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Assume we have rank $r$ real matrix $M\in\{0,1\}^{n\times n}$ (not constrained to symmetric).

Assume $W\in\{0,1\}^{n\times n}$ is rank $1$ real matrix.

Case $1$: $M+W\in\{0,1\}^{n\times n}$.

Could there be tight upper, lower bounds on $rank(M+W)-rank(M)$ depending on only $r$?

Case $2$: $M-W\in\{0,1\}^{n\times n}$.

Could there be tight upper, lower bounds on $rank(M-W)-rank(M)$ depending on only $r$?


Above can be reduced to rank $2$ matrices $W$ with symmetric cases.

Matrix $$0\mbox{ }M'$$$$M\mbox{ }0$$ is symmetric and has rank $2r$.

Matrix $$0\mbox{ }W'$$$$W\mbox{ }0$$ is symmetric and has rank $2$.

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  • $\begingroup$ Btw, why the extremal-combinatorics tag? Is there an interesting application you have in mind? $\endgroup$ – Felix Goldberg Feb 3 '15 at 6:33
  • $\begingroup$ Motivation comes from combinatorics. All 0-1 objects as far as I know have connections there. $\endgroup$ – Turbo Feb 3 '15 at 6:34
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THe interlacing theorem tells us that for rank 1 $W$ the eigenvalues of $M \pm W$ are sandwiched between those of $M$ (see Corollary 4.3.9 here). Therefore $null(M)-1 \leq nullity(M \pm W) \leq null(M)+1$ and this is equivalent to $r(M)-1 \leq r(M \pm W) \leq r(M)+1$.

These bounds are tight. For the case of graphs there are some detailed studies of when the different cases obtain.

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  • $\begingroup$ @Turbo It's proved in the reference I gave (the standard one on the subject). $\endgroup$ – Felix Goldberg Feb 3 '15 at 6:34
  • $\begingroup$ @Turbo The magic works because it's a rank 1 update. It's possible to use interlacing for rank $k$ updates, but the bounds get progressively weaker, of course. The proof for rank $k$ is just an inductive application of Corollary 4.3.9 $k$ times (decompose $W$ as the sum of $k$ matrices of rank $1$). $\endgroup$ – Felix Goldberg Feb 3 '15 at 6:36
  • $\begingroup$ @Turbo Yes, all you need is Hermitianness of $M$ and $W$. But as I said, having $\{0,1\}$ helps to perhaps pinpoint precisely which of the three cases occurs. $\endgroup$ – Felix Goldberg Feb 3 '15 at 6:37
  • $\begingroup$ @Turbo In that case, things are open again.... :( If you want to discuss the specific case, feel free to email me. $\endgroup$ – Felix Goldberg Feb 3 '15 at 6:39
  • $\begingroup$ Problem reduces to rank $2$ in symmetric case. $\endgroup$ – Turbo Feb 3 '15 at 7:03

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