Proving span(w) = W makes it s subspace

[Emspak]

Homework Statement

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

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\inW

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\inW and condition (a) is valid.

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

since x\inW
(-1)x\inW 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 + ... + λ_nx_n | λ_i \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 + ... + α_nx_n) for α_i\in K, x\in W
y = (β₁x₁ + β₂x₂ + ... + β_nx_n) for β_i\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 βα_i\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 \leqn x_i\inE bit this implies that α_ix_i\inE
that would mean
\mathbf x\mathbf = \sum_{i=1}^n a_i x_i \in\ E and x\inE so span(W) \inE 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?

Thanks in advance.

 

[LCKurtz]

Homework Statement

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

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

 

[Emspak]

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

 

[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?

 

[LCKurtz]

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?

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.

 

[Emspak]

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

 

[HallsofIvy]

Homework Statement

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

As LCKurtz told you this is not true!

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\inW

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

(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.

First we show the bit about zero. is there a vector 0' \in W such that x + 0' = x for all x\in W.

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

Since x + 0 = x also then 0'=0 so 0\inW and condition (a) is valid.

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.

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

since x\inW
(-1)x\inW 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.

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.

We've shown that W is a subspace.

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

what is Span(W)?
Span(W) is the set of all linear combinations of W.
Span(W) = {λ₁x₁ + λ₂x₂ + ... + λ_nx_n | λ_i \in K} where K is the field.

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

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

let x,y\in span(W)

that means
x = (α₁x₁ + α₂x₂ + ... + α_nx_n) for α_i\in K, x\in W
y = (β₁x₁ + β₂x₂ + ... + β_nx_n) for β_i\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)

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

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

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

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

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 \leqn x_i\inE bit this implies that α_ix_i\inE
that would mean
\mathbf x\mathbf = \sum_{i=1}^n a_i x_i \in\ E and x\inE so span(W) \inE 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?

Thanks in advance.

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.

 

[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?

Someone help me please.

 

[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)x_n

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?

 

[HallsofIvy]

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.

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.

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

I'm not sure what you mean by this.

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?

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

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?

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

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)x_n

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

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.

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.

Is this any better?

 

[Emspak]

To be more clear about the step adding X and y:

Let x, y ∈ span(W).

x = (α₁x₁+ α₂x₂ + … +α_nx_n)
y = (β₁y₁+ β₂y₂ + … +β_ny_n

This can also be written as:

x + y = (α₁+β₁)x₁+(α₂+β₂)x₂ + … +(α_n+β_n)x_n)

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