What is the maximum number of distinct roots does the characteristic polynomial have? 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?
 A: 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).
A: 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.
