Let $n\geq 2$. Is it true that any $n\times n$ matrix with entries from a given ring (with identity) can be written as a sum of two invertible matrices with entries from the same ring ?

2$\begingroup$ You might want to consider the answer in the $n=1$ case as an example of how you could handle the general case. I've voted to close. $\endgroup$ – Ryan Budney Sep 5 '13 at 21:23

1$\begingroup$ Use upper triangular matrices with non zero diagonals $\endgroup$ – Alexandre Eremenko Sep 5 '13 at 22:05

2$\begingroup$ @Alexandre Eremenko: What if the ring is not a division ring? Additionally, over a field with two elements, the identity matrix can only be written as a sum of two invertible matrices as follows: $\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix} + \begin{pmatrix}0 & 1 \\ 1 & 1\end{pmatrix}$ $\endgroup$ – Ricardo Andrade Sep 5 '13 at 22:22

1$\begingroup$ I agree that my proposal does not work for all rings. $\endgroup$ – Alexandre Eremenko Sep 5 '13 at 22:28

2$\begingroup$ It is true for many rings. The terminology is somethig like "clean".There is also some USENET news posts on Sums of Invertible Matrices. A web search may help further. $\endgroup$ – The Masked Avenger Sep 6 '13 at 1:06
The answer is negative.
There is a nice theorem of M. Henriksen which says that If $n\geq 2$ then every element of $M_n(R)$ is a sum of three units also he proves that there are nonunit matrices in $\bf{M_2(\Bbb{Z}_2[x_1,x_2])}$ that can not be written as a sum of two units. You can find a copy of the article HERE
This may help. Any $2\times 2$ matrix is a sum of four units:
$$ \begin{pmatrix} x & y \\ z & t \end{pmatrix} = \begin{pmatrix} x & 1 \\ 1 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 1 & t \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ z & 1 \end{pmatrix} + \begin{pmatrix} 1 & y \\ 0 & 1 \end{pmatrix}$$

1$\begingroup$ If e.g. z and y are each the sum of two units, you can reduce this to two summands. $\endgroup$ – The Masked Avenger Sep 6 '13 at 17:17

$\begingroup$ +1 For the nice identity. $\endgroup$ – user38122 Sep 6 '13 at 18:47