Prove Cayley-Hamilton using the JCF.
Solution
Let \(A\in\Endo_k(V)\) with \(V\) finite-dimensional.
1. It is enough to prove the identity after extending scalars to an algebraic closure \(\overline{k}\).
Set \[ \overline{V}=V\otimes_k\overline{k}, \qquad \overline{A}=A\otimes1. \] The characteristic polynomial is unchanged by scalar extension, so \(\chi_{\overline A}(t)=\chi_A(t)\) viewed in \(\overline{k}[t]\). Moreover, the natural map \[ \Endo_k(V)\longrightarrow\Endo_{\overline{k}}(\overline V), \qquad T\longmapsto T\otimes1, \] is injective. Hence if \(\chi_A(A)\otimes1=0\), then \(\chi_A(A)=0\) already over \(k\).
2. Over \(\overline{k}\), choose a basis in which \(\overline A\) has Jordan form\[ J=\bigoplus_r J_{m_r}(\lambda_r). \]
The characteristic polynomial splits over \(\overline{k}\), so the Jordan canonical form theorem applies. Similarity preserves polynomial evaluation: if \(\overline A=PJP^{-1}\), then \[ p(\overline A)=Pp(J)P^{-1} \] for every polynomial \(p\).
3. On a Jordan block \(J_m(\lambda)\) one has\[ (J_m(\lambda)-\lambda I)^m=0. \]
Write \[ J_m(\lambda)=\lambda I+N, \] where \(N\) has \(1\) on the superdiagonal and \(0\) elsewhere. The matrix \(N\) shifts each basis vector at most one step along the Jordan chain, so \(N^m=0\). Thus \[ (J_m(\lambda)-\lambda I)^m=N^m=0. \]
4. The characteristic polynomial \(\chi_J\) annihilates every Jordan block of \(J\).
3. Since all factors in the polynomial \(\chi_J(J_m(\lambda))\) commute, this gives\[ \chi_J(J_m(\lambda))=0. \]Therefore \(\chi_J(J)=0\) block by block.
5. Hence \(\chi_A(A)=0\).
By 2 and 4, \[ \chi_{\overline A}(\overline A)=0. \] Using 1, \[ \chi_A(A)\otimes1=0 \] implies \(\chi_A(A)=0\) over the original field \(k\).