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Is is true that if $A \in M_n(\mathsf{k})$ is arbitrary, for some field $\mathsf{k}$, then there exists two sequences $(L_i)_{i=1}^{m}, (R_i)_{i=1}^{m} \subseteq M_n(\mathsf{k})$ such that

$$ \sum_{i=1}^{m} L_i A R_i = I_n, $$

where $I_n$ is the $n \times n$ identity matrix?


I think this is true, but the proof I have in mind is fairly abstract and I was wondering if there is a direct-ish way of proving this (if it is indeed true).


Edit: As @Michael Burr points out, I forgot the condition that $A \neq 0$.

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    $\begingroup$ Does $A$ need to be nonzero? You could use the SNF to reduce this to a simpler case. $\endgroup$ – Michael Burr May 29 at 16:43
  • $\begingroup$ @MichaelBurr Yup! Thanks for catching my error. And ah yeah that dramatically simplifies it. Thanks! $\endgroup$ – Adam Higgins May 29 at 16:46
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Yes, because if $A$ is a nonzero element of $M_n(k)$, then the ideal generated by $A$ has to be all of $M_n(k)$ because it is a simple ring.

The ideal generated by $A$ is $M_n(k)AM_n(K)$, whose elements have exactly the form you required.

If you want to re-prove it from first elements, then just note that you can isolate a nonzero entry of $A$, "cut down" the matrix to that one nonzero entry, and then scale the entry and move it to any other position you want by using elementary row operations.

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