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P is not a subspace of R3. Why?

  • Thread starter dcramps
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  • #1
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Homework Statement


P={(x,y,z)|x+2y+z=6}, a plane in R3. P is not a subspace of R3. Why?

Homework Equations


See below.

The Attempt at a Solution


I am really quite confused here.

My text says:
"A subset W of a vector space V is called a subspace of V if W is itself a vector space under the addition and scalar multiplication defined on V"
and goes on to say that the only axioms that need verification are:
(a) If u and v are vectors in W, then u+v is in W.
(b) If k is any scalar and u is any vector in W, then ku is in W.

So from here...I'm a bit confused. Where does this x+2y+z=6 come in to play? I'm really quite lost and cannot find any relevant examples in my text.
 

Answers and Replies

  • #2
446
1


Homework Statement


P={(x,y,z)|x+2y+z=6}, a plane in R3. P is not a subspace of R3. Why?

Homework Equations


See below.

The Attempt at a Solution


I am really quite confused here.

My text says:
"A subset W of a vector space V is called a subspace of V if W is itself a vector space under the addition and scalar multiplication defined on V"
and goes on to say that the only axioms that need verification are:
(a) If u and v are vectors in W, then u+v is in W.
(b) If k is any scalar and u is any vector in W, then ku is in W.

So from here...I'm a bit confused. Where does this x+2y+z=6 come in to play? I'm really quite lost and cannot find any relevant examples in my text.

Hi dcramps :smile:

I'm learning linear algebra by myself , and i'll try to help you to the best of my ability.
If P is a subspace of [tex]R^{3}[/tex] what should be satisfied?
[Hint: It must passes through what?]
 
  • #3
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This is a bit of a guess here, since I am a bit confused on the whole thing still, but I believe it must pass through the origin, and since it states x+2y+z=6, that is not satisfied...?
 
  • #4
33,162
4,846


Right, the origin is not in the plane, which means that set P does not include (0, 0, 0). There are a couple of other requirements for P to be a subspace of R3, but since this one failed, there's no need to check the others.
 
  • #5
446
1


Right, the origin is not in the plane, which means that set P does not include (0, 0, 0). There are a couple of other requirements for P to be a subspace of R3, but since this one failed, there's no need to check the others.
Just to make a clearer picture for dcramps , when we say it must pass through the origin it must satisfy the equation of having x,y,z=0 , hence the equation of your plane would be 0+0+0=6 => 0=6 which is absurd. Therefore , we could have the plane passing through the origin and fail to satisfy the criteria of a subspace. :biggrin:
 
  • #6
33,162
4,846


Just to make a clearer picture for dcramps , when we say it must pass through the origin it must satisfy the equation of having x,y,z=0 , hence the equation of your plane would be 0+0+0=6 => 0=6 which is absurd. Therefore , we could have the plane passing through the origin and fail to satisfy the criteria of a subspace. :biggrin:
I'm not sure this is a clearer picture...
 
  • #7
43
0


Both answers were helpful. I do have another question though, since the solution given did not mention the origin. The solution to the question I asked above is:

(x+x') + 2(y+y') + (z+z') = (x+2y+z) + (x'+2y'+z') = 6 + 6 = 12
thus
(x+x') + 2(y+y') + (z+z') is not in P, and so P is not a subspace of R3

I don't understand why it isn't. Because it's 12? Huh?
 
  • #8
statdad
Homework Helper
1,495
35


The text's solution shows that if you take two points that are on the plane and "add them" the result is not in the plane (because of the 12) - the set of points in the plane is not closed under addition.
 
  • #9
33,162
4,846


If (x, y, z) is in P, then x + 2y + z = 6.
If (x', y', z') is in P, then x' + 2y' + z' = 6.

Now let's check whether (x, y, z) + (x', y', z') is in P.

If so, then (x + x', y + y', z + z') is in P, since (x, y, z) + (x', y', z') = (x + x', y + y', z + z').

(x + x', y + y', z + z') is in P provided that x + x' + 2y + 2y' + z + z' = 6. But x + x' + 2y + 2y' + z + z' = x + 2y + z + x' + 2y' + z' = 6 + 6 = 12, from previous work.
So x + x' + 2y + 2y' + z + z' [itex]\neq[/itex] 6, hence (x + x', y + y', z + z') is NOT in P.

If (x, y, z) + (x', y', z') = (x + x', y + y', z + z') is in P. If so, then x + x' + 2(y + y') + z + z' = 6.

But we've already seen that

(x, y, z) + (x', y', z')
 

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