbb{R}^n}=1$ achieves the specific bound. Let $$i^*=arg\max_{i=1,2,\ldots,m}\sum_{j=1}^n{|a_{ij}|}$$ i.e. $i^*$ is the index of the row of $A$ with the largest sum of absolute values of its elements. Then, if we select $\xi_j=sign(a_{i^*j}$) we have that $\|x\|_{\mathbb{R}^n}=1$ and $$\bigg|\sum_{j=1}^{n}{a_{i^*j}\xi_j}\bigg|=\sum_{j=1}^n{|a_{i^*j}|}=\max_{i=1,2,\ldots,m}\sum_{j=1}^n{|a_{ij}|}$$ Thus the vector $x^*:=[sign(a_{i^*1}),\ldots,sign(a_{i^*n})]^T$ achieves your bound and therefore $\max_{i=1,2,\ldots,m}\sum_{j=1}^n{|a_{ij}|}$ is indeed the induced matrix norm.

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  • $\begingroup$ your answer is just fantastic! But can we show this in some other way also, I wonder? $\endgroup$ – Saaqib Mahmood Mar 29 '15 at 12:41
  • $\begingroup$ You can prove it also as a limiting case of the Cauchy-Schwatz inequality. $\endgroup$ – RTJ Mar 29 '15 at 13:22

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