Let \(k\) be a field, let \(V\) be a \(k\)-vector space with \(0<\dim_k V<\infty\), and let \(T\colon V\to V\) be \(k\)-linear. Regard \(V\) as a \(k[x]\)-module with \(x\) acting by \(T\). The following are equivalent:
The minimal polynomial of \(T\) equals the monic characteristic polynomial \(\det(x\id_V-T)\).
The invariant factor decomposition of the \(k[x]\)-module \(V\) has exactly one invariant factor.
The rational canonical form of \(T\) consists of a single block.
There is a basis of \(V\) in which the matrix of \(T\) is a companion matrix.
There exists \(v\in V\) with \(\spanof_k\theset{T^j v \suchthat j\geq 0} = V\).
If \(T\) has \(\dim_k V\) distinct eigenvalues in \(k\), then \(T\) satisfies these conditions, but the converse fails. The characteristic polynomial is then a product of distinct linear factors, each of which divides the minimal polynomial, so the two polynomials are equal. For the converse, let \(n=\dim_k V\geq 2\) and let \(T\) have matrix the Jordan block \(J_n(\lambda)\) for some \(\lambda\in k\). Its minimal and characteristic polynomials are both \((x-\lambda)^n\), but \(\lambda\) is its only eigenvalue.