Prove Cayley-Hamilton in the following way. Let \(V=\spanof\ts{\vector v_1,\ldots,\vector v_n}\) and define \[ \Fil_iV=\spanof\ts{\vector v_1,\ldots,\vector v_i}, \qquad \Fil_0V=\ts{0}. \] Suppose the matrix of \(A\) in this basis is upper triangular, with diagonal entries \(\lambda_1,\ldots,\lambda_n\). Show that \[ A(\Fil_iV)\subseteq\Fil_iV, \qquad (A-\lambda_iI)\Fil_iV\subseteq\Fil_{i-1}V. \] Deduce that \[ (A-\lambda_1I)\cdots(A-\lambda_nI)V=0 \] and hence \(\chi_A(A)=0\). Finally, deduce Cayley-Hamilton for an arbitrary operator over an arbitrary field by extending scalars to an algebraic closure and triangularizing there.
Solution
1. For every \(i\), the flag subspace \(\Fil_iV\) is \(A\)-invariant.
Because the matrix of \(A\) in the ordered basis \(v_1,\ldots,v_n\) is upper triangular, for each \(j\) one has \[ Av_j\in\spanof\ts{v_1,\ldots,v_j}=\Fil_jV. \] Thus if \(j\le i\), then \(Av_j\in\Fil_iV\). Since \(\Fil_iV\) is spanned by \(v_1,\ldots,v_i\), it follows that \[ A(\Fil_iV)\subseteq\Fil_iV. \]
2. For every \(i\), one has\[ (A-\lambda_iI)\Fil_iV\subseteq\Fil_{i-1}V. \]
For \(j<i\), both \(Av_j\) and \(\lambda_i v_j\) lie in \(\Fil_{i-1}V\), hence so does \((A-\lambda_iI)v_j\). For \(j=i\), upper triangularity gives \[ Av_i=\lambda_i v_i+w \] for some \(w\in\Fil_{i-1}V\). Therefore \[ (A-\lambda_iI)v_i=w\in\Fil_{i-1}V. \] Since \(v_1,\ldots,v_i\) span \(\Fil_iV\), the inclusion follows.
3. The product of the linear factors annihilates \(V\):\[ (A-\lambda_1I)\cdots(A-\lambda_nI)V=0. \]
By 2, \[ (A-\lambda_nI)\Fil_nV\subseteq\Fil_{n-1}V. \] Applying successively the factors with indices \(n-1,n-2,\ldots,1\) gives \[ (A-\lambda_1I)\cdots(A-\lambda_nI)\Fil_nV \subseteq\Fil_0V=0. \] Since \(\Fil_nV=V\), the displayed product is the zero operator.
4. For an upper-triangular matrix,\[ \chi_A(t)=\prod_{i=1}^n(t-\lambda_i), \]so \(\chi_A(A)=0\).
3. Hence \(\chi_A(A)=0\).
5. Cayley-Hamilton holds for every endomorphism of a finite-dimensional vector space over an arbitrary field.
Let \(A\in\Endo_k(V)\) and let \(\overline{k}\) be an algebraic closure of \(k\). Extend scalars: \[ \overline{V}=V\otimes_k\overline{k}, \qquad \overline{A}=A\otimes1. \] Over \(\overline{k}\), the characteristic polynomial of \(A\) splits, so \(\overline{A}\) admits an upper-triangular matrix in some basis. By 4, \[ \chi_{\overline{A}}(\overline{A})=0. \] Scalar extension does not change the characteristic polynomial coefficients, so \[ \chi_{\overline{A}}(t)=\chi_A(t) \] viewed in \(\overline{k}[t]\). Thus \[ \chi_A(A)\otimes1=0 \] in \(\Endo_{\overline{k}}(\overline{V})\). The natural map \[ \Endo_k(V)\longrightarrow\Endo_{\overline{k}}(V\otimes_k\overline{k}), \qquad T\longmapsto T\otimes1, \] is injective because \(k\to\overline{k}\) is a field extension. Therefore \(\chi_A(A)=0\) already over \(k\).