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!

  • 1
    $\begingroup$ Does this help ? $\endgroup$ Apr 19, 2021 at 1:22
  • $\begingroup$ It does actually, thanks ! $\endgroup$
    – Rhaena
    Apr 19, 2021 at 9:34

1 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 $v\not = 0$.

  • $\begingroup$ 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 $\endgroup$
    – Rhaena
    Apr 19, 2021 at 9:21
  • $\begingroup$ Thanks! I will edit that. $\endgroup$ Apr 19, 2021 at 19:26

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