Show case when minimal polynomial coincides with its characteristic polynomial

Question

As is introduced in the title, I'm stuck on the following problem:

Considering a linear endomorphism $φ$ of an $n$-dimensional vector space $V$ having $n$ pairwise distinct eigenvalues, I would like to show that the minimal polynomial of $φ$ coincides with its characteristic polynomial.

I don't know much about the minimal polynomial but I've seen on this post Simple proof of when minimal polynomial coincides with the characteristic polynomial that a characteristic and minimal polynomial of a matrix coincides iff the set $\{I,A,A^2,...,A^{n−1}\}$ are linearly independent.

I guess we could represent $\varphi$ with an $n\times n$ matrix, but how can I connect the proof of the link above with the given $n$ pairwise distinct eigenvalues? Or is there another interesting way to show this?

Thanks in advance for your help!

Answer 1

You can show that all eigenvalues of $A$ must be roots of the minimal polynomial. Let $f(t)$ be its characteristic polynomial and $\mu(t)$ its minimal polynomial. Since $\mu(t) \ | \ f(t)$, it follows that the $\mu(t)=f(t)$.

Now we show that all eigenvalues of $A$ must be roots of $\mu(t)$. This follows immediately by the observation that in general, if $\lambda$ is an eigenvalue of $A$ then $g(\lambda)$ is an eigenvalue of $g(A)$. Applying this, let $\lambda$ be an eigenvalue of $A$ and $v$ is a corresponding eigenvector. Since $\mu(A)=0$, we have $$0=\mu(A)v=\mu(\lambda)v$$ Hence $\mu(\lambda)=0$ as $v\not = 0$.