From e271872cf62156a294a9f8fae007694dee09581a Mon Sep 17 00:00:00 2001
From: Yingjie Wang <phywyj@gmail.com>
Date: Fri, 10 Apr 2026 03:24:26 -0400
Subject: [PATCH] update: auto commit

---
 report.tex | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/report.tex b/report.tex
index 061a82d..1ec71a6 100644
--- a/report.tex
+++ b/report.tex
@@ -921,7 +921,7 @@ $}
   \begin{frame}{Appendix}{Why can't we fully automate the workflow using Mathematica?}
     The correct way is to check the geometric multiplicity for each eigenvalues. Say we get eigenvalues $\lambda_1, ..., \lambda_k$ with the algebraic multiplicity $m_1, ..., m_k$. Under the assumption that all these eigenvalues are different from each other, we check $\dim\ker(A - \lambda_i I)$ for each $i$. $\dim\ker(A - \lambda_i I) \leq m_i$, so we just have to find the condition for $\dim\ker(A - \lambda_i I) \geq m_i$. $\dim\ker(A - \lambda_i I) = n - \rank(A - \lambda_i I)$, so we check $\rank(A - \lambda_i I) \leq n - m_i$. The criterion is that any $(n-m_i+1)\times(n-m_i+1)$ minor of $A - \lambda_i I$ is zero.
 
-    Nice, this is a computable condition! But physically we expect most modes to have the same characteristic speed $\shift{^\xi}$, resulting a algebraic multiplicity $\sim 22$, while the size of matrix is 55, then the number of minors to check is
+    Nice, this is a computable condition! But physically we expect most modes to have the same characteristic speed $\shift{^\xi}$, resulting a algebraic multiplicity $\sim 22$, while the size of matrix is 55, then the number of minors to check for $\lambda = \shift{^\xi}$ is
     \begin{equation*}
       \binom{55}{34}^2 = 708507400701023142362023505625 = 7.085\times 10^{29}.
     \end{equation*}