Which of the subsets of R^3 is a subspace of R^3.

[physics=world]

1. Which of the subsets of R³ is a subspace of R³.

a) W = {(x,y,z): x + y + z = 0}

b) W = {(x,y,z): x + y + z = 1}

I was wondering if my answer for A is correct.

Homework Equations

3.

A) W = {(x,y,z): x + y + z = 0}

Since, x + y + z = 0. Then, the values for all the variables have to be zero. Therefore, the only vector in W is the zero vector. So, W is nonempty and a subset of R³.

Furthermore, because W is closed under addition and scalar multiplication, it is a subspace of R³.

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:
c is any real number

c(x₁, y₂, z₃)
= (cx₁, cy₂, cz₃) Where x₁, y₂, z₃ = 0

= (0,0,0)

 

[Simon Bridge]

Since, x + y + z = 0. Then, the values for all the variables have to be zero.

That just means (0,0,0) is one member of the set. Another member would be (1,1,-2) since 1+1-2=0 as well.

 

[LCKurtz]

1. Which of the subsets of R³ is a subspace of R³.

a) W = {(x,y,z): x + y + z = 0}

b) W = {(x,y,z): x + y + z = 1}

I was wondering if my answer for A is correct.

Homework Equations

3.

A) W = {(x,y,z): x + y + z = 0}

Since, x + y + z = 0. Then, the values for all the variables have to be zero. Therefore, the only vector in W is the zero vector. So, W is nonempty and a subset of R³.

What about (-2,1,1)?

Furthermore, because W is closed under addition and scalar multiplication, it is a subspace of R³.

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:
c is any real number

c(x₁, y₂, z₃)
= (cx₁, cy₂, cz₃) Where x₁, y₂, z₃ = 0

= (0,0,0)

You need to redo the closure arguments. The results of the operations have to satisfy x+y+z=0.

 

[physics=world]

If I wrote the first part like this:

Since, x + y + z = 0. Then, the values for all the variables could equal to zero. Therefore, the zero vector is in W. So, W is nonempty and a subset of R3.

would it work?

and I might need some help on the closure arguments.

 

[Mark44]

If I wrote the first part like this:

Since, x + y + z = 0. Then, the values for all the variables could equal to zero. Therefore, the zero vector is in W. So, W is nonempty and a subset of R3.

Restate the sentence that starts with "Then, the values ..." as If we let x = y = z = 0, then the zero vector is in this set, and W is nonempty.

You don't need "and a subset of R³." This is given in the problem statement.

would it work?

and I might need some help on the closure arguments.

Assume that u and v are two arbitrary vectors in your set. Show that u + v must also be contained in this set. The key here is arbitrary. You can't pick and choose specific vectors.

Assume that k is an arbitrary scalar. Show that ku is in the set.

 

[HallsofIvy]

Suppose (x, y, z) satisfies x+ y+ z= 0 and (a, b, c) satisfies a+ b+ c= 0. What can you say about (x+ a, y+ b, c+ d) and (ax, ay, az)?

 

[physics=world]

Suppose (x, y, z) satisfies x+ y+ z= 0 and (a, b, c) satisfies a+ b+ c= 0. What can you say about (x+ a, y+ b, c+ d) and (ax, ay, az)?

There closed under addition and scalar multiplication.

 

[physics=world]

Assume that u and v are two arbitrary vectors in your set. Show that u + v must also be contained in this set. The key here is arbitrary. You can't pick and choose specific vectors.

Would this work?Let u = (u₁, u₂, u₃) and
Let v = (v₁, v₂, v₃)

u + v = (u₁, u₂, u₃) + (v₁, v₂, v₃)

= (u₁ + v₁, u₂ + v₂, u₃ + v₃)

where x = u₁ + v₁, y = u₂ + v₂, z = u₃ + v₃.

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

 

[Mark44]

No, this won't work. Your u and v vectors are just arbitrary vectors in R³. You need to start with vectors in set W.

Your OP gives a precise but terse definition of W. With this definition, one can distinguish between vectors in W and vectors in R³ that aren't in W.

To get your intuition working, can you come up with specific vectors in W? Note that in proving that a set is a subspace, you can't use specific vectors, but this exercise might help you understand a little better what you need to do in your proof.

 

[physics=world]

to get your intuition working, can you come up with specific vectors in w?.

(0,0,0)

(1,1,-2)

 

[Simon Bridge]

Back to basics.
W is the set of all triples (x,y,z) which satisfy the condition C: x+y+z=0.

You know that (0,0,0) is a member of W because it satisfies the condition that x+y+z=0
Therefore W is not empty. So far so good...

It should be easy to find other members ... how many members are there altogether?
hint: more than 4 - see if you can come up with a few more, i.e. is (4,-1,7) a member of W?
This is the exercise suggested by Mark44.

If z=0, what possible values of x and y will work?
What if z=1? z=-1?, z=1/2,1/3,10?
If (x,y,z) were plotted on x,y,z axes, what geometric object would the members of W make?

You cannot use specific vectors to prove the subspace...

u=(a,b,c) and v=(e,f,g) are two representative members of W,
this means that a+b+c=0 and e+f+g=0

W is closed under addition if w=u+v is also a member of W.
How would you go about showing this is the case?
hint: find the components of w in terms of the components of u and of v, then see if they satisfy the condition C.