# Prob. 14, Sec. 2.7, in Kreyszig's Functional Analysis Book: How to find the norm of this bounded linear operator?

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

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

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

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

What is $$\lVert T \rVert$$?

Here we are using the following definition for $$\lVert T \rVert$$: $$\lVert T \rVert \colon= \sup \left\{ \frac{\lVert T(x)\rvert_{\mathbb{R}^m}}{\lVert x \rVert_{\mathbb{R}^n}} \colon x \in \mathbb{R}^n, x \neq \mathbf{0}_{\mathbb{R}^n} \right\}.$$

My effort:

For any $$x \colon= (\xi_1, \ldots, \xi_n ) \in \mathbb{R}^n$$, we have \begin{align} \lVert T(x) \rVert_{\mathbb{R}^m} &= \sum_{i=1}^m \left\lvert \sum_{j=1}^n \alpha_{ij} \xi_j \right\rvert \\ & \leq \sum_{i=1}^m \left( \sum_{j=1}^n \left\lvert \alpha_{ij} \xi_j \right\rvert \right) \\ &= \sum_{i=1}^m \left( \sum_{j=1}^n \left( \left\lvert \alpha_{ij} \right\rvert \left\lvert \xi_j \right\rvert \right) \right) \\ &= \sum_{j=1}^n \left( \sum_{i=1}^m \left( \left\lvert \alpha_{ij} \right\rvert \left\lvert \xi_j \right\rvert \right) \right) \\ &= \sum_{j=1}^n \left( \left\lvert \xi_j \right\rvert \left( \sum_{i=1}^m \left\lvert \alpha_{ij} \right\rvert \right) \right) \\ &\leq \sum_{j=1}^n \left( \left\lvert \xi_j \right\rvert \max_{k= 1, \ldots, n} \left( \sum_{i=1}^m \left\lvert \alpha_{ik} \right\rvert \right) \right) \\ &= \sum_{j=1}^n \left( \left\lvert \xi_j \right\rvert \right) \max_{k= 1, \ldots, n} \left( \sum_{i=1}^m \left\lvert \alpha_{ik} \right\rvert \right) \\ &= \lVert x \rVert_{\mathbb{R}^n} \max_{k= 1, \ldots, n} \left( \sum_{i=1}^m \left\lvert \alpha_{ik} \right\rvert \right). \end{align} If $$x$$ is not the zero vector in $$\mathbb{R}^n$$, then upon dividing by the norm of $$x$$ and then taking the supremum of the quantity on the left hand side, we obtain $$\lVert T \rVert \leq \max_{k= 1, \ldots, n} \left( \sum_{i=1}^m \left\lvert \alpha_{ik} \right\rvert \right).$$

Is it true that $$\lVert T \rVert = \max_{k= 1, \ldots, n} \left( \sum_{i=1}^m \left\lvert \alpha_{ik} \right\rvert \right)?$$ If so, then how to show this?

• What if you let $x = e_i = (0, \ldots, 1, \ldots, 0)$ where the 1 occurs at the $i$-th position?
– M.B.
Commented Mar 29, 2015 at 10:41
• @M.B. fantastic idea!! Commented Mar 29, 2015 at 16:45

Suppose the maximum $\max_{k=1,\ldots,n}\sum_{i=1}^{m}|\alpha_{ik}| = M$(say) is attained for $k = k_0$, i.e, $M = \sum_{i=1}^{m}|\alpha_{ik_0}|$. Then take the vector $x= (0,\ldots,1,\ldots,0)$ with $1$ at the $k_0$ position. Notice that $\|Tx\|_{\mathbb{R^m}} = M$. Since there is a vector such that this supremum is attained, it follows that $\|T\| = M$.