Determine whether W is a subspace of the vector space

[physics=world]

1. Determine whether W is a subspace of the vector space.

W = {(x,y,z): x ≥ 0}, V = R³

I am not sure if I am doing this right.


2. Test for subspace.

Let these conditions hold.

1. nonempty
2. closed under addition
3. closed under scalar multiplication



3.

Testing for closure under addition:

Let a = (a₁, a₂, a₃) and
Let b = (b₁, b₂, b₃)

a + b = (a₁, a₂, a₃) + (b₁, b₂, b₃)

= (a₁ + b₁, a₂ + b₂, a₃ + b₃)

where x = a₁ + b₁, y = a₂ + b₂, z = a₃ + b₃.

= (x,y,z) Closure under addition.


Testing for closure under scalar multiplication:

(5,y,z) is in W, but
(-1)(5,y,z)
= (-5,-y,-z) is not in W.

Not closed under scalar multiplication.



Therefore, W is not a subspace of R³

 

[Mark44]

1. Determine whether W is a subspace of the vector space.

W = {(x,y,z): x ≥ 0}, V = R³

I am not sure if I am doing this right.


2. Test for subspace.

Let these conditions hold.

1. nonempty
2. closed under addition
3. closed under scalar multiplication



3.

Testing for closure under addition:

Let a = (a₁, a₂, a₃) and
Let b = (b₁, b₂, b₃)

a + b = (a₁, a₂, a₃) + (b₁, b₂, b₃)

= (a₁ + b₁, a₂ + b₂, a₃ + b₃)

where x = a₁ + b₁, y = a₂ + b₂, z = a₃ + b₃.

= (x,y,z) Closure under addition.


Testing for closure under scalar multiplication:

(5,y,z) is in W, but
(-1)(5,y,z)
= (-5,-y,-z) is not in W.

Not closed under scalar multiplication.



Therefore, W is not a subspace of R³

Looks OK, except one of the things you should check is whether the set contains the zero vector. Your set is not a subspace because it's not closed under scalar multiplication. It is closed under addition, and does contain the zero vector.

One nit. You wrote "Let these conditions hold." You can't assume that these conditions are met. When you verify that a set is actually a subspace of the vector space it belongs to, you have to check that these conditions hold.

 

[physics=world]

How do I check for the zero vector?

I know that x can equal to zero, but how about y and z?

 

[Mark44]

Unlike x, which is zero or larger, y and z are arbitrary - there are no restrictions on them. The vector <0, 0, 0> belongs to the set.