# Sum of invertible matrices

I was posed with the question to prove that:

Every square matrix can be written as the sum of 2018 invertible matrices.

My attempt. Since 2018 seemed like a weird number to begin with, my guess was to write the first 2016 with half of them as the identity matrix and other half as $$-1$$ times the identity matrix so that they cancel out. As for the remaining two matrix I thought I could write one as a symmetric matrix and the other as an skew symmetric matrix. But, now i can see that not every symmetric matrix or skew symmetric matrix is invertible. So, can anyone help me out as to how to proceed?

• You need to specify the ring or field that the matrix elements are taken from. The statement is false, for instance, for 1x1 matrices over GF(2). – user1551 Oct 6 '18 at 6:30

Hint. If the given matrix is $$A\in \mathbb{C}^{n \times n}$$ then for a sufficiently large $$\lambda>0$$, $$A-\lambda I$$ is invertible (why?) and $$A=(A-\lambda I)+\lambda I.$$ Now it remains to write $$\lambda I$$ as the sum of $$2017$$ invertible matrices.
• So for any $m \ge 2$, any $A \in M_n(\mathbb{C})$ can written as the sum of $m$ invertible matrices. The same this true in $M_n(\mathbb{R})$. What about other fields? – Paul Frost Oct 6 '18 at 9:30