yields8.E. The definition(x, x,) · (y., ya) = x,y.is not an inner product on R'. Why?S.F. If x- (x, X, )€R', define ||x||, by...Prove that x x, is a norm on R'.

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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yields
8.E. The definition
(x, x,) · (y., ya) = x,y.
is not an inner product on R'. Why?
S.F. If x- (x, X, )€R', define ||x||, by
...
Prove that x x, is a norm on R'.

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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