# Proving span(w) = W makes it s subspace

1. Jun 8, 2013

### Emspak

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

Show that a subset W of vector space V is a subspace of V iff span(W) = V

3. The attempt at a solution

OK, I am trying to see if my reasoning is correct or if I am overthinking this.

To show this is a subspace three things have to be true.

(a) 0 $\in$ W,
(b) vectors x + y $\in$ W if x,y$\in$ W
(c) cx $\in$ W whenever c $\in$ W and x$\in$W

All this means the subspace has to be closed under addition and multiplication and contain 0.

First we show the bit about zero. is there a vector 0' $\in$ W such that x + 0' = x for all x$\in$ W. Since x + 0 = x also then 0'=0 so 0$\in$W and condition (a) is valid.

Is there an additive inverse of the vectors in W that lies in W?

since x$\in$W
(-1)x$\in$W because of the axiom that says an additive inverse exists and that for all scalars in a given field multiplied by a vector are in the vector space.

We've shown that W is a subspace. what is Span(W)?
Span(W) is the set of all linear combinations of W.
Span(W) = {λ1x1 + λ2x2 + ... + λnxn | λi $\in$ K} where K is the field.

We know span(W) ≠ $\emptyset$ because if span(W) = $\emptyset$ then span(W) = {0}

let x,y$\in$ span(W)

that means
x = (α1x1 + α2x2 + ... + αnxn) for αi$\in$ K, x$\in$ W
y = (β1x1 + β2x2 + ... + βnxn) for βi$\in$ K, y$\in$ W

which also means that x + y $\in$ span(W)
also if: $$\mathbf x\mathbf = \sum_{i=1}^n a_i x_i \in\ span(W)$$ and β$\in$ K, y$\in$ W then $$β\mathbf x\mathbf = \sum_{i=1}^n (βα_i) x_i \in\ span(W)$$

since from the axioms we see that scalars distribute and βαi$\in$ span(W)

span(W) is a subspace, but is it the smallest subspace of V containing W?

Let E $\subseteq$ V be a subspace containing W.

let x be a vector in span(W) and use the relation above. $$\mathbf x\mathbf = \sum_{i=1}^n a_i x_i \in\ span(W)$$

$\forall$ 1 $\leq$ i $\leq$n xi$\in$E bit this implies that αixi$\in$E
that would mean
$$\mathbf x\mathbf = \sum_{i=1}^n a_i x_i \in\ E$$ and x$\in$E so span(W) $\in$E as well.

so yes, span(W) the smallest subspace of V in W

Now what I want to know is if there's some silly egregious error I have made here. If you saw this on a test would you mark it right, wrong, or mostly right but missing some fundamental point, you know?

2. Jun 8, 2013

### LCKurtz

W= {(1,0)} is a subset of $V = R^2$. Span(W)={(a,0)}$=R^1\ne R^2$

3. Jun 8, 2013

### Emspak

So the short answer is I was waaaay overthinking this, huh?

4. Jun 8, 2013

### Emspak

Or was yours a better statement of the problem itself? (I'm just starting linear algebra and I won't claim super-fluency with the notation just yet)

as i read your reply, it's the vector (1,0) in the set W is a subset of V in R^2, and span(W) is the set of vectors (a,0) in R^1 which isn't the same as R^2 and is a subspace thereof. Is that right?

5. Jun 8, 2013

### LCKurtz

Yes. But I was thinking you might have copied the problem incorrectly and the right side of the iff statement should be span(W) = W, not span(W) = V. Check that.

6. Jun 8, 2013

### Emspak

You are right, I did copy it incorrectly. I'm curious if the rest of the problem was done right anyway tho.

7. Jun 8, 2013

### HallsofIvy

As LCKurtz told you this is not true!

(a) could be replace with "W is non-empty" If it is non-empty there exist some v in W and since it is "closed under scalar multiplication" it follows that (-1)v= -v is in W. Then since W is closed under addition it follows that v+ (-v)= 0 is in W. And, of course, if 0 is in W then W is non-empty.

Is there this a statement or a question? If a statement, how do you prove it?

I can see no reason to distinguish between "0" and "0'". Your proof makes no sense because you have not shown that there is "0' in W such that x+ 0'= x for all x $\in$ W.
If there exist a vector v in W (We are not actually told that W is non-empty but "span(W)" for empty W makes no sense) span(W), the set of all linear combinations, contains 0(v)= 0.

The axiom (or part of the definition of "vector space") says that an additive inverse exist in the vector space. It does NOT say that it is in subset W.

?? No, that's not even what you were asked to prove!

??? That makes no sense! If span(W) = $\emptyset$ then it can't be span(W) = {0} because that's NOT empty.

Which shows that span(W) is closed under addition of vectors, not that it is equal to V.

I am completely confused now as to what you are trying to prove!

You seem to have lost track of what you were trying to prove.
Assuming that the proposition was actually "subset W of V is a subspace if and only span(W)= W" then you were to prove two things: "If W is a subspace of V then span(W)= W" and "if span(W)= W then W is a subspace of V" but I don't see anywhere that you have even claimed that those are true! Certainly, "span(W) is the smallest subspace of V that contains W" is true but has nothing to do with what you are trying to prove.

Last edited by a moderator: Jun 8, 2013
8. Jun 8, 2013

### Emspak

OK. Now I am thanks to HallsofIvy, lost completely.

Let's try this again.

I have to prove subset W of V is a subspace if and only if span(W)= W

I went through the steps in the bloody lecture. So what is it that I am missing here? I was trying to show each of the three conditions that my prof said have to hold true. Don't anyone be all mysterious about it, please. I am the dumbest, most retarded student you have ever met-- make that assumption, please.

I want to prove the above proposition. Yes?

So. First I have to show that 0 is in W. Yes or no? that was proposition (or whatever you want to call it I am not sure the proper term) (a)

then I have to show (b). Is that correct or not? Yes? No? What?

Then I have to show that (c) holds. Is this the case? Is it not?

Our prof said that one step is showing there is an additive inverse. Is this the case or not?

9. Jun 8, 2013

### Emspak

OK. Going over some other sources. Here's what I have:

Span(W) is the collection of all the linear combinations of vectors in W. I want to show that it is a subspace of V.

further, that Span(W) = W and that's the only way it is a subspace.

first, 0 multiplied by any vector in W is going to be 0. It's a linear combination of all the vectors in W, meaning essentially that when I multiply any vector in W I get zero. the zero vector is thus in span(W). Is this correct or not?

Second step: let x be a vector in span(W). That means further that any scalar multiple of x is in the span as well. And it is still a linear combination of vectors in W. Is this also the case? Am I wrong here?

Now let x and y be vectors in W. They can both be written as vectors multiplied by scalars, and when you separate the scalars out you get s scalar multiplied by a single vector, like (a+b)x= (a1+b1)x1+ .... (a1+b1)xn

with that we see that the addition is a linear combination. which means that x+y are in span(W). So SPan(W) is closed under addition. That makes it a subspace of V.

Is this any better?

10. Jun 8, 2013

### HallsofIvy

No, you don't! That's true for any set W. You want to prove that, if span(W)= W then W is a subspace. Of course, because span(W) is a subspace, that is almost trivial. The hard part is the other way: if W is a subspace, then span(W)= W.

I'm not sure what you mean by this.

Yes, span(W) is always contains the 0 vector.

Yes if x is in W then any multiple of x is in span(W).

I don't understand this. What happened to "y"? And what are "a1", "x1", etc.?

This is always true and NOT what you were asked to prove! You were asked to prove:
1) If span(W)= W then W (NOT span(W)) is a subspace.
2) if W is a subspace then Span(W)= W.

11. Jun 8, 2013

### Emspak

Let x, y ∈ span(W).

x = (α1x1+ α2x2 + … +αnxn)
y = (β1y1+ β2y2 + … +βnyn

This can also be written as:

x + y = (α11)x1+(α22)x2 + … +(αnn)xn)

and that is still a linear combination of the two vectors x and y. Is that not correct? Am I missing something?