Let (: ) be an inner product in the vector space V. Given an isomorphismT : U + V. Score[u, v] = (Tu, Tv), for any u, V E U. Check that Jis an in-house product.Note:From the internal product (:) define a new "internal product (with the mentioned conditions)the inner product axioms must be verified in this new function (u, v] = (Tu,Tv)i [uiv]=[viu]i.i [urr,w] =آس، کا[uiw] +[uiw]%3Dw. [uiu] 70Yu[uiu] =0 u=0using the fact that T is an isomorphism

Answered: Let (,) be an inner product in the…

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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Related questions

Question

Let (: ) be an inner product in the vector space V. Given an isomorphismT : U + V. Score
[u, v] = (Tu, Tv), for any u, V E U. Check that Jis an in-house product.
Note:
From the internal product (:) define a new "internal product (with the mentioned conditions)
the inner product axioms must be verified in this new function (u, v] = (Tu,Tv)
i [uiv]=[viu]
i.
i [urr,w] =
آس، کا
[uiw] +[uiw]
%3D
w. [uiu] 70
Yu
[uiu] =0 u=0
using the fact that T is an isomorphism

Expert Solution

Step 1

Given $<,>$ is an inner product on vector space $V$ over real numbers. Therefore for all $u, v, w \in V$ and $a \in ℝ$ :

$\begin{array}{rcl} <u, v> & = & <v, u> \\ <u + v, w> & = & <u, w> + <v, w> \\ <a u, v> & = & a <u, v> \\ <u, u> & \geq & 0 \end{array}$

and $<u, u> = 0$ if $u = 0$ .

Given $T : U \to V$ is an isomorphism therefore $T (a u + v) = a T (u) + T (v)$ for all $u, v \in U$ and $a \in ℝ$ .

Define $[u, v] = <T (u), T (v)>$ for all $u, v \in U$ . Now to show that $[,]$ is an in-house product on $U$ , it is required to show that $[,]$ satisfies inner product axiom.

(i)

Let $u, v \in U$ be arbitrary elements. Now consider $[u, v]$ . Since $<,>$ is an inner product, therefore using $<T u, T v> = <T v, T u>$ :

$\begin{array}{rcl} [u, v] & = & <T u, T v> \\ = & <T v, T u> \\ = & [v, u] \end{array}$

Hence $[u, v] = [v, u]$ .

(ii)

Let $u, v, w \in U$ be arbitrary elements. Using the property of linear transformation:

$\begin{array}{rcl} [u + v, w] & = & <T (u + v), T (w)> \\ = & <T (u) + T (v), T (w)> \\ = & <T (u), T (w)> + <T (v) + T (w)> \\ = & [u, w] + [v, w] \end{array}$