
Linear Algebra Linear Algebra Math Forum 
 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.

May 16th, 2015, 12:54 PM  #2  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
Last edited by Country Boy; May 16th, 2015 at 12:57 PM.  
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 
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).

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. 

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 