# Sum of left and right products of any square matrix can be the identity.

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 $$A \neq 0$$.

• Does $A$ need to be nonzero? You could use the SNF to reduce this to a simpler case. – Michael Burr May 29 at 16:43
• @MichaelBurr Yup! Thanks for catching my error. And ah yeah that dramatically simplifies it. Thanks! – Adam Higgins May 29 at 16:46

## 1 Answer

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.