# Is the following set a subspace?

1. Oct 4, 2011

### bjohnson2001

1. The problem statement, all variables and given/known data

x belongs to the vector space R^6.

Is (x1-x2)^4 + x3^6 = 0 a subspace?

2. Relevant equations

Since we already know x is a vector space we only need to check:

1. The existence of the zero vector

3. The attempt at a solution

1. (0,0,0) = 0 √ check

2. x = (-4,-4,0), y = (3,3,0)

x+y = (-1,-1,0) satisfies equation √ check

3. -1*(2,2,0) = (-2,-2,0) satisfies equation √ check

All of this seems straightforward, I'm just having a real problem with the exponents in the expression. I know that vector spaces deal with linear combinations of vectors. Does it matter that the expression of the subspace itself is not linear?

Does that somehow change the rules or can a subspace be any crazy function we want? (as long as it consists of vectors and scalars)

2. Oct 4, 2011

### Staff: Mentor

x is a vector in R6. x is not a vector space, but R6 is.
No, your checks for 2 and 3 are wrong. To verify closure under vector addition, you need to show that x + y is in the given set whenever x and y are. And similarly for scalar multiplication. You don't get to pick specific vectors.

Instead, work with two arbitrary vectors x and y. Since x and y are assumed to belong to your set, it must be true that (x1 - x2)4 + x36 = 0. It must also be true that (y1 - y2)4 + y36 = 0. Now find out whether x + y also satisifies this equation.

3. Oct 4, 2011

### bjohnson2001

Oh good call, x is definitely a vector. So we know that x is a vector in R6 so linear combinations of x inherit the same properties of the original vector space. Since the subspace expression is not linear, does that mean the properties are not inherited and we have to go back and reprove all axioms of vector addition and scalar multiplication? Or is it OK for a subspace expression to be non-linear?

4. Oct 4, 2011

### Dick

Try and describe all members of R^6 (x1,x2,x3,x4,x5,x6) that satisfy the equation. Judging by your examples, you've already guessed that x3 must be zero and x1 must equal x2. {x4,x5,x6} can be anything right? Can you prove that? Does that sound like a subspace?

5. Oct 4, 2011

### bjohnson2001

Thats right, x4, x5 and x6 are all free variables and can be any value since they aren't constrained by the set. I think thats OK, if we were in R3 and only placed limits on x and y I would imagine that z could have any value like an infinite plane.

I'm not sure if that should sound normal for a subspace (not too confident in this area), but it sounds good to me! I hope you can correct me if I have got this all wrong

6. Oct 4, 2011

### Dick

Right. But can you show x1=x2 and x3=0? If so then any element of the solution to that equation should be (x1,x1,0,x4,x5,x6) where x1, x4, x5 and x6 can take any values. It should be easy to show that is a subspace.

7. Oct 5, 2011

### bjohnson2001

OK that sounds good, so its alright that the expression of the set is not linear? Usually I read proofs dealing with linear combinations of vectors where we can assume a number of things. I am not sure that the same properties are inherited when we square the difference of two vectors. Or is it OK as long as the vectors are the same vectors defined in R6?

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Thank you for your help, this is very useful to me!

Last edited: Oct 5, 2011
8. Oct 5, 2011

### Dick

It's 'often' true that the solutions to a nonlinear equation don't form a subspace. But it's not always true. You need to find ALL the solutions and check that they satisfy the conditions of a subspace. Did you check that for {x1,x1,0,x4,x5,x6}? Just checking particular examples doesn't prove it.

9. Oct 5, 2011

### bjohnson2001

I did and it satisfies all conditions of a subspace, I can conclude that it must be a subspace. Making the connection that x1=x2 and x3=0 was something I noticed for my example sets but didn't think to generalize for all x. This makes the proof much more complete. Thank you again for your help!

10. Oct 5, 2011

### Dick

Yes, it's a subspace, even though the defining function looks nonlinear.