### Transcription:8.E. The definition \[(x_1, x_2) \cdot (y_1, y_2) = x_1 y_1 \]is **not** an inner product on $ \mathbb{R}^2 $. Why?8.F. If $ x = (x_1, x_2, \ldots, x_n) \in \mathbb{R}^n $, define $ \|x\|_1 $ by \[\|x\|_1 = |x_1| + |x_2| + \cdots + |x_n|.\]Prove that $ x \mapsto \|x\|_1 $ is a norm on $ \mathbb{R}^n $.

Answered: 8.F. Ifx= (1, I, --- ,)€R", define x,…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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### Transcription:

8.E. The definition

\[
(x_1, x_2) \cdot (y_1, y_2) = x_1 y_1
\]

is **not** an inner product on $ \mathbb{R}^2 $. Why?

8.F. If $ x = (x_1, x_2, \ldots, x_n) \in \mathbb{R}^n $, define $ \|x\|_1 $ by

\[
\|x\|_1 = |x_1| + |x_2| + \cdots + |x_n|.
\]

Prove that $ x \mapsto \|x\|_1 $ is a norm on $ \mathbb{R}^n $.

Expert Solution

Step 1

Given that $x = (x_{1}, x_{2}, . ., x_{n})$ and ${||x||}_{1} = |x_{1}| + |x_{2}| + . . . + |x_{n}|$ .

The objective is to find the whether ${||x||}_{1} = |x_{1}| + |x_{2}| + . . . + |x_{n}|$ is norm.

To prove that ${||x||}_{1} = |x_{1}| + |x_{2}| + . . . + |x_{n}|$ is norm then it is enough to show that,

${||x + y||}_{1} \leq {||x||}_{1} + {||y||}_{1}$

${||a x||}_{1} = |a| {||x||}_{1}$

$||x|| = 0$ then $x = 0$

First

$\begin{array}{rcl} {||x + y||}_{1} & = & |x_{1} + y_{1}| + |x_{2} + y_{2}| + . . . + |x_{n} + y_{n}| \\ \leq & |x_{1}| + |y_{1}| + |x_{2}| + |y_{2}| + . . + |x_{n}| + |y_{n}| \\ = & (|x_{1}| + |x_{2}| + . . . + |x_{n}|) + (|y_{1}| + |y_{2}| + . . . + |y_{n}|) \\ = & {||x||}_{1} + {||y||}_{1} \end{array}$

Hence ${||x + y||}_{1} \leq {||x||}_{1} + {||y||}_{1}$ is satisfied.

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