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A matrix $D$ is said to have the NSP order of $\Omega$ if $$\forall v \neq 0 \text{ with } Dv = 0 \\ \sum_{j \in T} |v_j|< \sum_{j \notin T} |v_j| \text{ for all } T\in \{1, 2, \cdots, M\} \text{ with } \#T \leqslant \Omega $$

In a research paper, I learned that if any two columns of $D$ are a multiple of each other, the NSP has surely violated already for $\Omega = 2$

I don't understand what does it mean the NSP will be violated, does it mean the matrix will not have the NSP of any order if any two columns are correlated?

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Sure: Say the $j$ 'th column of $D$ is $d_j$ and the $j '$ th column is $\lambda d_j$ for some $\lambda\in\mathbb{C}$ . Then $v$ where $v_j=-\lambda$ and $v_{j'}=1$ and all other entries are $0$ is in the null space of $D$. Now take $T=\{j,j'\}$ and You will see that the null space property is violated for $\Omega=2$ and therefore the matrix does not have the NSP for all $\Omega\geq 2$.

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  • $\begingroup$ Thank you! So if two columns of $D$ are correlated, is it not even guaranteed to have the NSP of order 1? Since that depends on $\lambda$. $\endgroup$ Sep 12, 2018 at 8:41
  • $\begingroup$ Let me think:If $|\lambda|\leq 1$ the the NSP is violated by $T=\{j'\}$ and if $|\lambda|>1$ the NSP is violated by $T=\{j\}$. Yes I think Youŕe right it is even not NSP for $\Omega=1$. $\endgroup$ Sep 12, 2018 at 8:47
  • $\begingroup$ Thank you! This is very helpful! $\endgroup$ Sep 12, 2018 at 8:48
  • $\begingroup$ You're welcome! $\endgroup$ Sep 12, 2018 at 8:49

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