# Show case when minimal polynomial coincides with its characteristic polynomial

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?

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 $$v\not = 0$$.
• Well I didn't know that was true in general but I can see why thanks to your little demonstration so thank you ! Btw you forgot the 0 in $\mu(\lambda) = 0$ in the last line but editing your post for one character is not possible Apr 19, 2021 at 9:21