Proper Subspace Can be Multiplied to Be Within Other Subspace Lemma

Lemma

1. With $V$ as a finite dimensional Vector Space
2. With $T\in \mathcal{L}(V)$ such that the minimal polynomial for $T$ is a product of linear factors:
   1. $P(x)=(x-c_1{)}^{d_1}\dots (x-c_k{)}^{d_k}$
3. With $W$ as a Proper Subspace of $V$ invariant under $T$
4. Then, there exists a vector $\alpha \in V$ such that:
   1. $\alpha \in W$
   2. $(T-qI)\alpha \in W$ for some $c_i\in \{c_1,\dots ,c_k\}$

Intuition

There is a vector in $W$, that allows it to be multiplied by a $\lambda$ so that it is within the other subspace

Proof

1. Let $g$ be the $T$-Conductor of $S_T(\beta ,W)$
2. As $g$ divides the minimal polynomial of $T$, there exists $T_1,\dots ,T_k$ where $t_i\in \{d_1,\dots ,d_i\},\forall i$ such that: $g=(x-c_1{)}^{r_1}\dots (x-c_k{)}^{r_k}$ With at least one $r_i\ne 0$
3. We redefine $g={\Pi }_{i=1}^n(x-{\lambda }_i)$ where $n={\sum }_{i=1}^k{r}_i$, and consider $\beta$
4. Then, consider the sequence $\{g_1,\dots ,g_n\}$ where $g_j={\Pi }_{i=1}^j(x-\lambda )$
5. As $\beta \notin W$, there must exist a $m\in \{1,\dots ,m\}$ such that:
   1. $g_{m-1}(T)\beta \notin W$
   2. $g_m(T)\beta \in W$
6. Letting $\alpha =g_{m-1}(T)v$ gives our result as $\alpha \notin W$, but $(T-{\lambda }_{m}I)\alpha \in W$