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

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And isn't multiplying by a negative scalar just flipping the direction of the vector, which doesn't change the fact that it is still in the form <1x, 2x, 3x, ..., nx>?In summary, a vector space V = Rn and a set S = {(x,2x,3x,...,nx) | x is a real number} is closed under scalar multiplication. This means that when a vector in S is multiplied by a scalar, the resulting vector will still be in S. This is because every vector in S is in the form <1x, 2x, 3x, ..., nx> for some real number x, and multiplying by a scalar simply changes the magnitude and
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Miike012
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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?
 
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  • #2
Miike012 said:
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?
Miike012 said:
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?
 
  • #3
well is -4v in S?

if i multiply some vector in S by a negative scalar will that new vector still be in S?
 
  • #4
Miike012 said:
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>?
 

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

What is a subspace in a vector space?

A subspace is a subset of a vector space that satisfies the three properties of a vector space: closure under vector addition, closure under scalar multiplication, and containing the zero vector.

What does it mean for a subspace to be closed under scalar multiplication?

A subspace is closed under scalar multiplication if multiplying any vector in the subspace by a scalar results in another vector that is also in the subspace.

Can any subset of a vector space be considered a subspace?

No, a subset must satisfy the three properties of a vector space to be considered a subspace. If it does not satisfy all three properties, it cannot be considered a subspace.

How do you determine if a subset is a subspace?

To determine if a subset is a subspace, you must check if it satisfies the three properties of a vector space: closure under vector addition, closure under scalar multiplication, and containing the zero vector. If it satisfies all three properties, it is a subspace.

What is the importance of closed under scalar multiplication in a subspace?

The property of being closed under scalar multiplication is important because it ensures that the subspace remains a vector space. This allows for the use of standard vector operations and properties within the subspace.

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