= Solution
If $\alpha v=\lambda v$ with $v\ne0$, then $m_\alpha(\alpha)v=m_\alpha(\lambda)v$. The left side vanishes, so $m_\alpha(\lambda)=0$.
Conversely, if $m_\alpha(\lambda)=0$, write $m_\alpha(X)=(X-\lambda)q(X)$. Then $(\alpha-\lambda I)q(\alpha)=0$. If $\alpha-\lambda I$ were invertible, multiplying by its inverse would imply $q(\alpha)=0$, contradicting the minimal degree. Thus $\alpha-\lambda I$ is not invertible. In finite dimension this means it has a nonzero kernel, furnishing an <eigenvector> with <eigenvalue> $\lambda$. Hence \b[the <eigenvalues> are exactly the roots of the <minimal polynomial>], with the scalar-field convention stated in part (a).
Back to article page
