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Let $f$ be a field, $A\in Mat_{n,n}(f)$ s.t. $A$ is not decomposable. I want to show (directly!) that the characteristic polynomial $p_A$ of $A$ is equal to the minimal polynomial $m_A$ of $A$.

$A$ is decomposable $\Leftrightarrow\exists V,W: f^n=V \oplus W$ with $V$ and $W$ are $A$-invariant and nonzero.

$V$ subspace of $f^n$ is $A$-invariant $\Leftrightarrow A(V)\subset V$

gofa
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  • What do you mean by directly ? – Rene Schipperus Dec 04 '17 at 22:58
  • Perhaps look at https://math.stackexchange.com/questions/81467/when-are-minimal-and-characteristic-polynomials-the-same for help. – Luke Peachey Dec 04 '17 at 23:13
  • @ReneSchipperus Well I can't tell you rigorously what I mean, but the proof shouldn't use something as strong as the jordan normal form. I would guess that this is straightforward to proof, but I'm obviously missing something... – gofa Dec 04 '17 at 23:18
  • If the field is complete then its easy, but if its general it seems about as hard as the rational form theorem. – Rene Schipperus Dec 04 '17 at 23:22

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