# 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 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. – Beni Bogosel Dec 12 '13 at 10:56
• ... but the characteristic polynomial of a multiple of the identity matrix has only one distinct root. – Robert Israel Dec 12 '13 at 10:58

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.