$A$ is diagonalizable if the minimal and characteristic polynomial of $A$ are equal.

Question

Problem: The square real matrix $A$ is diagonalizable if the minimal and characteristic polynomial of $A$ are equal.

I think I've got this figured out if I can say that the characteristic polynomial has multiplicity 1 for all of its roots, but I don't know why that's true from what we're supposing here. Can anyone bridge that gap for me? Or is there an easier way to go about it that I'm not seeing?

As has been noted, the stated result is false. It is true that the dimension of each eigenspace must be one (see here for a proof that this is equivalent to characteristic and minimal polynomial being equal), but the equality does not imply that the matrix is diagonalizable. — Arturo Magidin, Dec 14 '20 at 01:37

ahulpke · Answer 1 · 2020-12-14T02:20:38.397

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As stated, this is not true. Take $A=\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right)$. Then minimal and characteristic polynomial of $A$ are both $(x-1)^2$ (and thus equal), but $A$ is not diagonalizable.

A criterion for diagonalizability (over a suitable extension) is that the minimal polynomial is squarefree.

edited Dec 14 '20 at 02:20

answered Dec 14 '20 at 01:33

ahulpke

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... and factors into linear terms over the field we are working on. – Arturo Magidin Dec 14 '20 at 01:35

score 0 · Answer 2 · 2020-12-14T01:39:21.480

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It isn't true. The characteristic and minimal polynomial of $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ are both $x^2$.

edited Dec 14 '20 at 01:39

answered Dec 14 '20 at 01:26

$A$ is diagonalizable if the minimal and characteristic polynomial of $A$ are equal.

2 Answers2