Suggest edit — T twiddle

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title: "T twiddle"
type: concept
related: [Isomorphism, Injectivity, Parallel Linear Algebra, Null Space]
source: https://www.jemoka.com/posts/kbht_twiddle/
confidence: high
status: active
---

Suppose \(T \in \mathcal{L}(V,W)\). Define a \(\widetilde{T}: V / (null\ T) \to W\) such that:
\begin{align} \widetilde{T}(v+ null\ T) = Tv \end{align}
so \(\widetilde{T}\) is the map that recovers the mapped result from an affine subset from the null space of the map.
\(\widetilde{T}\) is well defined Same problem as that with operations on quotient space. We need to make sure that \(\widetilde{T}\) behave the same way on distinct but equivalent representations of the same affine subset.
Suppose \(u,v \in V\) such that \(u+null\ T = v+null\ T\). Because two affine subsets parallel to \(U\) are either equal or disjoint, we have that \(u-v \in null\ T\). This means that \(Tu-Tv = 0 \implies Tu= Tv\). So applying \(\widetilde{T}\) on equivalent representations of the same affine subset would yield the same result, as desired. \(\blacksquare\)
properties of \(\widetilde{T}\) it is a linear map TBD proof. Basically just like do it inheriting operations from the operations on quotient space.
it is injective We desire here that \(null\ \widetilde{T} = \{0\}\) which will tell us that \(\widetilde{T}\) is injective.
Suppose some \(v + null\ T\) is in the null space of \(\widetilde{T}\). So, we have that:
\begin{equation} \widetilde{T}(v+null\ T) = Tv = 0 \end{equation}
So, we have that \(v \in null\ T\). Now, this means that \(v-0 \in null\ T\). Because two affine subsets parallel to \(U\) are either equal or disjoint, \(v + null\ T = 0 + null\ T\) WLOG \(\forall v+null\ T \in null\ \widetilde{T}\). This means that \(null\ \widetilde{T}=\{0\}\), as desired.
its range is equal to the map&rsquo;s range \begin{equation} range\ \widetilde{T} = range\ T \end{equation}
by definition of everything.
\(V / null\ T\) is isomorphic to \(range\ T\) &hellip;.is this the point of this whole thing?
Shown by the two sub-results above, and that injectivity and surjectivity implies invertability.