- #1
Miike012
- 1,009
- 0
Vector space V = Rn, S = {(x,2x,3x,...,nx) | x is a real number}
I know that it is closed under addition, but to be closed under scalar mult...
say r is a scalar, then
Say we multiply S by -4
Then -4S = (-4x,-8x,...-4nx)
Because the problem never stated that the number infront of x should be pos or neg, all it said was that "x is a real number" so when I mult S by a number r x is still a real number... does this explain why S is closed under scalar mult? or not exactly?
I know that it is closed under addition, but to be closed under scalar mult...
say r is a scalar, then
Say we multiply S by -4
Then -4S = (-4x,-8x,...-4nx)
Because the problem never stated that the number infront of x should be pos or neg, all it said was that "x is a real number" so when I mult S by a number r x is still a real number... does this explain why S is closed under scalar mult? or not exactly?