We first compute the dimension of
\(W\text{.}\) By
Definition 7.5.4, the subspace
\(W\) is spanned by
\begin{equation*}
e,\;(T-\lambda\unit_V)(e),\;(T-\lambda\unit_V)^2(e),\ldots,(T-\lambda\unit_V)^{m-1}(e).
\end{equation*}
Hence, \(\dim_FW\leq m\text{.}\) We show that \(\left\{(T-\lambda\unit_V)^i(e)\right\}_{i=0}^{m-1}\) is linearly independent. Suppose that
\begin{equation*}
a_0e+a_1(T-\lambda\unit_V)(e)+\cdots+a_{m-1}(T-\lambda\unit_V)^{m-1}(e)=0.
\end{equation*}
Since \((T-\lambda\unit_V)^m(e)=0\text{,}\) we get that \((T-\lambda\unit_V)^r(e)=0\) for any \(r\geq m\text{.}\) Hence, the linear map \((T-\lambda\unit_V)^{m-1}\) evaluated at the above expression gives
\begin{equation*}
a_0(T-\lambda\unit_V)^{m-1}(e)=0.
\end{equation*}
As \((T-\lambda\unit_V)^{m-1}(e)\neq 0\text{,}\) we have \(a_0=0\text{.}\) Therefore we are left with
\begin{equation*}
a_1(T-\lambda\unit_V)(e)+\cdots+a_{m-1}(T-\lambda\unit_V)^{m-1}(e)=0.
\end{equation*}
Now we apply \((T-\lambda\unit_V)^{m-2}\) to the above expression, and argueing as above, we obtain \(a_1=0\text{.}\) Continuing in this way we get that \(a_i=0\) for all \(i\text{.}\) This shows that \(\left\{(T-\lambda\unit_V)^i\right\}_{i=0}^{m-1}\) is a maximal linearly independent subset of \(W\text{,}\) i.e., it is a basis of \(W\text{.}\) Thus \(\dim_FW=m\text{.}\)
We now show that \(W\) is invariant under \(T\text{.}\) We rename the basis vectors of \(W\) obtained above as follows.
\begin{equation*}
w_1=e,\;w_2=(T-\lambda\unit_V)(e),\;w_3=(T-\lambda\unit_V)^2(e),\;\ldots,\;w_m=(T-\lambda\unit_V)^{m-1}(e)
\end{equation*}
Therefore we have
\begin{equation*}
(T-\lambda\unit_V)(w_i)=w_{i+1}\quad\text{for}\;1\leq i\leq m-1\quad\text{and}\quad (T-\lambda\unit_V)(w_m)=0.
\end{equation*}
In other words we have
\begin{equation*}
T(w_i)=\lambda w_i+w_{i+1}\in W\quad\text{for}\;1\leq i\leq m-1\quad\text{and}\quad T(w_m)=\lambda w_m\in W.
\end{equation*}
This shows that \(W\) is invariant under \(T\text{.}\)
