Prove that 0v = 0 for an arbitaryvector v belong to V

[zeion]

Homework Statement

Suppose V is a vector space over F.
Prove that 0v(vector) = 0(vector) for an arbitrary vector v belong to V

Homework Equations

Vector additive identity

The Attempt at a Solution

Let v = (v1,...,vn) belong to V, then 0v = 0(v1,...,vn) = (0,...,0).
Now (v1, ... ,v2) + (0,...,0) = (v1,...,vn), therefore (0,..,0) is the zero vector in V.

Is this ok?

 

[Pinu7]

There are probably many ways to do this. However, you are not close. A vector space does not necessarily have to be represented by euclidean vectors of the form (a,b). A vector space could even be the set of all differentiable functions. Anyway you have to use the full and general definition of a vector space: http://en.wikipedia.org/wiki/Vector_spaceUse these facts:

1+(-1)=0 (1 being the identity)
v+(-v)=0

Also, your last step should be:
(1-1)v=0
0v=0

Good luck!

 

[zeion]

Okay so you're saying that my representation of V is not general and only apply to euclidean vectors right?

Prove that 0v = 0(vector)

Given
1+(-1)=0,
then 1+(-1)v=0(vector)
(1-1)v = 0(vector)
0v = 0(vector)

 

[HallsofIvy]

Okay so you're saying that my representation of V is not general and only apply to euclidean vectors right?

Prove that 0v = 0(vector)

Given
1+(-1)=0,
then 1+(-1)v=0(vector)
(1-1)v = 0(vector)
0v = 0(vector)

And why is that last line true? Again, you are assuming what you want to prove!

What you need is, rather, 0v= (1+ (-1))v= 1v+ (-1)v= what?

 

[Pinu7]

And why is that last line true? Again, you are assuming what you want to prove!

Oooops! For some reason, I was proving something completely different, completely forgeting the OP.((-1)v=-v, I think)

The last line is SUPPOSED to be (0+0)v=0v in which the theorem follows immediately.

 

[D H]

Given
1+(-1)=0,
then 1+(-1)v=0(vector)
(1-1)v = 0(vector)
0v = 0(vector)

How does the second line follow from the first?