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. \(\blacksquare\).