Question

show that R x R with the standard addition
(x_1,y_1 )+ (x_2,y_2 )=(x_1+x_2,y_1+y_2)
and the nonstandard scalar multiplication
c(x,y)=(cx,0)
is not a vector space.

Answer 1

this set does not follow the property that 1u=u...
suppose if we have any vector v=(v_1,v_2) from R x R then we know that 1v=v if this is really a vector space, however since tthe scalar multiplication here is defined as c(x,y)=(cx,0), then we have,
\[1v=1(v_1,v_2)=(1(v_1),0)=(v_1,0)\neq(v_1,v_2)=v\]
now we have shown that in this set

\[1v \neq v\]
and so since this set doesn't have a property of a vector space which is 1v=v, then it is not a vector space