State and prove the Cayley-Hamilton Theorem. (You may use the Schur Decomposition Theorem.)
Solution
1. Cayley–Hamilton Theorem. Let \(A\in M_n(\mathbb C)\) and let\[ p_A(t)=\det(tI_n-A) \]be its characteristic polynomial. Then\[ p_A(A)=0. \]
We prove this using Schur decomposition, as allowed by the problem.
2. By Schur decomposition there is a unitary matrix \(U\) such that\[ T=U^*AU \]is upper triangular. If its diagonal entries are \(\lambda_1,\ldots,\lambda_n\), then\[ p_A(t)=p_T(t)=\prod_{j=1}^n(t-\lambda_j). \]
Similarity preserves characteristic polynomials: \[ \det(tI-U^*AU) =\det\bigl(U^*(tI-A)U\bigr) =\det(tI-A). \] For an upper-triangular matrix, the determinant is the product of the diagonal entries, so \[ \det(tI-T)=\prod_{j=1}^n(t-\lambda_j). \]
3. Let\[ V_j=\operatorname{span}\{e_1,\ldots,e_j\}, \qquad V_0=0. \]Then \(T(V_j)\subseteq V_j\), and\[ (T-\lambda_j I)V_j\subseteq V_{j-1} \qquad(1\le j\le n). \]
Upper triangularity means that for each \(j\), \[ Te_j\in \operatorname{span}\{e_1,\ldots,e_j\}=V_j, \] so every \(V_j\) is \(T\)-invariant. Moreover the coefficient of \(e_j\) in \(Te_j\) is the diagonal entry \(\lambda_j\), hence \[ (T-\lambda_jI)e_j\in V_{j-1}. \] For \(i<j\), both \(Te_i\) and \(\lambda_j e_i\) lie in \(V_{j-1}\), so \[ (T-\lambda_jI)V_{j-1}\subseteq V_{j-1}. \] Together these statements give \((T-\lambda_jI)V_j\subseteq V_{j-1}\).
4. The operator\[ \prod_{j=1}^n(T-\lambda_jI) \]is zero.
All factors are polynomials in \(T\), hence commute. We may therefore apply them in the order \[ (T-\lambda_nI),\ (T-\lambda_{n-1}I),\ldots,(T-\lambda_1I). \] Starting with \(V_n=\mathbb C^n\), 3 gives successively \[ V_n\xrightarrow{T-\lambda_nI}V_{n-1} \xrightarrow{T-\lambda_{n-1}I}V_{n-2} \longrightarrow\cdots\longrightarrow V_1\xrightarrow{T-\lambda_1I}V_0=0. \] Thus the product annihilates every vector in \(\mathbb C^n\).
5. Hence \(p_T(T)=0\), and therefore \(p_A(A)=0\).
By 2 and 4, \[ p_T(T)=\prod_{j=1}^n(T-\lambda_jI)=0. \] Since \(A=UTU^*\), polynomial evaluation commutes with similarity: \[ p_A(A)=p_T(A)=U\,p_T(T)\,U^*=0. \] This proves the Cayley–Hamilton theorem.