Here's the statement that I am to prove/disprove:
If the minimum polynomial of $A$ is $A^5 = 5 A^2$, then $A$ is not diagonalizable.
First, I'm confused about how this can actually be a minimum polynomial. From my understanding, the minimum polynomial is a specific polynomial of least degree which $A$ satisfies. But, couldn't I just multiply both sides of that equation by $A^{-2}$? Then the minimum polynomial would be $A^3 - 5I = 0$, which is of lesser degree. That's really confusing for me!!
Anyway, I continue with this problem focusing on the given polynomial. I know that $A$ is at least 5x5, because the minimum polynomial is of degree 5. However, there are only two distinct roots of the minimum polynomial:
$A^5 - 5A^2 = 0 \Longleftrightarrow A^2 (A^3 - 5I) = 0 \Longrightarrow$ eigenvalues: $0,\,5^{1/3}$.
Is that correct? If there were 5 distinct roots, and I knew $A$ were exactly 5x5, then this question would be cake, and it's definitely diagonalizable. Instead, all I know is that there are only 2 eigenvalues, which does not necessarily mean that these are no more than 2 eigenvectors, so $A$ may or may not be diagonalizable (e.g., when looking at roots of $I$).
Any help is appreciated.
Thanks!