Simple Vector Proof

  • #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?
 

Answers and Replies

  • #2
Karlx
75
0
Hi iamalexalright.

What is the definition of the zero vector ?
 
  • #3
iamalexalright
164
0
It's the identity element: leaves a vector unchanged when you have vector addition:
v + 0 = 0 + v = 0
 
  • #4
Karlx
75
0
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
iamalexalright
164
0
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
Karlx
75
0
Apply the definition of vector zero to the vector [tex]v=r\vec{a}[/tex]
 
  • #7
iamalexalright
164
0
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
Karlx
75
0
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
iamalexalright
164
0
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
capandbells
96
0
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...)
 

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