Injectivity Implies Null Space is 0 and vice versa

Injective $⟺$ Null space is $\{0\}$

Let $T$ in $\mathcal{L}(V,W)$. $T$ is injective if and only if $\text{null }T=\{0\}$.

Proof. Suppose $T$ is injective. We want to prove that $\text{null }T=\{0\}$. We already know that $\{0\}\subseteq \text{null }T$ (since linear maps take 0 to 0). To prove the inclusion in the other direction, suppose $v\in \text{null }T$. Then

$$T(v)=0=T(0)$$

Because $T$ is injective, the equation above implies that $v=0$. Thus we can conclude that $\text{null }T=\{0\}$, as desired.

To prove the implication in the other direction, now suppose $\text{null }T=\{0\}$. We want to prove that $T$ is injective. To do this, suppose $u,v\in V$ and $Tu=Tv$. Then

$$0=Tu-Tv=T(u-v)$$

Thus, $u-v$ is in $\text{null }T$, which equals $\{0\}$. This means $u-v=0$ or $u=v$. Hence, $T$ is injective, as desired.