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I was going through this PDF and was reminded of an issue I always run into as outlined below.

A subspace for $R^n$ is any collection S of vectors in $R^n$ such that

  1. The zero vector 0 is in S.

  2. If u and v are in S, then u+v is in S [closed under addition].

  3. If u is in S and c is scalar, then cu is in S [closed under multiplication].

Property 1 is only needed to ensure that S is non-empty; for non-empty S, property 1 follows from property 3, 0a = 0.

So, when do we know we have to check if the set is nonempty? I just don't understand when we have to check for zero vector(Property 1). I just check for all three each time.

ChocolateAndMath
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    the presence of $\vec 0$ follows from the other two, if you have non-trivial elements in $S$, as $\vec v+(-\vec v)=\vec 0$ must be in $S$ if $\vec v$ is. – lulu Oct 23 '17 at 18:13
  • Of course. But by that logic, we always only check for the second and third properties. In what case do we check for the first property is my question. – ChocolateAndMath Oct 23 '17 at 18:15
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    Not sure I understand the question. It is often clear that $\vec 0$ is not in whatever subset you are considering, and in that situation that's all you need to point out to settle the matter. In other settings, it suffices to look at the last two conditions as they imply the first (if the set is non-empty). – lulu Oct 23 '17 at 18:24

3 Answers3

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Once you have shown (2) and (3) you can either complete the proof by show that $S$ is nonempty, or by showing that $0\in S$. No matter whether you do one or the other, you then have enough to conclude that $S$ is a subspace.

There's no situation where you have to do it one way or the other. It's purely up to you, depending on what you find is easiest in the situation.

As @lulu notes in a comment, pointing out that $0\notin S$ can be a quick and easy way to show/discover that something is not a subspace.

Note that it is an extremely rare situation in mathematics to be given a particular subset of a vector space and being asked to find out whether it's a subspace or not. Essentially the only case where this happens is in exercises in introductory linear algebra courses -- and then the point is not that you need to be particularly proficient at this rather artificial task, but simply to give you a chance to develop some intuition about the meaning of the word "subspace" (and do some work with the definition such that you'll remember it later).

hmakholm left over Monica
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  • Right. I understand the motivation behind it now. But, I guess this question came in my head as my professor was reviewing a question we had on a quiz. In that question, he mentioned that we must show that the zero vector belongs to the subspace and subsequently check for the other properties. It may be an isolated case for that question. – ChocolateAndMath Oct 23 '17 at 18:32
  • I remember showing that it is closed under addition and scalar multiplication correctly but, lost points for not showing that the zero vector is in the subspace. Because, as pointed out by @lulu that if it is closed under addition and scalar multiplication, it follows that the set is nonempty. I can post that question and the correction that Professor made to make things clearer if needed. – ChocolateAndMath Oct 23 '17 at 18:32
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    @ChocolateAndMath: In a classroom setting it is entirely possible that the teacher insists that a task be carried out with a particular method for learning purposes and not because that particular method is the only mathematically valid one. – hmakholm left over Monica Oct 23 '17 at 18:34
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    @ChocolateAndMath: Also, if you merely prove closure under addition and scalar multiplication, but don't show either that it is nonempty or that it contains zero, then you certainly don't have enough for a proof that it's a subspace. – hmakholm left over Monica Oct 23 '17 at 18:35
  • Right. That's what I was looking for. It all clicks now and I understand now why I was getting the sarcastic berating from other contributors. Got it now what the professor meant! – ChocolateAndMath Oct 23 '17 at 18:38
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Often noting/showing that the zero-vector in the set is the easiest way to show that the set (potential subspace) is nonempty.

paw88789
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You should verify that it is non-empty when

  1. You are feeding your proof to a computer, or a similarly literal-minded being, or

  2. This is your first time working with vector spaces and need to make it clear that you know and understand all the parts of the definition, or

  3. You've defined a fairly exotic subspace (or addition operation) and it's not that obvious which element acts as the zero vector, or that it's in your subspace.

Math is a human activity, and lots of it happens based on non-precise conventions.

JonathanZ
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