January 8th, 2017, 06:34 AM  #1 
Newbie Joined: Jan 2017 From: Pakistan Posts: 4 Thanks: 0  R2 is a vector space
How to verify that R2 is a vector space? "Absolute honesty isn't the most diplomatic, nor the safest form of communication with emotional beings" 
January 8th, 2017, 07:08 AM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,300 Thanks: 664  
January 13th, 2017, 04:22 AM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,576 Thanks: 667  Quote:
To prove this forms a vector space, you prove that all of the parts of the definition of a "vector space" are true: 1) Commutativity: U+ V= V+ U. 2) Associativity of addition: U+ (V+ W)= (U+ V)+ W. 3) Existence of an additive identity: There exist a vector, O, such that, for all vectors, V, O+ V= V. 4) Existence of additive inverse: For any vector, V, there exist a vector, U, such that U+ V= O. 5) Distributivity of vector sums: For any vectors, U, V, and scalar a(U+ V)= aU+ aV. 6) Scalar sum identity. If "e" is the scalar field identity, then, for any vector V, eV= V To show that $\displaystyle R^2$ is a vector space you must show that each of those is true. For example, if U= (a, b) and V= (c, d), where a, b, c, and d are real numbers, then U+ V= (a+ c, b+ d). Since addition of real numbers is "commutative", that is the same as (c+ a, d+ b)= (c, d)+ (a, b)= V+ U so (1), above, is true. Last edited by Country Boy; January 13th, 2017 at 04:24 AM.  
March 7th, 2017, 12:40 PM  #4 
Member Joined: Jul 2014 From: Amherst, MA Posts: 32 Thanks: 6 Math Focus: Calculus, Differential Geometry, Physics, Topology 
I believe you also need distributivity with scalars, and associativity with scalars, that is: $\displaystyle (a+b)v=av + bv $ if $\displaystyle a,b$ scalar, $\displaystyle v$ vector. and, $\displaystyle (ab)v = a(bv)$ 
March 9th, 2017, 10:21 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,576 Thanks: 667 
Yes, I forgot those! Thanks.


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