Exercise 8.C.12

Proof. Suppose $V$ has a basis consisting of eigenvectors of $T$. Let ${\lambda }_1,\dots ,{\lambda }_m$ denote the distinct eigenvalues of $T$. Then, it is easy to see that

by applying the left side of the equation to each of the basis vectors, because the parentheses commute. By 8.46, $(z-{\lambda }_1)\cdots (z-{\lambda }_m)$ is a polynomial multiple of the minimal polynomial of $T$. This polynomial has no repeated zeros. Hence the minimal polynomial has no repeated zeros.

Conversely, suppose the minimal polynomial of $T$, call it $p$, has no repeated zeros. By 8.23, $T$ has a basis of generalized eigenvectors. Let $v$ be one vector in this basis. Then $v\in G(\lambda ,T)$ for some eigenvalue $\lambda$ of $T$. By 8.49, $\lambda$ is zero of $p$. Thus, we can write $p(z)=(z-\lambda )q(z)$ for some polynomial $q$ with $q(\lambda )\ne 0$. We have

Note that $(T-\alpha )\widehat{v}\ne 0$ for all nonzero $\widehat{v}\in G(\lambda ,T)$ and all $\alpha \in 𝔽$ with $\alpha \ne \lambda$. This implies that $(T-\lambda )v=0$, since $G(\lambda ,T)$ is invariant under every polynomial of $T$ and so $(T-\mathit{\lambda I})v\ne 0$ implies $q(T)(T-\mathit{\lambda I})\ne 0$ (because we can factor $q(T)$ and $\lambda$ is not a zero of $q$). Therefore $v$ is an eigenvector of $T$ and the basis of $V$ consisting of generalized eigenvectors of $T$ actually consists of eigenvectors of $T$. □