1
$\begingroup$

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?

$\endgroup$
2
  • 3
    $\begingroup$ What if you let $x = e_i = (0, \ldots, 1, \ldots, 0)$ where the 1 occurs at the $i$-th position? $\endgroup$
    – M.B.
    Commented Mar 29, 2015 at 10:41
  • $\begingroup$ @M.B. fantastic idea!! $\endgroup$ Commented Mar 29, 2015 at 16:45

1 Answer 1

1
$\begingroup$

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 $\|T\| = M$.

$\endgroup$
2
  • $\begingroup$ Indeed!! Thanks very much. $\endgroup$ Commented Mar 30, 2015 at 17:14
  • 1
    $\begingroup$ Never Mind...Happy to Help!!! $\endgroup$ Commented Mar 30, 2015 at 17:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .