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Let $F$ be a field and $M_n(F)$ be the $n\times n$ matrix algebra. Given a matrix $A\in M_n(F)$, under what condition there exists a matrix $B$ such that $A$ and $B$ generate $M_n(F)$? (Is the condition where the characteristic polynomial of $A$ has $n$ distinct roots in a splitting field a sufficient condition?)

REMARK: Let $D = \sum_1^{n-1}e_{i,i+1} + e_{n,1}$, suppose we know that if $A$ is diagonal then $A$ and $D$ genarate $M_n(F)$(for any field $F$), hence in general if the characteristic polynomial of $A$ has $n$ distinct roots in a splitting field $E/F$, we therefore can pick P $\in$ $M_n(E/F)$ sunch that $PAP^{-1}$ is diagonal and with $D$ they generate $M_n(E/F)$ but $P^{-1}DP$ might $\notin$ $M_n(F)$, if $PD = DP$ then $P^{-1}DP = D$ then $A$ and $D$ generate $M_n(F)$. but the condition $PD = DP$ seems redundant.

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Presumably, the answer is yes. Note first the theorem of Burnside, which says that (over an algebraically closed field) the only irreducible matrix sub-algebra of $M_n$ is $M_n$ itself,as long as $n > 1,$ which seems to indicate that if $A$ and $B$ have no eigenspaces in common, then $A$ and $B$ generate all of $M_n.$ In your case, we can assume that $A$ is diagonal, so finding a $B$ which fixes no coordinate subspace is easy (for example, a rotation about a random vector).

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  • $\begingroup$ thanks!!! but I don't know what is "first the theorem of Burnside" and "irreducible matrix sub-algebra ", I will check them out. $\endgroup$ – sunya hu Mar 28 '17 at 5:46
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That if $A$ and $B$ have no eigenspaces in common, then they generate all of $M_n(F)$ is not true. Because they may still have non trivial invariant subspace in common. This implies that the algebra generated by $A$ and $B$ is not simple. So they can not generate full matrix algebra which is simple.

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