Eigenvalue criterion using the minimal polynomial ($m_T(\lambda )=0⟺\lambda \text{ is an eigenvalue of }T$)

For a linear operator on a finite-dimensional vector space, an eigenvector gives $m_T(T)v=m_T(\lambda )v=0$, proving the forward implication from eigenvalue to root. Conversely write $m_T(X)=(X-\lambda )q(X)$. If $T-\lambda I$ were invertible, the equation $(T-\lambda I)q(T)=0$ would imply $q(T)=0$, contradicting minimal degree. Thus $T-\lambda I$ has a nonzero kernel of a linear map. The equivalence concerns roots in the scalar field; roots in an extension describe eigenvalues after extending scalars.