= Eigenvalue criterion using the minimal polynomial
{title2=$m_T(\lambda)=0\iff\lambda\text{ is an eigenvalue of }T$}
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.
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