# Commutation when minimal and characteristic polynomial agree

Hello I am studying for the qualifying exam in algebra and I am having trouble solving this seemingly easy problem. If A is a matrix whose minimal polynomial and characteristic polynomial agree, and B commutes with A then B is a polynomial in A.

I have shown that the dimension of the subspace of polynomials in A must be equal to the dimension of the underlying vector space. Clearly this subspace is contained in the subspace of matrices that commute with A. So if I can show the latter must have dimension less than or equal to the dimension of V, I'll be done. But I don't see how to show that.

Or is there an easier way?

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Have you asked the question you intended? Where does $B$ come from? –  Geoff Robinson Aug 13 '11 at 22:18
The dimension in question (i.e. that of $K[A]$, where $K$ is the underlying field) is always equal to the degree of the minimal polynomial. In your case it's equal to (not less than) the dimension of the underlying vector space. (I agree with Geoff. I'm ignoring $B$ in this comment.) –  Pierre-Yves Gaillard Aug 13 '11 at 22:18
@geoff Whoops. Yes B must commute with A. pierre. That's what I meant to write. Fixed both errors now. –  Joe Aug 13 '11 at 22:20

## 2 Answers

Hint: 1) Show that there is a vector $e$ such that $(e,Ae,A^2e,\dots,A^{n-1}e)$ is a basis (this is another standard exercice).

Write $Be = \sum_{i=0}^{n-1} a_i A^i e$.

2) Show that $B$ and $\sum_{i=0}^{n-1} a_i A^i$ agree on the basis.

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I don't have enough reps to comment, but could you (user10676) or someone else please explain the previous answer a bit further?

I'm not sure what $e$ should be and, even if such a vector $e$ exists, I have a hard time understanding why that implies that $B$ is a polynomial in $A$. How do we use the fact that $AB=BA$?

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