Let $V$ be a finite-dimensional vector space over $\mathbb{C}$, and let $\alpha \in$ End$(V)$ with minimal polynomial $m(x)\in\mathbb{C}[x]$. Show that $\lambda$ is an eigenvalue of $\alpha$ iff $m(\lambda)=0$

Suppose $\lambda$ is an eigenvalue of $\alpha$. The minimal polynomial must divide the characteristic polynomial, by definition. Thus, the minimial and characteristic polynomial must have the same roots, i.e. $m(\lambda)=0$.

The converse is basically the same argument...

Is there a more formal way to prove this result?


I'm not sure how you jump to "the minimial and characteristic polynomial must have the same roots" so immediately.

If $m(\lambda)=0$ then indeed $\chi(\lambda)=0$ since $m(x)$ divides $\chi(x)$, the characteristic polynomial of $\alpha$. But the roots of $\chi$ are the eigenvalues.

The converse is more subtle, and really depends on $m(x)$ being the minimal polynomial rather than just being any factor of $\chi(x)$. If $\lambda$ is an eigenvalue, then $\chi(\lambda)=0$ (since $\alpha-\lambda 1_V$ is not injective). Hence by the factor theorem $\chi(x)=(x-\lambda)^rp(x)$ for some $r>1$ and some polynomial $p$ such that $p(\lambda)\neq 0$. Now $p(\alpha)v=p(\lambda)v\neq 0$ for an eigenvector $v\in E_\lambda(\alpha)$, so $p(\alpha)\neq 0$. Hence $m(x)$ is not a factor of $p(x)$, and the only way $m(x)$ to divide $\chi(x)$ is to have some positive power of $(x-\lambda)$ in its factorization. That gives $m(\lambda)=0$ as claimed.

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