# Determine whether a subset is a subspace

1. Jul 31, 2015

### 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 $f(x + y) \neq f(x) + f(y)$?

2. Aug 1, 2015

### 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.

3. Aug 1, 2015

### Fredrik

Staff Emeritus
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.

4. Aug 1, 2015

### Hugo S

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?

5. Aug 1, 2015

### Fredrik

Staff Emeritus
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.
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.

6. Aug 1, 2015

### micromass

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

7. Aug 1, 2015

### FactChecker

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)

8. Aug 1, 2015

### Fredrik

Staff Emeritus
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
$$f(x)+f(y)=1+1=2.$$