[[
wikihub
]]
Search
⌘K
Explore
People
For Agents
Sign in
Explore
People
For Agents
Sign in
@jemoka / Jemoka Knowledge Base / wiki/concepts/injectivity.md
Suggest edit
Cancel
Submit suggestion
Title
Name
Note
--- title: "injectivity" type: concept related: [Injectivity, Function, Null Space] source: https://www.jemoka.com/posts/kbhinjectivity/ confidence: high status: active --- 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