Proving Subspace: Vectors (x,y,z) in R^3 Satisfying x+y+z=0

[jeffreylze]

Homework Statement

Show that the following set of vectors are subspaces of R^m

The set of all vectors (x,y,z) such that x+y+z=0 of R^3 .

Then find a set that spans this subspace.

Homework Equations

The Attempt at a Solution

I managed to proof that the set of vectors is a subspace by showing that it is non-empty, closed under addition and scalar multiplication. However, I have no idea how to start on part b, how do I find a spanning set for that subspace? If I am not mistaken, I have to find linear combinations.

 

[Dick]

Ok, name one vector in the subspace. Can you find another one that's independent of the first? Can you find a third that's not a combination of those two?

 

[jeffreylze]

is (1,1,1) one of the vector? I am confused.

 

[Mark44]

x = -y - z
y = y
z = z

If you stare at this awhile, you might see two vectors staring back at you.

 

[Mark44]

is (1,1,1) one of the vector? I am confused.

Only if 1 + 1 + 1 = 0.

 

[jeffreylze]

oh, ok. Tell me if this is right. Since x = -y-z , y=y , z=z hence (-y-z , y , z) . So x(0,0,0) + y(-1,1,0) + z(-1,0,1) , So the spanning sets are (0,0,0) , (-1,1,0) , (-1,0,1) But the given answers don't include (0,0,0) . before all that, how do you know y=y and z=z ? I only know why x = -y-z .

 

[Mark44]

Well, your set spans the subspace, but it also does so if you remove (0, 0, 0).

How did I know that y = y and z = z? The two equations are obviously true, aren't they?

 

[jeffreylze]

Ah, you got me there. They are obviously true. I was complicating stuffs, now looking back, that seemed a stupid question. OK, thanks, that did help me understand spanning sets better. Cheers

 

[Mark44]

If it didn't seem like a stupid question then, but now it does, I guess that means you're getting smarter, which is a good thing.