> [!NOTE] Lemma > Let $V,W$ be [[Vector spaces|vector spaces]] over a [[Field (Algebra)|field]] $\mathbb{F}$ with zeros $0_{V}$ and $0_{W}$ respectively. Let $\varphi:V\to W$ be a [[Linear maps|linear map]]. Then $\varphi$ is [[Injection|injective]] if and only if its [[Kernel of Linear Map|kernel]] is given by $\ker{\varphi} = \{ 0_{V} \}.$ **Proof**: Suppose $\varphi$ is injective. Let $v\in \ker\varphi.$ Then by [[Linear Map Fixes Zero]], $\varphi(v)=0_{W}=\varphi(0_{V})$which gives $v=0_{V}$ by injectivity of $\varphi.$ Thus $\ker \varphi =\{ 0_{V} \}.$ Conversely, suppose $\ker \varphi=\{ 0_{V} \}.$ Let $v,w\in V$ such that $\varphi(v)=\varphi(w).$ Then $\varphi(v)-\varphi(w)=0_{W}$ which by linearity of $\varphi$ and [[Scalar Multiplication by -1 gives Additive Inverse in Real Vector Space]], gives $\varphi(v+(-1)w)= \varphi(v-w)=0_{W}$Thus $v-w\in \ker \varphi$ which implies $v-w=0_{V},$ therefore, $v=w.$ That is, $\varphi$ is injective.