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Let $V$ be a finitely-generated inner product space and let $\alpha $ be a normal endomorphism of $V$ . If the minimal polynomial of $\alpha$ is completely reducible, then it does not have multiple roots.

This is a theorem from this book http://www.springer.com/us/book/9789400726352. I have a problem in a part of the proof. In this book $\sigma_0$ is being used to denote the $0$-map and $\sigma_1$ is being used to denote the identity map.

Proof: Let $p(X)$ be the minimal polynomial of $\alpha$, which we assume is completely reducible. Assume that there exists a scalar $c$ and a polynomial $q(X)$ such that $p(X) = (X − c)^2q(X)$. Since $p(\alpha) = \sigma_0$, we have $(\alpha − c\sigma_1)^2q(\alpha) = \sigma_0$ and so $ker((α − c\sigma_1)^2q(\alpha)) = V $. Note that $\beta = \alpha − c\sigma_1$ is a normal endomorphism of $V$ . Let $v \in V$ and let $w = q(\alpha)(v)$. Then $\beta^2(w) = 0_V$ and so $\beta(w) \in Im(\beta) \cap Ker(\beta) = {0_V }$. Thus we see that $\beta q(\alpha)(v) = 0_V$ for all $v \in V$ and hence $α$ annihilates the polynomial $(X − c)q(X)$, contradicting the minimality of $p(X)$.

$\beta=\alpha-c\sigma_1$, I know why it's normal since $\beta\beta^*=\beta^*\beta$. What I don't understand is that why it says $Im(\beta) \cap Ker(\beta) = {0_V }$.

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Note that $\beta(v)=0\implies \beta^*(v)=0$ by normality, because $\|\beta^*(v)\|^2=\langle \beta^*(v),\beta^*(v)\rangle=\langle \beta\beta^*(v),v\rangle=\langle \beta^*\beta(v),v\rangle=\langle \beta(v),\beta(v)\rangle=\|\beta(v)\|^2$.

Also note that $\mathrm{ker}(\beta^*\beta)=\mathrm{ker}(\beta)$, not depending on normality.

If $v$ is in the kernel and image of $\beta$, then $\beta(v)=0$ and there exists $w$ such that $\beta(w)=v$. Then $\beta^*\beta(w)=\beta^*(v)=0$ by the first paragraph, hence $\beta(w)=0$ by the second paragraph. Thus $v=\beta(w)=0$.