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title: "Axler 3.A"
source: https://www.jemoka.com/posts/kbhaxler_3_a/
---

OMGOMGOMG its Linear Maps time! &ldquo;One of the key definitions in linear algebra.&rdquo;
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&rsquo;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&rsquo;t usually make sense, we call the &ldquo;composition&rdquo; operation on Linear Maps their &ldquo;product&rdquo; and show that this product is associative, distributive, and has an identity New Definitions Linear Map &mdash; additivity (adding &ldquo;distributes&rdquo;) and homogeneity (scalar multiplication &ldquo;factors&rdquo;) \(\mathcal{L}(V,W)\) 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?
