# A0 = 0 (0 of a vector space)

1. Jul 17, 2013

### 1MileCrash

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

Prove that a0 = 0

2. Relevant equations

3. The attempt at a solution

Let V be a vector space on a field F. Let x be a member of V and a be a member of F.

Consider that the 0 vector is the unique vector such that

x + 0 = x

Now, apply a scalar multiplication by a to both sides of the equation. Because scalar multiplication is distributive in all vector spaces,

ax + a0 = ax

Thus, we see that a0 has the same property of the 0 vector in V. Since the 0 vector in V is also unique, it must be the case that

a0 = 0

QED

Can I do that?

2. Jul 17, 2013

### LCKurtz

I don't think you quite have it. Have you shown that for ALL y you have y + a0 = y?

3. Jul 17, 2013

### 1MileCrash

But if

x + 0 = x

is true for all x (which it is by definition of the zero vector) then

ax + a0 = ax

is also true for all x, and it is true for all a (because scalar multiplication is distributive in all vector spaces), so then it is always true, so it is true for any ax (or y).

Right?

Last edited: Jul 17, 2013
4. Jul 17, 2013

### LCKurtz

It may be trivial, but you haven't shown that any y can be written in the form ax.

5. Jul 17, 2013

### 1MileCrash

Oh, ok.

So, I need to justify that ax is also an element of V? Am I understanding that correctly?

If so, then can I just say that in all vector spaces, for each element a in F and each element x in V, there is a unique element ax in V, thus, let ax = y, which is a known element of V (and then the result follows)?

Last edited: Jul 17, 2013
6. Jul 17, 2013

### LCKurtz

You have shown that a0 works as an additive identity on all elements of the form ax. If every y in V can be written that way you have it as an additive identity on all of V, which is what you want. Think about $y = a(\frac 1 a y)$.

As an alternative approach to the problem think about a(0+0) in the first place.

7. Jul 17, 2013

### 1MileCrash

Ok, instead of saying "the result follows" I'll just write it out.

"for each element a in F and each element x in V, there is a unique element ax in V, thus, let ax = y, which is a known element of V (and then the result follows)?"

ax = y
=>
x = (1/a)y

(1/a) is the multiplicative inverse of a and is therefore a member of F. y is already known as a member of V. Therefore any element x of V can be expressed as a scalar multiplication of a member of the field and another member of V.

That's what I was getting at, would you say that is adequate?

Thanks again.

8. Jul 17, 2013

### Zondrina

Using LC's hint :

a0 = a(0+0) = a0 +a0

So you have :

a0 = a0 + a0

Hmm what property of vector spaces would help you here now? There aren't too many to consider so I'm sure you will see it any moment.

9. Jul 17, 2013

### 1MileCrash

I'd rather just complete the attempt I've done myself, this is the proof my book gives.

10. Jul 17, 2013

### LCKurtz

No, I don't think so. At the very least, it is confusing. Summarizing your argument so far you have shown that for any $a\in F$ and ${\bf x} \in V$ that $a{\bf x} + a{\bf 0} = a{\bf x}$. What you need to show is that for any ${\bf y}\in V$ you have ${\bf y} + a{\bf 0} = {\bf y}$. You have to start with $\bf y$. You can't start with $a{\bf x}$ and call it $\bf y$. Look at my suggestion in post #6 again.

11. Jul 17, 2013

### Zondrina

Well you know that $a \in F$ and I would recommend writing $0$ as $0_V$ as to be more explicit.

So really you want to show $a0_V = 0_V$.

$F$ is a field of scalars. Since $V$ is a vector space, it is closed under addition and scalar multiplication. You know that $\forall v \in V$ and $\forall a \in F$ that $av \in V$ because $V$ is closed.

Now, we've shown there are basically infinitely many vectors of the form $av$ inside of $V$. So if you want to proceed your way from before, add $av$ to both sides :

$a0_V + av = 0_V + av$

Now because $0_V$ is the unique vector such that $0_V + v = v, \forall v \in V$ ( Prove this if you havent! ) we can proceed :

$a0_V + av = av$

I'll let you finish that one if you want to. Your argument in your last post wouldn't work though. All you did was set y=ax and say y was in V. LC stated if you can show every $y \in V$ can be written as some $ax$, then you know that $a0$ will work perfectly as an identity.

12. Jul 17, 2013

### 1MileCrash

Why can't I call ax, y? It's just an element of V.

Looking at y=a((1/a)y) makes me fall back to the same reasoning. If y is an element of V then so is (1/a)y, so a((1/a)y) is an expression for y that is a scalar multiplication of an element of F and an element of V. But that's exactly what I did before. What is wrong with this reasoning?

13. Jul 18, 2013

### verty

1MileCrash: I think you can't say for each $a$ because you are given $a$. Given $a$, you must show that a$0_v = 0_v$. I can see a much quicker proof using components.

-edit- Hmm, I suppose this component idea is limited (without choice) to finite-dimensional vector spaces, better to prove this in general.

Last edited: Jul 18, 2013
14. Jul 18, 2013

### LCKurtz

Because you have to start with an arbitrary ${\bf y}\in V$. You have all the pieces but your argument should be written like this:

Suppose ${\bf y}\in V$ and $a \in F$.

Let ${\bf x} =\frac 1 a{\bf y}$

${\bf x} + {\bf 0} = {\bf x}$

$a{\bf x} + a{\bf 0} = a{\bf x}$

$a\frac 1 a{\bf y} +a{\bf 0} = a\frac 1 a{\bf y}$

${\bf y} +a{\bf 0} ={\bf y}$

Therefore $a{\bf 0}$ is the unique additive identity.

15. Jul 18, 2013

### 1MileCrash

Oh, I see, that comes out very nicely. I just thought that since x was an arbitrary vector, then so is ax or y. I don't see a difference between letting ax=y and letting x=(1/a)y.

Thanks again.

16. Jul 18, 2013

### 1MileCrash

EDIT: nevermind, got it backwards.

17. Jul 19, 2013

### verty

There is still a mistake here. Given a, we aren't told that it isn't 0.

Gotta keep your apples in order. If you start with a given $a \in F$, you must be faithful to that.

18. Jul 19, 2013

### 1MileCrash

I have no idea what you are saying. Could you explain how I have not done that? Or what that means?

All I did was define a to be an element of F so that I could use it in an expression without the reader not knowing what it is. What do you mean by "starting" with it?