Homework Statement
{(x1,x2)T| |x2|=|x2|}
So my first thought is we would have to check for both cases (x1,x1) and (-x1,-x1)
a=(x1,x1)T b=(v1,v1)T
βa=(βx1,βx1)T for the case where a<0 βa(-βx1,-βx1)T
thus it is closed under scalar multiplication.
a+b=(x1+v1,x1+v1) for the case...