Prove: r\vec{0} = \vec{0} in Vector Space Over F

[iamalexalright]

Homework Statement

V is a vector space over F.
Prove $$r\vec{0} = \vec{0}$$
for all r in F and v in V.

Homework Equations

Properties of a vector space:
1. associativity
2. commutativity
3. zero vector
4. additive inverse
5. scalar multiplication:
1u = u (and a few others)I worked this similar to another proof:
$$r\vec{0} = r(\vec{0} + \vec{0})$$

$$r\vec{0} = r\vec{0} + r\vec{0}$$

$$r\vec{0} + (-r\vec{0}) = (r\vec{0} + r\vec{0}) + (-r\vec{0})$$

$$r\vec{0} + (-r\vec{0}) = r\vec{0} + (r\vec{0} + (-r\vec{0}))$$

$$\vec{0} = r\vec{0} + \vec{0}$$

$$\vec{0} = r\vec{0}$$

There may be other issues but my main one is that I assume:
$$r\vec{0} + (-r\vec{0}) = \vec{0}$$
where in our definitions we only give(ie, not multiplied by a scalar):
$$\vec{v} + (-\vec{v}) = \vec{0}$$

Now if I try to prove that rv + -rv = 0 I have to use the proof of the original problem !

Am I going about this the wrong way?

 

[Karlx]

Hi iamalexalright.

What is the definition of the zero vector ?

 

[iamalexalright]

It's the identity element: leaves a vector unchanged when you have vector addition:
v + 0 = 0 + v = 0

 

[Karlx]

That's right.
The zero vector is the identity element for the sum.
So, it is unique and satisfies the condition v + 0 = 0 + v = v, for all v $$\in$$V.

This definition plus scalar multiplication plus associativity is all you need to proof that $$r\vec{0}$$=$$\vec{0}$$

 

[iamalexalright]

I don't see it : /

If I start with the condition:
v + 0 = 0 + v = v

multiply by scalar r and start to simplify I don't get to where I need to be : /

Any more hints?

 

[Karlx]

Apply the definition of vector zero to the vector $$v=r\vec{a}$$

 

[iamalexalright]

Could this work?
$$\vec{a} = \vec{a} + \vec{0}$$
multiply by scalar r
$$r\vec{a} = r\vec{a} + r\vec{0}$$
Since ra is in V, let v = ra
$$\vec{v} = \vec{v} + r\vec{0}$$
add the additive inverse of v
$$\vec{v} + (-\vec{v}) = (\vec{v} + r\vec{0}) + (-\vec{v})$$
associativity
$$\vec{v} + (-\vec{v}) = (\vec{v} + (-\vec{v}) + r\vec{0}$$
additive inverse
$$\vec{0} = \vec{0} + r\vec{0}$$
identity element for sum
$$\vec{0} = r\vec{0}$$

 

[Karlx]

You got it !
But no need to go beyond the second expression you wrote.

1: $$r\vec{a}=r(\vec{a}+\vec{0})=r\vec{a}+r\vec{0}$$
2: $$r\vec{a}=r(\vec{0}+\vec{a})=r\vec{0}+r\vec{a}$$

From 1) and 2) you see that $$r\vec{0}$$ is such a vector that leaves unchanged the vector $$r\vec{a}$$, which is an arbitrary vector of the vector space V.

But there is only one vector that satisfies this condition: the $$\vec{0}$$ vector. So $$r\vec{0}=\vec{0}$$, as we wanted to prove.

 

[iamalexalright]

Ah, I see what you were getting at now. I forgot about the lefthand/righthand side (I know there is a better name for this...).

Thanks for the help Karlx!

 

[capandbells]

Ah, I see what you were getting at now. I forgot about the lefthand/righthand side (I know there is a better name for this...).

Thanks for the help Karlx!

It's usually called the cancellation rule, and it's one of the first things most linear algebra books prove (for obvious reasons...)