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Let $ A\in {\mathbb{F} }^{n\times n} $ be a fixed matrix. The set of all matrices that commute with A forms a subring of ${\mathbb{F} }^{n\times n}$.

Is any subring of ${\mathbb{F} }^{n\times n }$ (which contains the identity) of the above form?

Thanks.

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Do you require your subrings to contain the identity matrix? –  Arturo Magidin Jan 7 '11 at 21:24
    
This might help: mathoverflow.net/questions/3270/… –  PEV Jan 7 '11 at 21:26
    
@Arturo - Thanks for your comment, I edited the question. –  Pandora Jan 7 '11 at 21:33
    
I've added a negative response to the obvious generalization. –  Arturo Magidin Jan 7 '11 at 21:51
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up vote 5 down vote accepted

Not if $n\gt 1$. Consider the subring of all scalar multiples of the identity matrix (this is isomorphic to $\mathbb{F}$ itself).

Since the collection of all matrices that commute with $A$ always includes $A$ itself, if the subring of all scalar multiples of the identity were of the form $C(A)$ for some $A$, then $A$ would necessarily be a scalar multiple of the identity; but scalar multiples of the identity are central in $\mathbb{F}^{n\times n}$, so the centralizer of such an $A$ would contain more than just the scalar multiples of the identity.

You might ask more generally whether every subring of $\mathbb{F}^{n\times n}$ of the form $C(S)$ for some $S\subseteq \mathbb{F}^{n\times n}$, where $C(S) = \{M\in\mathbb{F}^{n\times n}\mid MA=AM\text{ for all }A\in S\}$. But even here the answer is still negative; though my example no longer works in this setting (taking $S=\mathbb{F}^{n\times n}$ gives the scalar multiples of the identity), Mariano's example still does. Note that $S\subseteq C(C(S))$; if $T$ is the set of upper triangular matrices, then considering the matrix $E_{ij}$ that has a $1$ in the $(i,j)$ entry, $i\leq j$, and zeros elsewhere, you have that $E_{ij}A$ is the matrix that has zeros everywhere except the $i$th row, where it has the $j$th row of $A$; while $AE_{ij}$ is the matrix that has zeros everywhere except the $j$th column, where it has the $i$th column of $A$. Thus, if $A\in C(T)$, then $A$ must be a scalar multiple of the identity; this means that if $T=C(S)$ for some $S$, then $S$ must be contained in the scalar multiples of the identity, and once again we obtain a contradiction.

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Let $T$ be subring of $M_2(\mathbb F)$ of upper triangular matrices. Is there an $A\in M_2(\mathbb F)$ such that $T$ is the set of matrices which commute with $A$?

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