1
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ 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). $\endgroup$
    – user1551
    Oct 6, 2018 at 6:30

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$
    – Paul Frost
    Oct 6, 2018 at 9:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .