Is a0 Always Equal to the Zero Vector in Any Vector Space?

[1MileCrash]

Homework Statement

Prove that a0 = 0

Homework Equations

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

QEDCan I do that?

 

[LCKurtz]

Homework Statement

Prove that a0 = 0

Homework Equations

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?

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

 

[1MileCrash]

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

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?

 

[LCKurtz]

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

 

[1MileCrash]

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

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)?

 

[LCKurtz]

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)?

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.

 

[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.

 

[STEMucator]

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.

 

[1MileCrash]

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.

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

 

[LCKurtz]

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.

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.

 

[STEMucator]

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.

 

[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?

 

[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.

 

[LCKurtz]

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?

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.

 

[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.

 

[1MileCrash]

EDIT: nevermind, got it backwards.

 

[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.

 

[1MileCrash]

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

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?