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

In summary, the zero vector is the identity element for the sum, and that's all you need to proof that r\vec{0}=\vec{0}
  • #1
iamalexalright
164
0

Homework Statement


V is a vector space over F.
Prove [tex]r\vec{0} = \vec{0}[/tex]
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:
[tex]r\vec{0} = r(\vec{0} + \vec{0})[/tex]

[tex]r\vec{0} = r\vec{0} + r\vec{0}[/tex]

[tex]r\vec{0} + (-r\vec{0}) = (r\vec{0} + r\vec{0}) + (-r\vec{0})[/tex]

[tex]r\vec{0} + (-r\vec{0}) = r\vec{0} + (r\vec{0} + (-r\vec{0}))[/tex]

[tex]\vec{0} = r\vec{0} + \vec{0}[/tex]

[tex]\vec{0} = r\vec{0}[/tex]

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

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?
 
Physics news on Phys.org
  • #2
Hi iamalexalright.

What is the definition of the zero vector ?
 
  • #3
It's the identity element: leaves a vector unchanged when you have vector addition:
v + 0 = 0 + v = 0
 
  • #4
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 [tex]\in[/tex]V.

This definition plus scalar multiplication plus associativity is all you need to proof that [tex]r\vec{0}[/tex]=[tex]\vec{0}[/tex]
 
  • #5
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?
 
  • #6
Apply the definition of vector zero to the vector [tex]v=r\vec{a}[/tex]
 
  • #7
Could this work?
[tex]\vec{a} = \vec{a} + \vec{0}[/tex]
multiply by scalar r
[tex]r\vec{a} = r\vec{a} + r\vec{0}[/tex]
Since ra is in V, let v = ra
[tex]\vec{v} = \vec{v} + r\vec{0}[/tex]
add the additive inverse of v
[tex]\vec{v} + (-\vec{v}) = (\vec{v} + r\vec{0}) + (-\vec{v})[/tex]
associativity
[tex]\vec{v} + (-\vec{v}) = (\vec{v} + (-\vec{v}) + r\vec{0}[/tex]
additive inverse
[tex]\vec{0} = \vec{0} + r\vec{0}[/tex]
identity element for sum
[tex]\vec{0} = r\vec{0}[/tex]
 
  • #8
You got it !
But no need to go beyond the second expression you wrote.

1: [tex]r\vec{a}=r(\vec{a}+\vec{0})=r\vec{a}+r\vec{0}[/tex]
2: [tex]r\vec{a}=r(\vec{0}+\vec{a})=r\vec{0}+r\vec{a}[/tex]

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

But there is only one vector that satisfies this condition: the [tex]\vec{0}[/tex] vector. So [tex]r\vec{0}=\vec{0}[/tex], as we wanted to prove.
 
  • #9
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!
 
  • #10
iamalexalright said:
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...)
 

What does "r\vec{0} = \vec{0} in Vector Space Over F" mean?

This means that when a scalar (r) multiplies a zero vector (r\vec{0}), the resulting vector is also a zero vector ( \vec{0} ). This is true for any vector space over the field (F).

Why is it important to prove r\vec{0} = \vec{0} in Vector Space Over F?

Proving this statement is important because it is a fundamental property of vector spaces and is used in many mathematical proofs and applications. It also helps to understand the properties of vector spaces and how they behave under scalar multiplication.

How is the statement r\vec{0} = \vec{0} in Vector Space Over F proven?

The statement can be proven by using the properties of vector spaces and scalar multiplication. It can also be proven by using the definition of the zero vector and the properties of scalar multiplication, such as the zero scalar property and the distributive property.

Can this statement be generalized to other vector spaces?

Yes, this statement can be generalized to any vector space over any field. This is because the properties of vector spaces and scalar multiplication hold true for any vector space over any field.

What are some real-world applications of this statement?

This statement is used in various fields such as physics, engineering, and computer science. In physics, it is used to describe the behavior of forces acting on objects. In engineering, it is used to solve problems involving forces and moments. In computer science, it is used in linear algebra and graphics programming.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
977
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
293
Replies
17
Views
509
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
Back
Top