A linear transformation is injective iff its kernel is trivial

up:: Linear Transformation

Let $T:U\to V$ be a linear transformation. Denote its Kernel by $\ker T$.

Injective $⟹$ Trivial kernel

Proof by contrapositive: Let $\ker T\ne \{0\}$. Then there is some other vector $v\in V$ such that

$$v\ne 0\mid T(v)=T(0)=0$$

Thus, $T$ cannot be Injective, since $v$ and $0$ have the same value.

Trivial kernel $⟹$ Injective

Let $\ker T=\{0\}$. Let $u,v\in V$ such that $T(u)=T(v)$. Then, since it is a linear transformation, we have that

$$T(u-v)=0⟹u-v\in \ker T⟹u-v=0⟹u=v$$

Thus, $T$ is injective.