An injective function is one which is one-to-one: that it maps distinct inputs to distinct outputs. constituents A function T: V \to W requirements T is injective if Tu = Tv implies u=v. additional information injectivity implies that null space is \{0\} Proof: let T \in \mathcal{L}(V,W); T is injective IFF null\ T = \{0\}. given injectivity Suppose T is injective. Now, we know that 0, because it indeed gets mapped by T to 0, is in the null space of T. Because linear maps take 0 to 0, T0=0. Now, because T is injective, for any v that Tv = 0 = T 0 implies v=0. So 0 is the only thing that an injective T can map to 0, and it is indeed in the null space, so the null space is just \{0\}. given null\ T=\{0\} Suppose we have some Tu = Tv, we desire to proof that u=v to show that T is injective. Given Tu=Tv, we have that Tu-Tv. Given additivity, T(u-v) = 0. This makes (u-v) \in\ null\ T. Given only 0 is in the null space of T, u-v = 0, so u=v, as desired. \blacksquare. map to smaller space is not injective See map to smaller space is not injective