linear algebra Let (:) be an inner product in the vector space V. Given an isomorphismT : U - V. Put[u, v] = (Tu, Tv), for any u, V E U. Check thatl:lis 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 [uru,w] = [uw] +[viw]ii. Cauiu] =x [uiv]N. [uiu] 7OYu[uiu] =0 u=0

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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Question

linear algebra

Let (:) be an inner product in the vector space V. Given an isomorphismT : U - V. Put
[u, v] = (Tu, Tv), for any u, V E U. Check thatl:lis 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 [uru,w] = [uw] +[viw]
ii. Cauiu] =x [uiv]
N. [uiu] 7O
Yu
[uiu] =0 u=0

Branch of mathematics concerned with mathematical structures that are closed under operations like addition and scalar multiplication. It is the study of linear combinations, vector spaces, lines and planes, and some mappings that are used to perform linear transformations. Linear algebra also includes vectors, matrices, and linear functions. It has many applications from mathematical physics to modern algebra and coding theory.

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.