08 - First Isomorphism Theorem

First Isomorphism Theorem - V2

Let $T : V \to W$ be a linear map of vector spaces over $F$ then
$\overline{T} : V / Ker (T) v + Ker (T) \to Im (T) \mapsto T (v)$
is an isomorphism of vector spaces

Proof - Short

It follows from the first isomorphism theorem for groups that $T$ is an isomorphism of (abelian) groups.
As $T$ is also compatible with scalar multiplication then
$T is a linear isomorphism$

Proof - Detailed

Well-Defined

$v + Ker (T) = v^{'} + Ker (T) ⟹ v = v^{'} + u for some u \in Ker (T) ⟹ \overline{T} (v + Ker (T)) := T (v) = T (v^{'} + u) = T (v^{'}) =: \overline{T} (v^{'} + Ker (T))$
Linear
$\overline{T} (a (v + Ker (T)) + (v^{'} + Ker (T))) = \overline{T} (a v + v^{'} + Ker (T)) := T (a v + v^{'}) = a T (v) + T (v^{'}) =: a \overline{T} (v + Ker (T)) + \overline{T} (v^{'} + Ker (T))$
Surjective
$w \in Im T ⟹ \exists v \in V : T (v) = w ⟹ \overline{T} (v + Ker (T)) = T (v) = w ⟹ w \in Im \overline{T}$
Injective
$v + Ker (T) \in Ker (\overline{T}) ⟹ \overline{T} (v + Ker (T)) = T (v) = 0 ⟹ v \in Ker (T) ⟹ v + Ker (T) = 0 + Ker (T)$

Oxford Maths Notes

Explorer

08 - First Isomorphism Theorem - V2

Graph View

Backlinks