Determine whether a subset is a subspace

[Hugo S]

Determine whether the following subset is a linear subspace of ##F^3##.

## X = \left\{ (x_1, x_2, x_3) \in \mathbb{F^3}:x_1 x_2 x_3 = 0 \right\} ##

I know that I can simply provide a counterexample and show that the subset X above is not closed under addition -- namely, I can construct two vectors who are elements of X such that their sum is not an element of X (for instance, I could take the vectors (2, 0, 0) and (0, 3, 3) in X, whose sum (2, 3, 3) is clearly not in X. While I know such an argument would be entirely legitimate, I wondered if there were a more general way of demonstrating whether the subset is a subspace of ##F^3##.

For instance, could I construct an argument centered on the nonlinearity of the function ##f(x_1, x_2, x_3) = x_1 x_2 x_3 ##, showing that in general ##f(x + y) \neq f(x) + f(y) ##?

 

[FactChecker]

The standard way to prove it is a subspace would be to show that the subset satisfies all the properties of a space. Since the addition and scalar multiplication operations are inherited from the larger space, their properties are guaranteed. So I think that only the property of closure under addition and scalar multiplication need to be shown.

To prove that a subset is no a subspace, you will need to demonstrate a counterexample to the closure property under addition or under scalar multiplication.

 

[Fredrik]

So I think that only the property of closure under addition and scalar multiplication need to be shown.

Those two are not quite sufficient, since the empty set is closed under addition and scalar multiplication. You also need to prove that the subset is non-empty. The standard way is to prove that the zero vector is in the subset.

 

[Hugo S]

The standard way to prove it is a subspace would be to show that the subset satisfies all the properties of a space. Since the addition and scalar multiplication operations are inherited from the larger space, their properties are guaranteed. So I think that only the property of closure under addition and scalar multiplication need to be shown.

To prove that a subset is no a subspace, you will need to demonstrate a counterexample to the closure property under addition or under scalar multiplication.

Thank you for your response. So what you are saying then is that the only way to show that this particular set is not a linear subspace is to show a counterexample?

 

[Fredrik]

Thank you for your response. So what you are saying then is that the only way to show that this particular set is not a linear subspace is to show a counterexample?

You can certainly prove something that implies that a counterexample exists. But then you'd have to prove the implication as well.

This is my interpretation of your idea: Let ##f:\mathbb F^3\to\mathbb F##. Define ##S=\{x\in\mathbb F^3|f(x)=0\}##. Now the question is if the following claim is true:

S is a subspace if and only if f is linear.
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The implication "f is linear" ⇒ "S is a subspace" is just the theorem that says that the kernel of a linear transformation is a subspace. The problem is to prove the converse "S is a subspace" ⇒ "f is linear". If you can prove that, then you can solve this problem by proving that the f given in the problem isn't linear.

 

[micromass]

The problem is to prove the converse "S is a subspace" ⇒ "f is linear".

And this is not true, so I wouldn't bother trying to prove that.

 

[FactChecker]

So what you are saying then is that the only way to show that this particular set is not a linear subspace is to show a counterexample?

No. Sorry. I got sloppy. I'm sure that there are circumstances where other proofs are possible. In fact, they may be more common. Suppose the number of elements is wrong for a vector space (a set of finite positive number of elements can not be a vector space over the reals) . Or if one of the vector space properties is specifically ruled out. (0 not in the set)

 

[Fredrik]

And this is not true, so I wouldn't bother trying to prove that.

Thanks. I didn't try very hard to find a counterexample. Now I see one immediately. Let f(x) be the distance between x and the 1-axis. Then S is the 1-axis, which is a subspace, but f is not linear. For example, if x=(0,1,0) and y=(0,0,1), we have
$$f(x+y)=f((0,1,1))=\sqrt{2}$$ but
$$f(x)+f(y)=1+1=2.$$