Proving the Vector Space Property: cv = 0, v ≠ 0 → c = 0

[bjgawp]

I'm considering the problem: Given $$c \in \bold{F}$$, $$v \in V$$ where F is a field and V a vector space, show that $$cv = 0, v \neq 0 \ \Rightarrow \ c = 0$$

I've been wrapping my head around this one for a while now but I can't seem to get it. Proving that if cv = 0 and v $$\neq$$ 0 implies v = 0 is easy since we can simply multiply by $$c^{-1}$$ but in vector space, we don't have that kind of inverse for vectors seeing how we only have scalar multiplication.

 

[VeeEight]

But can you not simply multiply by the inverse of c to get a contradiction - that is v=0?

 

[HallsofIvy]

I'm considering the problem: Given $$c \in \bold{F}$$, $$v \in V$$ where F is a field and V a vector space, show that $$cv = 0, v \neq 0 \ \Rightarrow \ c = 0$$

I've been wrapping my head around this one for a while now but I can't seem to get it. Proving that if cv = 0 and v $$\neq$$ 0 implies v = 0 is easy since we can simply multiply by $$c^{-1}$$ but in vector space, we don't have that kind of inverse for vectors seeing how we only have scalar multiplication.

Since c is NOT a vector that doesn't matter!

 

[aPhilosopher]

When stuck, think about the components. It will get you more stuck or get you an easy proof.

 

[aPhilosopher]

A "fancy" way to do it if you don't like the coordinate way is to take the alternate definition of a vector space as an embedding of a field into the endomorphism ring of an abelian group. The map for the embedding is just:

$$\theta: F \rightarrow End(A), c \rightarrow cI$$

Where I is the identity matrix. Then if $$v \in A$$ what we normally write as $$cv$$ is just $$\theta(c)(v)$$ and you can use that to read off all of the vector space properties.

Then $$\theta(c)(v) = 0$$ means that $$v \in Ker(\theta(c))$$ and that should give you the result that you're looking for considering that $$\theta(c)$$ is invertible by virtue of being in a subfield of End(A).

EDIT: I should note that you can't use the map that I gave you in the definition because we're conflating the Mat_n(A) with End(A) but they are isomorphic. I gave it to you so that you would know what the map is. In reality, you would prove that the map induces a ring homomorphism as advertised. This is just a simple exercise. To take it as the definition of a vector space, you assume that the embedding is given.

 

[Esd]

Suppose cv=0, v =/= 0

then take a new scalar from F (call it a)

a*(cv)=a*0
(ac)(v)=0
ca(v)=0
c(av)=0

Then c * (the whole vector space spanned by v) = 0
Since this is a vector space, c must be 0 (I forget the name of this property).

 

[emyt]

the vector spaces are over fields, the components of a vector are elements of the field

 

[Landau]

I guess Vee's method is the simplest:

Assume, by contradiction, that $$cv=0$$ and $$v\neq 0$$, but $$c\neq 0$$. Then the inverse $$c^{-1}$$ exists, and multiplying $$cv=0$$ to the left with it we get $$c^{-1}cv=c^{-1}0$$, i.e. $$v=0$$, in contradiction with $$v\neq 0$$.