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January 8th, 2017, 06:34 AM   #1
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R2 is a vector space

How to verify that R2 is a vector space?

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January 8th, 2017, 07:08 AM   #2
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https://en.wikipedia.org/wiki/Vector_space#Definition
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January 13th, 2017, 04:22 AM   #3
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Quote:
Originally Posted by muneeb977 View Post
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 04:24 AM.
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March 7th, 2017, 12:40 PM   #4
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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,

$\displaystyle (ab)v = a(bv)$
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March 9th, 2017, 10:21 AM   #5
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Yes, I forgot those! Thanks.
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