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 January 8th, 2017, 07:34 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, 08:08 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,646 Thanks: 836 Thanks from muneeb977
January 13th, 2017, 05:22 AM   #3
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Quote:
 Originally Posted by muneeb977 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"
The first thing you should do is state what "R2" means! I suspect you mean "$\displaystyle R^2$" or "R^2", the set of all ordered pairs of real numbers with addition and scalar multiplication defined "coordinate wise": (a, b)+ (c, d)= (a+ c, b+ d) and a(b, c)= (ab, ac).

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 05:24 AM.

 March 7th, 2017, 01:40 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, 11:21 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,872 Thanks: 766 Yes, I forgot those! Thanks.

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