Suppose T \in \mathcal{L}(V), and U \subset V, an invariant subspace under T. Then:

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