Show that if \(q\) is the minimal polynomial of a linear transformation \(\phi: E\to E\) with \(\dim_k E = n\) then \(\deg q \leq n\).
Solution
Choose an ordered basis of \(E\), let \(A \in M_n(k)\) be the matrix of \(\phi\), and let \(p(x) = \det(x I_n - A) \in k[x]\) be the characteristic polynomial.
1. \(p\) is monic of degree \(n\) and \(p(\phi) = 0\).
Expanding the determinant, the only term of degree \(n\) in \(x\) is the product of the diagonal entries, so \(p\) is monic of degree \(n\). By the Cayley–Hamilton theorem \(p(A) = 0\), hence \(p(\phi) = 0\).
2. Q.E.D.
The minimal polynomial \(q\) is the monic generator of the ideal \(\{f \in k[x] : f(\phi) = 0\}\). By step 1, \(p\) lies in this ideal, so \(q \mid p\). Since \(p \neq 0\), \(\deg q \le \deg p = n\).