State/prove the Cayley–Hamilton theorem.
Solution
Cayley–Hamilton. If \(A\in M_n(F)\) and \[ \chi_A(t)=\det(tI-A), \] then \[ \chi_A(A)=0. \]
1. It suffices to prove the theorem after extending scalars to an algebraic closure \(\overline F\).
The matrix \(\chi_A(A)\) has entries in \(F\). If it becomes the zero matrix after applying the injective map \(F\hookrightarrow\overline F\), then it was already zero over \(F\).
2. Over \(\overline F\), choose a basis in which \(A\) is upper triangular with diagonal entries \(\lambda_1,\dots,\lambda_n\).
Over an algebraically closed field, every linear operator admits a complete invariant flag, equivalently an upper-triangular matrix form.
3. Let \(F_i\) be the span of the first \(i\) basis vectors. Then\[ (A-\lambda_iI)F_i\subseteq F_{i-1}. \]
Upper triangularity gives \(AF_i\subseteq F_i\), and on the one-dimensional quotient \(F_i/F_{i-1}\) the induced map is multiplication by the \(i\)th diagonal entry \(\lambda_i\).
4. Therefore\[ (A-\lambda_1I)\cdots(A-\lambda_nI)=0. \]
Apply the factors from right to left. By 3, the factor \(A-\lambda_nI\) sends \(F_n\) into \(F_{n-1}\), then \(A-\lambda_{n-1}I\) sends that into \(F_{n-2}\), and so on, eventually into \(F_0=0\).
5. Since\[ \chi_A(t)=\prod_{i=1}^n(t-\lambda_i), \]we obtain \(\chi_A(A)=0\).
For an upper-triangular matrix the characteristic polynomial is the product of \(t\) minus the diagonal entries. Combine this with 4 and then descend using 1.