Let R" = {(a,, a, . .. , a) l a, E R}. Show that the mapping o: (a,,Az, ... , a) → (-q,, -a, · .the group R" under componentwise addition. This automorphism iscalled inversion. Describe the action of o geometrically.-a) is an automorphism of

Given: ℝn=a1.....,anai∈ℝWe need to show that the mapping ϕa1,.....,an→-a1,.....,-an is an…

Answered: Let R" = {(a,, a, . .. , a) l a, E R}.…

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 R" = {(a,, a, . .. , a) l a, E R}. Show that the mapping o: (a,,
Az, ... , a) → (-q,, -a, · .
the group R" under componentwise addition. This automorphism is
called inversion. Describe the action of o geometrically.
-a) is an automorphism of

Expert Solution

Step 1

Given: $ℝ^{n} = \{(a_{1} . . . . ., a_{n}) a_{i} \in ℝ\}$
We need to show that the mapping $ϕ (a_{1}, . . . . ., a_{n}) \to (- a_{1}, . . . . ., - a_{n})$ is an automorphism of the group $ℝ^{n} .$ Also, the geometrical action of $ϕ$ needs to be described.

Step 2

It is clear that $ϕ \circ ϕ$ is the identity, so $ϕ$ has an inverse and must be a bijection. Now, we have -
$ϕ (0, . . . . ., 0) = (0, . . . . ., 0) and {(a_{1}, . . . ., a_{n})}^{- 1} = (- a_{1}, . . . ., - a_{n})$ , so :
$ϕ ({(a_{1}, . . ., a_{n})}^{- 1}) = ϕ (- a_{1}, . . ., - a_{n}) = (a_{1}, . . ., a_{n}) = {(- a_{1}, . . ., - a_{n})}^{- 1} = ϕ {(a_{1}, . . ., a_{n})}^{- 1}$
Finally,
$ϕ ((a_{1}, . . ., a_{n}) + (b_{1}, . . ., b_{n})) = ϕ (a_{1} + b_{1}, . . ., a_{n} + b_{n}) = (- (a_{1} + b_{1}), . . ., - (a_{n} + b_{n})) = (- a_{1}, . . ., - a_{n}) + (- b_{1}, . . ., - b_{n}) = ϕ (a_{1}, . . ., a_{n}) + ϕ (b_{1}, . . ., b_{n})$
This proves that $ϕ$ is an automorphism of $ℝ^{n}$ under component wise addition.