distinguishing is equivalent to double nullspace property

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Proposition

$G$ distinguishes $k$ -sparse if and only if $G$ has the $2 k$ -NSP

Proof

(⟸)

(via contrapositive)

Suppose $G$ does not distinguish $k$ -sparse vectors. Then there is some $x \neq x^{'}$ with $| | x | |_{0}, | | x^{'} | |_{0} \leq k$ such that $G x = G x^{'}$ . Note that $| | x - x^{'} | |_{0} \leq 2 k$ , but

G x - G x^{'} = G (x - x^{'}) \neq 0 $ $ T h u s $ G $ d o e s * n o t * h a v e t h e $ 2 k $ N S P .

(⟹)

Now, suppose $G$ does distinguish $k$ sparse vectors. Suppose $| | x | |_{0} \leq 2 k$ and $x \neq 0$ . Then there exist $x^{'}, x^{″}$ $k$ -sparse such that $x = x^{'} - x^{″}$ . Then

0 \neq G x^{'} - G x^{″} = G (x^{'} - x^{″}) = G x

ie $G$ has the $2 k$ NSP.

\begin{matrix} ◼ \end{matrix}

Review

#flashcards/math/rmt

Proposition

$G$ {1||distinguishes $k$ -sparse vectors||action} if and only if $G$ {2||has the $2 k$ -nullspace property||characteristic}

References

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distinguishing is equivalent to double nullspace property

Review

References

See Also

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