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1 Broader vector spaces
- Doesn't have to be physics vectors
- maybe it's like matrices
- or linear maps themselves
2 The Linear Map 0
A linear map \(S = 0\) is a map where \(Su = 0 \forall u\).
3 Axler 3.A ex7 (w/ Vienna + Mason)
Let \(w = Tv\).
3.1 If \(v = 0\) then
\[Tv = 0\] By Axler 3.11 (Maps take 0 to 0). Thus, \(\lambda\) can be anything in \(\mathbb F\).
3.2 Otherwise,
\(\frac{1}{v} \in \mathbb F\) because the field has multiplicative inverses for all elements except 0. \[ Tv = w = \left( w \frac{1}{v} \right)v \] Let \(\lambda = w \frac{1}{v}\), then \[ \lambda v = w \frac{1}{v} v = w \] which is in \(\mathbb F\) because \(w, \frac{1}{v} \in \mathbb F\) and fields are closed under multiplication.
4 Axler 3.A ex10 (w/ Vienna + Mason)
The additivity of a linear map \(T\) requires \(T(u+v) = Tu + Tv\). Because \(U \subset V, U \neq V\), there must be some element \(v \in V\) yet \(v \notin U\).
For some element \(u \in U\), \[Tu + Tv = Su + 0 = Su\] Yet \(u+v \notin U\) because if it were, then \((u+v)+(-v) = v\) would be in \(U\). Thus, \[T(u+v) = 0\]
Because for some \(u\) \(Su\neq 0\), additivity does not hold over \(T\) and thus the map is not linear.