distinguishing is equivalent to double nullspace property

[[concept]]

Topics

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}
 
 

Proposition

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

Proof

$(⟸)$ (via contrapositive) Suppose $G$ does not distinguish $k$-sparse vectors. Then there is some $x\ne x^{\prime }$ with $||x|{|}_0,||x^{\prime }|{|}_0\le k$ such that $Gx=G{x}^{\prime }$. Note that $||x-x^{\prime }|{|}_0\le 2k$, but $Gx-G{x}^{\prime }=G(x-x^{\prime })\ne 0$ Thus $G$ does not have the $2k$ NSP.

$(⟹)$ Now, suppose $G$ does distinguish $k$ sparse vectors. Suppose $||x|{|}_0\le 2k$ and $x\ne 0$. Then there exist $x^{\prime },x^{\prime \prime }$ $k$-sparse such that $x=x^{\prime }-x^{\prime \prime }$. Then $0\ne G{x}^{\prime }-G{x}^{\prime \prime }=G(x^{\prime }-x^{\prime \prime })=Gx$ ie $G$ has the $2k$ NSP.

\tag*{$\blacksquare$}

Review

rmt

Proposition

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

References

References

See Also

Mentions

Mentions

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