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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.

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

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The proof should look very similar to the one for group homomorphisms.

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.

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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.

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