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

Injective $⟹$ Trivial kernel

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

v \neq = 0 ∣ 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.