|January 8th, 2017, 07:34 AM||#1|
Joined: Jan 2017
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 13th, 2017, 05:22 AM||#3|
Joined: Jan 2015
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|
Joined: Jul 2014
From: Amherst, MA
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.
$\displaystyle (ab)v = a(bv)$
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