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

1. Nov 10, 2013

### 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.

2. Relevant 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.

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.

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)

2. Nov 10, 2013

### Simon Bridge

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.

3. Nov 10, 2013

### LCKurtz

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

4. Nov 10, 2013

### 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.

5. Nov 10, 2013

### Staff: Mentor

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.
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.

6. Nov 10, 2013

### HallsofIvy

Staff Emeritus
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)?

7. Nov 10, 2013

### physics=world

There closed under addition and scalar multiplication.

8. Nov 10, 2013

### physics=world

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.

9. Nov 10, 2013

### Staff: Mentor

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.

10. Nov 10, 2013

### physics=world

(0,0,0)

(1,1,-2)

11. Nov 10, 2013

### 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.