Closed Under Scalar Multiplication: Subspace Question Vector Space V = Rn

[Miike012]

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?

 

[Mark44]

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

S is a set of vectors. You're not multiplying S by a scalar; you're multiplying an arbitrary vector in S by a scalar.

Let v be a vector in S, which means that v = a<1, 2, 3, ..., n> for some scalar a. Is kv also in S?

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?

 

[Miike012]

well is -4v in S?

if i multiply some vector in S by a negative scalar will that new vector still be in S?

 

[Mark44]

well is -4v in S?

if i multiply some vector in S by a negative scalar will that new vector still be in S?

Why wouldn't it be? Every vector in S is in the form <1x, 2x, 3x, ..., nx> for some real number x. Isn't this the same as x <1, 2, 3, ..., n>?