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let the matrix $A\in M_{2}(C)$,and $A\neq aI$,and define $$S=\{B\in M_{2}(C)|AB=BA\}$$,if $X,Y\in S$ show that $$XY=YX$$

My try: since $X,Y\in S$,then $$AX=XA$$ $$AY=YA$$ then How can prove $$XY=YX$$ Thank you

math110
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  • If $X$ and $Y$ commute with $A$ then they also commute with $A^{-1}$... – Giovanni De Gaetano Jan 02 '14 at 08:35
  • @GiovanniDeGaetano it is true that one can assume without loss of generality that $A$ is invertible, but I'm not really sure where you were going with that? – Dustan Levenstein Jan 02 '14 at 09:01
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    For a generalization, it is worth noting that if $A$ is an $n\times n$ matrix whose minimal polynomial is of degree $n$, then $S$ will be the collection of polynomials in $A$, and the result follows from $p(A)q(A)=(pq)(A)=q(A)p(A)$. See http://math.stackexchange.com/q/326293/9863 – Aaron Jan 02 '14 at 09:25

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By a change of basis, we may assume that $A$ is in its Jordan form.

If $A$ is a diagonal matrix, since it is not a multiple of $I$, the two diagonal entries are distinct. Thus every $B$ that commutes with $A$ is a diagonal matrix. Since diagonal matrices commute, the result follows.

If $A$ is not diagonalisable, then it is a Jordan block. Thus every $B$ that commutes with $A$ is an upper triangular Toeplitz matrix. Since upper triangular Toeplitz matrices commute, the result follows.

user1551
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