Suppose \(T \in \mathcal{L}(V)\), and \(U \subset V\), an invariant subspace under \(T\). Then:
\begin{equation} (T / U)(v+U) = Tv+U, \forall v \in V \end{equation}
where \(T / U \in \mathcal{L}(V / U)\)
“if you can operator on \(V\), you can operator on \(V / U\) in the same way.” Yes I just verbed operator.
quotient operator is well-defined
Why is this not possible for any subspace of \(V\)? This is because we need \(T\) to preserve the exact structure of the subspace we are quotienting out by; otherwise our affine subset maybe squished to various unexpected places. The technical way to show that this is well-defined leverages the property of two affine subsets being equal:
Suppose \(v +U = w+U\), we desire that \(T / U (v+U) = T / U (w+U)\). That is, we desire that \(Tv +U = Tw +U\).
If \(v+U = w+U\) , then, \(v-w \in U\). Now, this means that \(T(v-w) \in U\) only because \(U\) is invariant under \(T\) (otherwise it could be sent to anywhere in \(V\) as \(T \in \mathcal{L}(V)\) not \(\mathcal{L}(U)\)). Therefore, \(Tv-Tw \in U\), and so \(Tv +U = Tw+U\), as desired. \(\blacksquare\)