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?