Proof. We assume \(m>1\) since for \(m=1\) the statement is trivial. By definition, we need to show that \[(f_{\mathbf{J}_m(\lambda)}-\lambda\mathrm{Id}_{\mathbb{K}^m})^m(\vec{e}_m)=0_{\mathbb{K}^m}\qquad \text{and} \qquad (f_{\mathbf{J}_m(\lambda)}-\lambda\mathrm{Id}_{\mathbb{K}^m})^{m-1}(\vec{e}_m)\neq 0_{\mathbb{K}^m}.\] By definition, we have \(\mathbf{J}_m(\lambda)-\lambda\mathbf{1}_{m}=\mathbf{J}_m(0)=\sum_{i=1}^{m-1}\mathbf{E}_{i,i+1}.\) We use induction to show that for \(1\leqslant k\leqslant m-1,\) we have \[\tag{12.3} (\mathbf{J}_m(0))^k=\sum_{i=1}^{m-k}\mathbf{E}_{i,i+k}.\] For \(k=1\) the statement is obviously correct and hence anchored.

Proof. (i) We only need to show that the vectors \(\{u_1,\ldots,u_m\}\) are linearly independent as by definition, \(\{u_1,\ldots,u_m\}\) is a generating set for \(Z(g_{\lambda},v).\) Write \(g_{\lambda}=f-\lambda\mathrm{Id}_V\) then \[(u_1,\ldots,u_m)=(g_{\lambda}^{m-1}(v),g_{\lambda}^{m-2}(v),\ldots,g_{\lambda}(v),v).\] Suppose we have scalars \(\mu_1,\ldots,\mu_m\) such that \[\tag{12.6} 0_V=\mu_1u_1+\cdots+\mu_mu_m=\mu_1g_{\lambda}^{m-1}(v)+\mu_2g_{\lambda}^{m-2}(v)+\cdots+\mu_{m-1}g_{\lambda}(v)+\mu_m v.\] Since by assumption \(g_{\lambda}^m(v)=0_V\) we have \(g_{\lambda}^{k}(v)=0_V\) for all \(k\geqslant m.\) Applying \(g_{\lambda}\) \((m-1)\)-times to (12.6) thus gives \[\mu_1g_{\lambda}^{2m-2}(v)+\mu_2g_{\lambda}^{2m-3}(v)+\cdots +\mu_{m-1}g_{\lambda}^m(v)+\mu_{m}g_{\lambda}^{m-1}(v)=\mu_{m}g_{\lambda}^{m-1}(v)=0_V.\] By assumption \(g_{\lambda}^{m-1}(v)\neq 0_V,\) hence we conclude that \(\mu_{m}=0.\) Therefore, (12.6) becomes \[\mu_1u_1+\cdots+\mu_mu_m=\mu_1g_{\lambda}^{m-1}(v)+\mu_2g_{\lambda}^{m-2}(v)+\cdots+\mu_{m-1}g_{\lambda}(v)=0_V.\] Applying \(g_{\lambda}\) \((m-2)\)-times to the previous equation we conclude that \(\mu_{m-1}=0\) as well. Continuing in this fashion it follows that \(\mu_1=\mu_2=\cdots=\mu_m=0,\) hence the vectors \(\{u_1,\ldots,u_m\}\) are linearly independent, as claimed.

(ii) Since \(\{u_1,\ldots,u_m\}\) is a basis of \(Z(g_{\lambda},v),\) it is sufficient to show that for all \(1\leqslant i\leqslant m\) the vector \(f(u_i)\) is a linear combination of \(\{u_1,\ldots,u_m\}.\) By construction, we have \[\begin{aligned} (f-\lambda\mathrm{Id}_V)(u_1)&=g_{\lambda}^m(v)=0_V,\\ (f-\lambda\mathrm{Id}_V)(u_2)&=g_{\lambda}^{m-1}(v)=u_1,\\ (f-\lambda\mathrm{Id}_V)(u_3)&=g_{\lambda}^{m-2}(v)=u_2,\\ &\vdots\\ (f-\lambda\mathrm{Id}_V)(u_m)&=g_{\lambda}(v)=u_{m-1} \end{aligned}\] Equivalently, we have \[f(u_1)=\lambda u_1,\quad f(u_2)=u_1+\lambda u_2,\quad f(u_3)=u_2+\lambda u_3,\quad \ldots \quad f(u_m)=u_{m-1}+\lambda u_m,\] which shows the claim.

(iii) Previously we showed that \(f(u_1)=\lambda u_1,\) hence the first column vector of \(\mathbf{M}(\hat{f},\mathbf{b},\mathbf{b})\) is \(\lambda \vec{e}_1.\) For \(2\leqslant i \leqslant m,\) we have \(f(u_i)=1 u_{i-1}+\lambda u_i\) and hence the \(i\)-th column vector of \(\mathbf{M}(\hat{f},\mathbf{b},\mathbf{b})\) is given by \(\vec{e}_{i-1}+\lambda \vec{e}_i.\) This shows that \(\mathbf{M}(\hat{f},\mathbf{b},\mathbf{b})=\mathbf{J}_{m}(\lambda).\)

Proof. There exists an integer \(m \in \mathbb{N}\) such that \(\mathcal{E}_f(\lambda)=\operatorname{Ker}((f-\lambda\mathrm{Id}_V)^m).\) Therefore, for all \(v \in \mathcal{E}_f(\lambda)\) we have \((f-\lambda\mathrm{Id}_V)^m(v)=0_V\) which shows that \(g^m=o,\) as claimed.

Proof. We use induction on the dimension of the vector space \(V.\) For \(\dim V=1\) the only nilpotent endomorphism is the zero endomorphism \(o : V \to V\) and we can take \(\ell=1,\) \(m_1=1\) and \(\mathbf{b}=(v)\) for any non-zero vector \(v \in V.\) The statement is thus anchored.

Since \(u_1,\ldots,u_k \in U=\operatorname{Im}(g),\) there exist vectors \(v_1,\ldots,v_k\) such that \(u_i=g(v_i)\) for all \(1\leqslant i\leqslant k.\) Set \(m_i=n_i+1\) for \(1\leqslant i\leqslant k\) and consider the set \[\begin{gathered} S=\{g^{m_1-1}(v_1),g^{m_1-2}(v_1),\ldots,g(v_1),v_1,g^{m_2-1}(v_2),g^{m_2-2}(v_2),\ldots,g(v_2),v_2,\ldots\\ \ldots,g^{m_{k}-1}(v_k),g^{m_{k}-2}(v_k),\ldots,g(v_k),v_k\}. \end{gathered}\] We claim \(S\) is linearly independent. Suppose we can find a linear combination \(w\) of the elements of \(S\) that gives the zero vector. Applying \(g\) to this linear combination, we obtain a linear combination of the elements of \[\begin{gathered} \{g^{m_1}(v_1),g^{m_1-1}(v_1),\ldots,g^2(v_1),g(v_1),g^{m_2}(v_2),g^{m_2-1}(v_2),\ldots,g^2(v_2),g(v_2),\ldots\\ \ldots,g^{m_{k}}(v_k),g^{m_{k}-1}(v_k),\ldots,g^2(v_k),g(v_k)\} \end{gathered}\] that gives the zero vector. Equivalently, we obtain a linear combination of the elements of \[\begin{gathered} \{g^{m_1-1}(u_1),g^{m_1-2}(u_1),\ldots,g(u_1),u_1,g^{m_2-1}(u_2),g^{m_2-2}(u_2),\ldots,g(u_2),u_2,\ldots\\ \ldots,g^{m_{k}-1}(u_k),g^{m_{k}-2}(u_k),\ldots,g(u_k),u_k\} \end{gathered}\] that gives the zero vector. Equivalently, we obtain a linear combination of the elements of \[\begin{gathered} \{h^{n_1}(u_1),h^{n_1-1}(u_1),\ldots,h(u_1),u_1,h^{n_2}(u_2),h^{n_2-1}(u_2),\ldots,h(u_2),u_2,\ldots\\ \ldots,h^{n_{k}}(u_k),h^{n_{k}-1}(u_k),\ldots,h(u_k),u_k\} \end{gathered}\] that gives the zero vector. Here we use that \(m_i=n_i+1\) for \(1\leqslant i\leqslant k\) and that \(h=g\) on \(\operatorname{Im}(g).\) The tuple \(\mathbf{c}\) is an ordered basis of \(U,\) hence all the coefficients in this linear combination must vanish, except the coefficients before each vector \(h^{n_i}(u_i),\) since \(h^{n_i}(u_i)=0_V\) for all \(1\leqslant i\leqslant k.\) The initial linear combination \(w\) thus simplifies to become \[\mu_1g^{m_1-1}(v_1)+\mu_2g^{m_2-1}(v_2)+\cdots+\mu_{k}g^{m_k-1}(v_k)=0_V.\] for some scalars \(\mu_1,\ldots,\mu_k.\) It remains to argue that these scalars are all zero. The previous equation is equivalent to \[\mu_1 h^{n_1-1}(u_1)+\mu_2h^{n_2-2}(u_2)+\cdots+\mu_kh^{n_k-1}(u_k)=0_V.\] Using the linear independence of the elements of \(\mathbf{c}\) again, we conclude that \(\mu_1=\cdots=\mu_k=0,\) as desired.

Observe that by construction, the vectors \(v_1,\ldots,v_k\) satisfy \(g^{m_1}(v_1)=g^{m_2}(v_2)=\cdots=g^{m_k}(v_k)=0_V.\)

Finally, the first \(m_1\) vectors of \(\mathbf{b}\) are \(y_i=g^{m_1-i}(v_1)\) for \(1\leqslant i\leqslant m_1\) and we have \(g(y_1)=0_V\) and \(g(y_i)=y_{i-1}\) for \(2\leqslant i\leqslant m_1.\) This contributes the Jordan block \(\mathbf{J}_{m_1}(0)\) to the matrix representation of \(g\) with respect to \(\mathbf{b}.\) The remaining blocks arise by considering the vectors \(g^{m_k-i}(v_k)\) for \(2\leqslant k\leqslant \ell\) and where \(1\leqslant i\leqslant m_k.\)