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Let $A$ be a $3\times 3$ matrix with real entries which commutes with all $3\times 3$ matrices with real entries. What is the maximum number of distinct roots that the characteristic polynomial of $A$ can have?

First let me clear what is the matrix that commutes with all $3\times 3$ matrices?

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    $\begingroup$ A characteristic polynomial of a $n \times n$ matrix has degree $n$ so it has at most $n$ solutions. A matrix that commutes with all matrices is a multiple of the identity matrix. $\endgroup$ – Beni Bogosel Dec 12 '13 at 10:56
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    $\begingroup$ ... but the characteristic polynomial of a multiple of the identity matrix has only one distinct root. $\endgroup$ – Robert Israel Dec 12 '13 at 10:58
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Any $n \times n$ matrix $A \in M_{n}(K)$ over any field $K$ (in fact over any integral domain) that commutes with all matrices in $M_n(K)$ will be a scalar multiple of the identity matrix. To see this, you just need to commute $A$ with $n^2$ matrices $E^{i,j}, 1 \le i, j \le n$ whose entries are all zero except at the $i^{th}$ row and $j^{th}$ column and see what happens.

The characteristic polynomial of $A = \lambda I_n$ is given by $$\det( x I_n - A ) = (x-\lambda)^n$$ It is a simple power and has one and only one distinct root.

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Suppose that the characteristic polynomial has more than one distinct real root (so it has three real roots, possibly identical). Write the matrix in a base where it's diagonal, and find another matrix which does not commute with it (hint: a rotation).

Suppose that the characteristic polynomial has only one real root (and two complex ones). Write the matrix in a base where it's a rotation on two axes composed with an expansion on the third axis, and find another matrix that does not commute with it (hint: an expansion on one of the rotated axes).

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  • $\begingroup$ If we work over the reals only, a diagonal matrix may not always be achievable. $\endgroup$ – Hagen von Eitzen Dec 12 '13 at 11:43
  • $\begingroup$ I'll edit, thanks. $\endgroup$ – rewritten Dec 12 '13 at 12:18

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