My Math Forum  

Go Back   My Math Forum > College Math Forum > Linear Algebra

Linear Algebra Linear Algebra Math Forum


Thanks Tree1Thanks
  • 1 Post By v8archie
Reply
 
LinkBack Thread Tools Display Modes
May 16th, 2015, 10:21 AM   #1
Newbie
 
Joined: May 2015
From: Utah

Posts: 3
Thanks: 0

Need help with Closure under scalar multiplication

I am having difficulty proving this axiom.

My last response from my instructor was this: Support for law 6 is not sufficient.

The requirements are to solve and explain each step of the axiom. This is what I have so far, I hope someone can help me fix my errors.
  • Closure under scalar multiplication
  • If X is any vector in V and r is any real scalar, then rX ∈ V.
  • r(a, b) ∈ V substituting, coordinate form of vectors.
  • (r(a), r(b)) Scalar multiplication

  • r(a) by the closure property (r) and (a) are real number. Therefore (r) and (a) are closed under real number multiplication.
  • r(b) by the closure property (r) and (b) are real number. Therefore (r) and (a) are closed under real number multiplication.
  • ((r)(a), (r)(b) we know is in R^2 because (r) is in R^2, and X is in R^2
  • Coordinate X are real numbers, and is in R^2
  • Therefore r(X) is in R^2
bgi11in is offline  
 
May 16th, 2015, 12:54 PM   #2
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

Quote:
Originally Posted by bgi11in View Post
I am having difficulty proving this axiom.

My last response from my instructor was this: Support for law 6 is not sufficient.

The requirements are to solve and explain each step of the axiom. This is what I have so far, I hope someone can help me fix my errors.
  • Closure under scalar multiplication
  • If X is any vector in V and r is any real scalar, then rX ∈ V.
  • r(a, b) ∈ V substituting, coordinate form of vectors.
  • "Coordinate for of vectors"? Are told that V is two dimensional?

    Quote:
  • (r(a), r(b)) Scalar multiplication

  • r(a) by the closure property (r) and (a) are real number. Therefore (r) and (a) are closed under real number multiplication.
  • ?? I don't know what you are saying here. (r) and (a) are specific numbers. It makes no sense to say that they are "closed under real number multiplication". That would be like saying that "4" and "18" are "closed under real number multiplication".

    Quote:
  • r(b) by the closure property (r) and (b) are real number. Therefore (r) and (a) are closed under real number multiplication.
  • Same thing.

    [quote]
  • ((r)(a), (r)(b) we know is in R^2 because (r) is in R^2, and X is in R^2[quote]
    No, in this problem, r is in R, not R^2. Once again, were you specifically told that X was in R^2? I see that nowhere in your statement of the problem which was just to "show closure under scalar multiplication".

    Quote:
  • Coordinate X are real numbers, and is in R^2
  • Therefore r(X) is in R^2
What was the exact statement of the problem? Was it specifically to show that R^2, with the usual definition of "scalar multiplication", is "closed under scalar multiplication"?

Last edited by Country Boy; May 16th, 2015 at 12:57 PM.
Country Boy is offline  
May 16th, 2015, 03:15 PM   #3
Newbie
 
Joined: May 2015
From: Utah

Posts: 3
Thanks: 0

That you for the replay. The assignment is to prove "Closure under scalar multiplication
If X is any vector in V and r is any real scalar, then rX ∈ V."
I have to break each step down and use the following to explain what I did.

Substitution of coordinate form for vector form
Substitution of vector form for coordinate form
Vector addition of coordinate form
Scalar multiplication of coordinate form
Real number associativity of addition or multiplication
Real number commutativity of addition or multiplication
Closure of real number addition or multiplication
Distributive property of real number multiplication over real number addition
Identity property of real number addition or multiplication
Inverse property of real number addition or multiplication
bgi11in is offline  
May 16th, 2015, 04:02 PM   #4
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,654
Thanks: 2632

Math Focus: Mainly analysis and algebra
So $\vec v \in V$ is an $n$-tuple $\langle a_1, a_2, \ldots, a_n\rangle$ of real numbers. (Substitution of vector form for coordinate form. Thus for $r \in \mathbb R$ we have $r\vec v = \langle ra_1, ra_2, \ldots, ra_n\rangle$ (scalar multiple of coordinate form). Since $r$ and all the $a_i$ are real, the $ra_i$ are also real (closure of the reals under multiplication). Thus $\vec w = \langle ra_1, ra_2, \ldots, ra_n\rangle$ is an $n$-tuple of reals and $\vec w = r\vec v$ is thus in $V$ (substitution of coordinate form for vector form).
Thanks from bgi11in
v8archie is offline  
May 16th, 2015, 04:12 PM   #5
Newbie
 
Joined: May 2015
From: Utah

Posts: 3
Thanks: 0

Thanks V8archie.
Yeah the instructor said my steps are correct however my explanation is incorrect.
bgi11in is offline  
Reply

  My Math Forum > College Math Forum > Linear Algebra

Tags
closure, multiplication, scalar



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Closure of the Closure of a Set jjpbq6 Real Analysis 3 March 6th, 2013 05:26 PM
show that the closure of A is subset of closure of B 450081592 Real Analysis 4 November 2nd, 2011 07:25 AM
scalar multiplication axioms waytogo Real Analysis 1 October 16th, 2011 12:30 AM
Scalar Multiplication in a complex vector Space Teddy Linear Algebra 3 December 11th, 2009 03:08 PM
set closure.. ElMarsh Algebra 3 October 3rd, 2009 07:01 AM





Copyright © 2019 My Math Forum. All rights reserved.