Suppose we have a matrix $M\in\Bbb Q^{n\times n}$ with no $0$ elements and we write as sum of two matrices $M_1$ and $M_2$ on following constraint.

$M_{1,ij}=M_{ij}$ and $M_{2,ij}=0$ or $M_{1,ij}=0$ and $M_{2,ij}=M_{ij}$ or $M_{1,ij}=M_{2,ij}=\frac{M_{ij}}2$.

If $M$ is of rank $1$ under what conditions on the split can we expect ranks of $M_1$ and $M_2$ to be bound by $O(\log\|M\|_2)$?

  • 2
    $\begingroup$ Why would you expect such a bound or any kind of relation between the rank of the $M_i$ and the norm of $M$? Scaling by a positive factor changes the norm and keeps the ranks of everything unchanged. $\endgroup$ – Arnaud Mortier Jan 10 '18 at 12:21

Take $M$ with all $M_{ij} = 1$, $M_1 = I$, $M_2 = M-I$. Since the eigenvalues of $M$ are $0$ and $n$, $M_1$ and $M_2$ both have rank $n$ if $n > 1$.

In the other direction, if $M_1 = M_2 = M/2$, $\text{rank}(M_1) = \text{rank}(M_2) = \text{rank}(M) = 1$.

  • $\begingroup$ is there no general conditions? $\endgroup$ – Brout Jan 10 '18 at 1:17
  • $\begingroup$ No. And this example can be generalized to the summands each having rank r. In general, for your situation you should not expect much difference between ranks of the two summands if the result is low rank. Gerhard "Invertible Matrices Generate A Lot" Paseman, 2018.01.10. $\endgroup$ – Gerhard Paseman Jan 10 '18 at 12:39

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