Let $V$ be a finite-dimensional complex vector spaces , and let $\varphi:V\rightarrow V$ be a linear operator such that
$$\begin{matrix} \mathbf{a_{1}},& \mathbf{a_{2}}, & \mathbf{a_{3}} & & & &\text{ is a basis for } \ker\varphi ,\\ \mathbf{a_{1}},& \mathbf{a_{2}},& \mathbf{a_{3}};& \mathbf{b_{1}},& \mathbf{b_{2}} & &\text{ is a basis for }\ker{\varphi}^{2} ,\\ \mathbf{a_{1}}, & \mathbf{a_{2}}, & \mathbf{a_{3}};& \mathbf{b_{1}},& \mathbf{b_{2}}; &\mathbf{c_{1}} &\text{ is a basis for }\ker{\varphi}^{3} . \end{matrix}$$ Obviously, $\mathbb{span}\{\varphi(\mathbf{b_{1}}),\varphi(\mathbf{b_{2}})\}\subseteq \mathbb{span}\{\mathbf{a_{1}},\mathbf{a_{2}},\mathbf{a_{3}}\}.$ Can we proof that $\mathbb{span}\{\varphi(\mathbf{c_{1}})\}\subseteq \mathbb{span}\{\mathbf{b_{1}},\mathbf{b_{2}}\} ?$
$$\varphi(\mathbf{c_{1}})\in \ker\varphi^2=\ker\varphi \bigoplus \mathbb{span}\{\mathbf{b_{1}},\mathbf{b_{2}}\},$$ $$\varphi(\mathbf{c_{1}})+\ker\varphi\in \ker\varphi^2 / \ker \varphi \text{(a quotient space);}$$ $$\ker\varphi^2/ \ker \varphi =\mathbb{span}\{\mathbf{b_{1}}+\ker \varphi,\mathbf{b_{2}}+\ker \varphi\},$$ $$\mathbb{span}\{\mathbf{b_{1}},\mathbf{b_{2}}\}\cong \ker\varphi^2/ \ker \varphi \text{(the symbol} \cong \text{means } \textit{isomorphism}).$$
But how to proof $$\varphi(\mathbf{c_{1}})=k_1\mathbf{b_{1}}+k_2\mathbf{b_{2}}?$$
Let us consider the matrix $$A=\begin{pmatrix} 1& 0& -1& 1& 0\\ -4& 1& -3& 2& 1\\ -2& -1& 0& 1& 1\\ -3& -1& -3& 4& 1\\ -8& -2& -7& 5& 4 \end{pmatrix}, \det(\lambda I_{5}-A)=(\lambda-2)^{5}.$$ Set $$B=A-2I_{5}=\begin{pmatrix} -1& 0& -1& 1& 0\\ -4& -1& -3& 2& 1\\ -2& -1& -2& 1& 1\\ -3& -1& -3& 2& 1\\ -8& -2& -7& 5& 2 \end{pmatrix},$$
then $$\ker{B}=\text{span}\{\mathbf{x_4}=\begin{pmatrix} 0\\ -1\\ 1\\ 1\\ 0 \end{pmatrix},\mathbf{x_{5}}=\begin{pmatrix} 0\\ 1\\ 0\\ 0\\ 1 \end{pmatrix}\};$$
$$\ker{B^2}=\text{span}\{\mathbf{x_4},\mathbf{x_{5}};\mathbf{x_{2}}=\begin{pmatrix} 0\\ 1\\ 0\\ 0\\ 0 \end{pmatrix},\mathbf{x_{3}}=\begin{pmatrix} -1\\ 0\\ 1\\ 0\\ 0 \end{pmatrix}\};$$
$$\ker{B^3}=\text{span}\{\mathbf{x_4},\mathbf{x_{5}};\mathbf{x_{2}},\mathbf{x_{3}};\mathbf{x_{1}}=\begin{pmatrix} 1\\ 0\\ 0\\ 0\\ 0 \end{pmatrix}\}.$$
$$\begin{matrix} \mathbf{x_{4}},& \mathbf{x_{5}}, & & & & &\text{ is a basis for } \ker{B} ,\\ \mathbf{x_{4}},& \mathbf{x_{5}};& \mathbf{x_{2}},& \mathbf{x_{3}}& & &\text{ is a basis for }\ker{B}^{2} ,\\ \mathbf{x_{4}}, & \mathbf{x_{5}}; & \mathbf{x_{2}},& \mathbf{x_{3}};& \mathbf{x_{1}} & &\text{ is a basis for }\ker{B}^{3} . \end{matrix}$$
$$\begin{array}{c|cc|cc} B^{2}(\mathbf{x_1})& B(\mathbf{x_2})& B(\mathbf{x_3})& \mathbf{x_4}& \mathbf{x_5}& \\ B(\mathbf{x_1})& \mathbf{x_2}& \mathbf{x_3}& & & \\ \mathbf{x_1}& & & & & \\ \end{array}$$
Obviously,$$\mathbb{span}\{B(\mathbf{x_{1}})\}\nsubseteqq \mathbb{span}\{\mathbf{x_{2}},\mathbf{x_{3}}\}\quad !$$