Subspace Proof (using addition and multiplication)

[erok81]

Homework Statement

Determine whether or not W is a subset of R4

W is the set of all vectors in R4 such that x1x2=x3x4

Homework Equations

Two methods.

u+v (addition)
cu (multiplication)

The Attempt at a Solution

I having trouble getting the hang of subspaces. I thought I was getting close to grasping it, until this problem.

u=(1,1,1,1) and v=(2,2,2,2) (is there a method to choosing these? I just chose points where the original statement still holds)

u+v=(3,3,3,3) which holds under addition.
cu=c(1,1,1,1) where c=2 (again, any method to this, or just choose something other than 0 or 1?)
=(2,2,2,2) which holds under multiplication.

Therefore it is a subspace. But...

The answer in the back states it is not a subspace, so I messed up somewhere. I think it has to do with choosing my u and v vectors.

 

[erok81]

Ah...it my choosing of u and v.

If I choose u=(-1,2,2,-1) and v=(1,4,1,4) they hold in the original statement, but not under additional.

Which brings me to my other questions - how do you choose those values? Clearly that is where my mistake lies.

 

[fzero]

Choose the most general vectors u = (u₁,u₂,u₃,u₄) and v = (v₁,v₂,v₃,v₄), with u₁u₂=u₃u₄ and v₁v₂=v₃v₄. The vectors you chose are too special (they satisfy a much stronger condition).

 

[erok81]

In my book they chose (1,1,1,1) for a few of their vectors...which values would you choose? The only thing I am worried about it this doesn't seem like a very good way to prove something as depending on your choice, it could prove/disprove the subspace.

Or is there a better way to do these without choosing arbitrary vector values?

 

[fzero]

If you're trying to prove a conjecture that a large set of objects has some property, you generally need to choose the most general representative of that set for your proof. If you are simply trying to disprove such a conjecture, a single counterexample might suffice.

Since this question was a "whether or not" issue, you were right to first look for some counterexamples, since that can be more efficient than trying to develop a line of argument that might prove the statement. Where you went wrong was to first choose vectors of the form (u,u,u,u), which satisfy u₁=u₂=u₃=u₄ instead of just u₁u₂=u₃u₄. Your second set of vectors was a better choice because they only satisfied the original condition.

 

[erok81]

Ok, I get it. My vector choice was too general and satisfied more than one statement. Which makes sense because I've done about 20 of these so far and the others worked fine...because I chose vectors more like my second choices. This was the first one I chose a (u,u,u,u) type vector.

Is there a more accurate method than the way I am doing or just be careful when choosing the vectors?

Thanks for the help as well.