[[
wikihub
]]
Search
⌘K
Explore
People
For Agents
Sign in
Explore
People
For Agents
Sign in
×
@jemoka / Jemoka Knowledge Base / raw/textbook/axler/kbhaxler_3_a.md
Suggest edit
Cancel
Submit suggestion
Title
Name
Note
--- title: "Axler 3.A" source: https://www.jemoka.com/posts/kbhaxler_3_a/ --- OMGOMGOMG its Linear Maps time! “One of the key definitions in linear algebra.” Key Sequence We define these new-fangled functions called Linear Maps, which obey \(T(u+v) = Tu+Tv\) and \(T(\lambda v) = \lambda Tv\) We show that the set of all linear maps between two vector spaces \(V,W\) is denoted \(\mathcal{L}(V,W)\); and, in fact, by defining addition and scalar multiplication of Linear Maps in the way you’d expect, \(\mathcal{L}(V,W)\) is a vector space! this also means that we can use effectively the \(0v=0\) proof to show that linear maps take \(0\) to \(0\) we show that Linear Maps can be defined uniquely by where it takes the basis of a vector space; in fact, there exists a Linear Map to take the basis anywhere you want to go! though this doesn’t usually make sense, we call the “composition” operation on Linear Maps their “product” and show that this product is associative, distributive, and has an identity New Definitions Linear Map — additivity (adding “distributes”) and homogeneity (scalar multiplication “factors”) \(\mathcal{L}(V,W)\) any polynomial map from Fn to Fm is a linear map addition and scalar multiplication on \(\mathcal{L}(V,W)\); and, as a bonus, \(\mathcal{L}(V,W)\) a vector space! naturally (almost by the same \(0v=0\) proof), linear maps take \(0\) to \(0\) Product of Linear Maps is just composition. These operations are: associative distributive has an identity Results and Their Proofs technically a result: any polynomial map from Fn to Fm is a linear map basis of domain of linear maps uniquely determines them Questions for Jana why does the second part of the basis of domain proof make it unique?