Proof. If \(\lambda\) is an eigenvalue of \(f\) then there exists a non-zero vector \(v \in \operatorname{Ker}(f-\lambda\mathrm{Id}_V)\) and hence \(\dim \mathcal{E}_f(\lambda)>0\) so that \(\mathcal{E}_f(\lambda)\neq \{0_V\}.\) Conversely, suppose \(\mathcal{E}_f(\lambda)\neq \{0_V\}\) so that there exists an integer \(m\) and a non-zero vector \(v \in V\) such that \((f-\lambda\mathrm{Id}_V)^m(v)=0_V.\) We may assume \(m\) to be the smallest such integer. Then, by assumption, \(w=(f-\lambda\mathrm{Id}_V)^{m-1}(v)\neq 0_V\) and \(w\) satisfies \(f(w)=\lambda w\) and hence is an eigenvector of \(f\) with eigenvalue \(\lambda.\)

Proof. By definition, the zero vector \(0_V\) is an element of \(\mathcal{E}_f(\lambda),\) hence \(\mathcal{E}_{f}(\lambda)\) is non-empty. Let \(t_1,t_2 \in \mathbb{K}\) and \(v_1,v_2 \in \mathcal{E}_{f}(\lambda).\) Then there exist \(k_1,k_2\) such that \((f-\lambda\mathrm{Id}_V)^{k_1}(v_1)=0_V\) and \((f-\lambda\mathrm{Id}_V)^{k_2}(v_2)=0_V.\) Take \(k\) to be the maximum of \(\{k_1,k_2\}.\) Then, using the linearity of \(f-\lambda\mathrm{Id}_V\) and its powers, we compute \[\begin{aligned} 0_V&=t_1(f-\lambda\mathrm{Id}_V)^{k-k_1}(0_V)+t_2(f-\lambda\mathrm{Id}_V)^{k-k_2}(0_V)\\ &=t_1(f-\lambda\mathrm{Id}_V)^{k-k_1}((f-\lambda\mathrm{Id}_V)^{k_1}(v_1))+t_2(f-\lambda\mathrm{Id}_V)^{k-k_2}((f-\lambda\mathrm{Id}_V)^{k_2}(v_2))\\ &=t_1(f-\lambda\mathrm{Id}_V)^k(v_1)+t_2(f-\lambda\mathrm{Id}_V)^k(v_2)=(f-\lambda\mathrm{Id}_V)^k(t_1v_1+t_2v_2) \end{aligned}\] so that \(t_1v_1+t_2v_2 \in \operatorname{Ker}((f-\lambda\mathrm{Id}_V)^k) \subset \mathcal{E}_f(\lambda)\) and hence \(\mathcal{E}_f(\lambda)\) is a subspace by Definition 3.21.

We now show that \(\mathcal{E}_f(\lambda)\) is stable under \(f.\) Let \(v \in \mathcal{E}_f(\lambda)\) so that there exists \(k\geqslant 0\) with \((f-\lambda\mathrm{Id}_V)^k(v)=0_V.\) Write \(w=f(v).\) Then we obtain \[\begin{aligned} (f-\lambda\mathrm{Id}_V)^{k}(w)&=(f-\lambda\mathrm{Id}_V)^{k}(f(v)-\lambda v+\lambda v)\\ &=(f-\lambda\mathrm{Id}_V)^k(f(v)-\lambda v)+\lambda (f-\lambda\mathrm{Id}_V)^{k}(v)\\ &=(f-\lambda\mathrm{Id}_V)^{k+1}(v)+\lambda (f-\lambda\mathrm{Id}_V)^k(v)=0_V. \end{aligned}\] Therefore \(w=f(v) \in \mathcal{E}_f(\lambda)\) and hence \(\mathcal{E}_f(\lambda)\) is stable under \(f.\)

Proof. Let \(\lambda_1,\ldots,\lambda_k\) be distinct eigenvalues of \(f\) and let \(n_i\) for \(1\leqslant i\leqslant k\) be such that \(\mathcal{E}_f(\lambda_i)=\operatorname{Ker}((f-\lambda_i\mathrm{Id}_V)^{n_i}).\) For \(1\leqslant i\leqslant k\) let \(v_i,\hat{v}_i \in \mathcal{E}_{f}(\lambda_i)\) be such that \[\tag{12.2} v_1+v_2+\cdots+v_k=\hat{v}_1+\hat{v}_2+\cdots+\hat{v}_k\] We want to show that \(w_i=v_i-\hat{v}_i=0_V\) for all \(1\leqslant i\leqslant k.\) For \(1\leqslant i\leqslant k\) consider the endomorphism \[g_i=(f-\lambda_1\mathrm{Id}_V)^{n_1}\circ \cdots\circ (f-\lambda_{i-1}\mathrm{Id}_V)^{n_{i-1}}\circ (f-\lambda_{i+1}\mathrm{Id}_V)^{n_{i+1}}\circ \cdots\circ (f-\lambda_k\mathrm{Id}_V)^{n_k}.\] Notice that \(g_i\) does not contain the mapping \((f-\lambda_i\mathrm{Id}_V)^{n_i}.\) For \(i \neq j\) the mapping \(g_i\) contains \((f-\lambda_j\mathrm{Id}_V)^{n_j}.\) Rearranging the mappings in \(g_i\) if necessary, we can assume that \(g_i=h\circ (f-\lambda_j\mathrm{Id}_V)^{n_j}\) for some endomorphism \(h.\) Rearranging does not change \(g_i\) since for all \(\mu_1,\mu_2 \in \mathbb{K}\) we have \[(f-\mu_1\mathrm{Id}_V)\circ (f-\mu_2\mathrm{Id}_V)=(f-\mu_2\mathrm{Id}_V)\circ (f-\mu_1\mathrm{Id}_V).\] Since \(w_j \in \mathcal{E}_{f}(\lambda_j)=\operatorname{Ker}((f-\lambda_j\mathrm{Id}_V)^{n_j})\) we thus conclude that \(g_i(w_j)=0_V.\)

By Lemma 12.6 the subspace \(\mathcal{E}_f(\lambda_i)\) is stable under \(f\) and hence it is also stable under \(f-\mu\mathrm{Id}_V\) for all \(\mu \in \mathbb{K}.\) This implies that \(\mathcal{E}_f(\lambda_i)\) is also stable under \(g_i.\) Write (12.2) as \[w_1+w_2+\cdots +w_k=0_V.\] Applying the endomorphism \(g_i\) to the previous equation and using that \(g_i(w_j)=0_V\) for \(i\neq j,\) we obtain that \(g_i(w_i)=0_V.\) Since for \(j\neq i\) none of the \(\lambda_j\) is a generalised eigenvalue of \(f|_{\mathcal{E}_f(\lambda_i)},\) the restriction of \(g_i\) to \(\mathcal{E}_f(\lambda_i)\) is invertible as an endomorphism of \(\mathcal{E}_f(\lambda_i).\) Since \(g_i(w_i)=0_V,\) this implies that \(w_i=0.\) Since \(i\) is arbitrary, we have \(w_1=w_2=\cdots=w_k=0_V,\) as desired.

Proof. Let \(U=\mathcal{E}_f(\lambda_1)\oplus \mathcal{E}_f(\lambda_2) \oplus \cdots \oplus \mathcal{E}_f(\lambda_k)\) and suppose that \(U\neq V.\) Then, by Corollary 6.11 there exists a complement \(U^{\prime}\) of \(U\) with \(\dim U^{\prime}\geqslant 1.\) Let \(\Pi : V \to U^{\prime}\) denote the projection onto \(U^{\prime}\) with kernel \(U\) and consider the endomorphism \(\hat{f}=\Pi \circ f|_{U^{\prime}} : U^{\prime} \to U^{\prime}.\) Since we work over the complex numbers and since \(\dim U^{\prime}\geqslant 1,\) Theorem 6.49 implies that \(\hat{f}\) admits an eigenvalue \(\mu.\) Let \(v \in U^{\prime}\) be a corresponding eigenvector of \(\hat{f}.\) Since \(U=\operatorname{Ker}\Pi\) is a complement of \(U^{\prime},\) we obtain \[f(v)=\mu v+u\] for some vector \(u \in U.\) We can write \(u=\sum_{i=1}^k u_i\) with \(u_i \in \mathcal{E}_{f}(\lambda_i).\) Now define \(g=f-\mu\mathrm{Id}_V : V \to V\) so that \[g(v)=\sum_{i=1}^k u_i.\] Suppose \(1\leqslant i\leqslant k\) is such that \(\lambda_i\neq \mu.\) By definition, \(\operatorname{Eig}_f(\lambda_i)\subset \mathcal{E}_f(\lambda_i),\) hence the restriction of \(g=f-\mu\mathrm{Id}_V\) to \(\mathcal{E}_f(\lambda_i)\) is invertible as an endomorphism of \(\mathcal{E}_f(\lambda_i),\) so there exists a vector \(v_i \in \mathcal{E}_f(\lambda_i)\) such that \(g(v_i)=u_i.\) If \(\lambda_i\neq \mu\) for all \(1\leqslant i\leqslant k,\) then we obtain \[g\left(v-\sum_{i=1}^kv_i\right)=0_V\] so that \(v-\sum_{i=1}^k v_i\) is an element of \(\operatorname{Ker}g=\operatorname{Ker}(f-\mu\mathrm{Id}_V)=\{0_V\},\) where the last equality follows since \(\mu\) is not an eigenvalue of \(f.\) We can therefore write \(v=\sum_{i=1}^k v_i \in U,\) but this contradicts the assumption that \(v \in U^{\prime}.\)

We conclude that we can find an integer \(i\) with \(1\leqslant i\leqslant k\) such that \(\lambda_i=\mu.\) After possibly renumbering the eigenvalues we can assume that \(\lambda_1=\mu\) and hence that \(\lambda_i\neq \mu\) for \(2\leqslant i\leqslant k,\) since the eigenvalues are distinct. So again for \(2\leqslant i\leqslant k\) we have vectors \(v_i \in \mathcal{E}_{f}(\lambda_i)\) such that \(g(v_i)=u_i.\) We thus have \[g\left(v-\sum_{i=2}^k v_i\right)=u_1.\] Since \(\mathcal{E}_f(\lambda_1)=\operatorname{Ker}((f-\lambda_1\mathrm{Id}_V)^{n_1})\) for some integer \(n_1\) and \(g=f-\lambda_1\mathrm{Id}_V,\) applying \(g^{n_1},\) we obtain \[g^{n_1+1}\left(v-\sum_{i=2}^k v_i\right)=g^{n_1}(u_1)=0_V,\] where the last equality uses that \(u_1 \in \mathcal{E}_f(\lambda_1).\) It follows that \(v-\sum_{i=2}^k v_i \in \mathcal{E}_f(\lambda_1)\) and hence that \(v \in U\) which is again a contradiction to the assumption that \(v \in U^{\prime}.\)