If you let $n$ be greater than or equal to $2$ and let $M$ be an $n×n$ matrix with minimal and characteristic polynomial equal to $(x-a)^n$ for some $a$ in the reals. Why would this not be diagonalizable? I've looked online and have mostly seen about Jordan blocks but this is out of the scope of my studies so I'm not sure what I'm supposed to be doing here. Any help would be appreciated!
2026-03-27 02:34:14.1774578854
$n× n$ matrix being non diagonalisable.
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If the characteristic polynomial is $(x-a)^n$, then the only eigenvalue is $a$. If the matrix were diagonalizable, then it must be $M=aI$ (diagonal matrix with all diagonal entries equal to $a$). But then the minimal polynomial can't be $(x-a)^n$ since $x-a$ is a lower degree polynomial with $M-aI=0$.