P is not a subspace of R3. Why?

[dcramps]

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 into play? I'm really quite lost and cannot find any relevant examples in my text.

 

[icystrike]

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 into play? I'm really quite lost and cannot find any relevant examples in my text.

Hi dcramps

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 R^{3} what should be satisfied?
[Hint: It must passes through what?]

 

[dcramps]

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...?

 

[Mark44]

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.

 

[icystrike]

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.

 

[Mark44]

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.

I'm not sure this is a clearer picture...

 

[dcramps]

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?

 

[statdad]

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.

 

[Mark44]

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' \neq 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')