Set R^(2) with the usual vector addition forms an abelian group

[ak123456]

Homework Statement

the set R^(2) with the usual vector addition forms an abelian group. For a belongs to R and x=(x1,x2) belongs to R^(2) we put a *x :=(ax1,0),this defines a scalar multiplication R*R^2 ---R^2 (a,x)---a*x.
determine which of the axioms defining a vector space hold for the abelian group R^2 with the scalar multiplication

Homework Equations

The Attempt at a Solution

I know that a*x1=ax1 but a*x2=0?? It confused me . And how to use an axiom to define it ?

 

[HallsofIvy]

Homework Statement

the set R^(2) with the usual vector addition forms an abelian group. For a belongs to R and x=(x1,x2) belongs to R^(2) we put a *x :=(ax1,0),this defines a scalar multiplication R*R^2 ---R^2 (a,x)---a*x.
determine which of the axioms defining a vector space hold for the abelian group R^2 with the scalar multiplication

Homework Equations

The Attempt at a Solution

I know that a*x1=ax1 but a*x2=0?? It confused me . And how to use an axiom to define it ?

You don't "use an axiom to define it"- it is already defined. And "a*x1" doesn't mean anything- a*(x1, x2) is defined as (ax1, 0). You need to show that this still obeys the axioms defining a vector space. What are those axioms?

 

[ak123456]

You don't "use an axiom to define it"- it is already defined. And "a*x1" doesn't mean anything- a*(x1, x2) is defined as (ax1, 0). You need to show that this still obeys the axioms defining a vector space. What are those axioms?

oh,i see . i have to use ax1 and 0 to show 1st and 2nd distributivity law,identity elements and the compatibility are all exist ?