# How to prove that a vector space homomorphism $\phi$ is injective if and only if $\ker (\phi)=\{0\}$?

A vector space homomorphism $$\phi: V\rightarrow W$$ is a map between two vector spaces which satisfies the rule

$$\phi(\lambda v + \mu v')=\lambda\phi(v)+\mu\phi(v').$$

Injectivity means

$$\forall {x,x'\in V}\phi(x)=\phi(x')\Rightarrow x=x'.$$

I have already proved the statement for group homomorphisms.

And $$\phi$$ is a group homomorphism under addition, i.e.,

$$\phi(v+w)=\phi(v)+\phi(w).$$

How can I make the transition to vector spaces; I hope it is clear what I am asking for.

I am confused because there are two operations under $$\phi$$: the scalar multiplication and the addition.

• The definition of the kernel is the same in both cases, though. – Randall Jan 30 '19 at 18:58
• A vector space is an abelian group (plus an additional structure)., so already proved. – Bernard Jan 30 '19 at 19:45

The proof is the same. A vector space is a group wrt the internal operation (i.e. $$(V,+)$$ is a group) and a vector space homomorphism is, in particular, a group homomorphism.
If $$\phi$$ is injective, then $$\phi(x)=0=\phi(0)$$ implies that $$x=0$$, so the kernel is $$\{0\}$$. Conversely, suppose the kernel is $$\{0\}$$ and that $$\phi(x)=\phi(y)$$. Then $$\phi(x)-\phi(y)=0$$. Using homomorphism property, this means $$\phi(x-y)=0$$ so that $$x-y$$ is in the kernel, which only consists of $$0$$. Thus $$x-y=0$$, or $$x=y$$, proving injectivity.
As you observed, if you take in your definition of $$\phi$$, $$\lambda=\mu=1$$, it follows that, in particular: $$\phi(x+y)=\phi(x)+\phi(y),\ \forall x,y\in V.$$ Therefore $$\phi:V\to W$$ is, in particular, a morphism between the groups $$V$$ and $$W$$. Applying your result for groups, it follows that $$\ker(\phi)$$ (seen as the kernel of the group morphism is $$0$$). The external operation does not intervene in this statement.