依题意知，子空间\(0,U_i=\langle \xi_i,\xi_{i+1},\cdots ,\xi_n\rangle\)是\(\varphi\)-不变子空间。下证\(V\)的\(\varphi\)-不变子空间有且只有以上这些。设\(U\neq 0\)是\(V\)的\(\varphi\)-不变子空间，记
\begin{equation*}
i_0=\min\{i\ |\ a_i\neq 0\mbox{且}a_i \xi_i+a_{i+1} \xi_{i+1}+\cdots +a_n \xi_n\in U\},
\end{equation*}
则\(U\subseteq\langle \xi_{i_0},\xi_{i_0+1},\cdots ,\xi_n\rangle\)。我们断言，\(U=\langle \xi_{i_0},\xi_{i_0+1},\cdots ,\xi_n\rangle\)。事实上，由\(i_0\)的定义知：存在\(a_{i_0},a_{i_0+1},\cdots ,a_n\in\mathbb{F}\)使得\(a_{i_0}\xi_{i_0}+a_{i_0+1}\xi_{i_0+1}+\cdots +a_n \xi_n\in U\)。由\((2)\)知\(\xi_n\in U\)，所以
\begin{equation*}
(a_{i_0}\xi_{i_0}+a_{i_0+1}\xi_{i_0+1}+\cdots +a_n \xi_n)-a_n \xi_n\in U.
\end{equation*}
记\(\beta_1\triangleq a_{i_0}\xi_{i_0}+a_{i_0+1}\xi_{i_0+1}+\cdots +a_{n-1} \xi_{n-1}\)，则\(\beta_1\in U\)。于是，
\begin{equation*}
\varphi(\beta_1)-a \beta_1-a_{n-1}\xi_n\in U,
\end{equation*}
即
\begin{equation*}
\gamma_1\triangleq a_{i_0}\xi_{i_0+1}+a_{i_0+1}\xi_{i_0+2}+\cdots +a_{n-2} \xi_{n-1}\in U,
\end{equation*}
再根据\(\varphi(\gamma_1)-a \gamma_1-a_{n-2}\xi_n\in U\)得
\begin{equation*}
\gamma_2\triangleq a_{i_0}\xi_{i_0+2}+a_{i_0+1}\xi_{i_0+3}+\cdots +a_{n-3} \xi_{n-1}\in U,
\end{equation*}
依此类推，有\(a_{i_0}\xi_{n-1}\in U\)。由\(a_{i_0}\neq 0\)可知\(\xi_{n-1}\in U\)。于是，
\begin{equation*}
\beta_2\triangleq\beta_1-a_{n-1}\xi_{n-1}= a_{i_0} \xi_{i_0}+a_{i_0+1}\xi_{i_0+1}+\cdots +a_{n-2}\xi_{n-2}\in U.
\end{equation*}
同理，我们有
\begin{equation*}
\varphi(\beta_2)-a\beta_2-a_{n-2}\xi_{n-1}\in U,
\end{equation*}
即
\begin{equation*}
\eta_1\triangleq a_{i_0}\xi_{i_0+1}+a_{i_0+1}\xi_{i_0+2}+\cdots +a_{n-3}\xi_{n-2}\in U.
\end{equation*}
再由\(\varphi(\eta_1)-a\eta_1-a_{n-3}\xi_{n-1}\in U\)，得
\begin{equation*}
\eta_2\triangleq a_{i_0}\xi_{i_0+2}+a_{i_0+1}\xi_{i_0+3}+\cdots +a_{n-4}\xi_{n-3}\in U.
\end{equation*}
依此类推，\(a_{i_0}\xi_{n-2}\in U\)。由\(a_{i_0}\neq 0\)可知\(\xi_{n-2}\in U\)。于是，
\begin{equation*}
\beta_3\triangleq\beta_2-a_{n-2}\xi_{n-2}= a_{i_0} \xi_{i_0}+a_{i_0+1}\xi_{i_0+1}+\cdots +a_{n-3}\xi_{n-3}\in U.
\end{equation*}
重复上述步骤，最后有\(a_{i_0}\xi_{i_0}\in U\)。由\(a_{i_0}\neq 0\)得\(\xi_{i_0}\in U\)。因此，
\begin{equation*}
U=\langle \xi_{i_0},\xi_{i_0+1},\cdots ,\xi_n\rangle .
\end{equation*}
