From b95f42e73e36375da79069dfbfac86e974a54766 Mon Sep 17 00:00:00 2001 From: Yu Cong Date: Wed, 26 Nov 2025 17:34:58 +0800 Subject: [PATCH] z --- notes.tex | 8 +++++--- 1 file changed, 5 insertions(+), 3 deletions(-) diff --git a/notes.tex b/notes.tex index 618d8cf..6926e72 100644 --- a/notes.tex +++ b/notes.tex @@ -70,14 +70,16 @@ Let $M$ be a binary matroid with binary representation $B\in \F_2^{n\times m}$. \end{lemma} \begin{proof} Let $M'$ be $M\left(\begin{bmatrix}B\\ \sigma\end{bmatrix}\right)$ and let the ground set be $E$. -$\lambda(M')\leq \lambda(M)$ (row space). -Let $F^*\in \argmin_{F\subset E} \frac{|E-F|}{r'(E)-r'(F)}$. +We have $\lambda(M')\leq \lambda(M)$ since the row space becomes larger. + +Let $F^*\in \argmin_{F\subset E} \frac{|E-F|}{r'(E)-r'(F)}$. Let $r$ be the rank of $M$ and let $r'$ be the rank of $M'$. +We can assume that $r(E)-r(F^*)\geq 1$ since otherwise we have $r'(E)-r'(F^*)\leq 1$ which implies the gap of $M'$ is 1. \begin{equation*} \begin{aligned} \sigma(M') &= \frac{|E-F^*|}{r'(E)-r'(F^*)}\\ &\geq \frac{|E-F^*|}{r(E)+1-r(F^*)}\\ - &\geq \frac{|E-F^*|}{2(r(E)-r(F^*)}\\ + &\geq \frac{|E-F^*|}{2(r(E)-r(F^*))}\\ &\geq \frac{\sigma(M)}{2} \end{aligned} \end{equation*}