From c0c4165b5de04498fc7928a424b37b5bdc406fe8 Mon Sep 17 00:00:00 2001
From: Yingjie Wang <phywyj@gmail.com>
Date: Thu, 9 Apr 2026 20:48:20 -0400
Subject: [PATCH] update: auto commit

---
 report.tex | 18 +++++++++++-------
 1 file changed, 11 insertions(+), 7 deletions(-)

diff --git a/report.tex b/report.tex
index a813424..bb67d56 100644
--- a/report.tex
+++ b/report.tex
@@ -289,13 +289,17 @@
     it's very clear that the principal symbol matrix is already symmetric, and thus the system is symmetrically hyperbolic, and well-posed in the $L^2$ sense.
 
     The eigenvalues are $\lambda_1 = \lambda_2 = \lambda_3 = 0$, $\lambda_4 = -1$ and $\lambda_5 = 1$. The eigenvectors are
-    \begin{equation}
-      \myvec{e}_1 = \mqty(1\\0\\0\\0\\0),
-      \myvec{e}_2 = \mqty(0\\0\\-\xi_3\\0\\\xi_1),
-      \myvec{e}_3 = \mqty(0\\0\\-\xi_2\\\xi_1\\0),
-      \myvec{e}_4 = \mqty(0\\1\\\xi_1\\\xi_2\\\xi_3),
-      \myvec{e}_5 = \mqty(0\\-1\\\xi_1\\\xi_2\\\xi_3).
-    \end{equation}
+    {
+      % \fontsize{small}
+      \begin{equation}
+        \myvec{e}_1 = \mqty(1,0,0,0,0)^T,
+        \myvec{e}_2 = \mqty(0,0,-\xi_3,0,\xi_1)^T,
+        \myvec{e}_3 = \mqty(0,0,-\xi_2,\xi_1,0)^T,
+        \myvec{e}_4 = \mqty(0,1,\xi_1,\xi_2,\xi_3)^T,
+        \myvec{e}_5 = \mqty(0,-1,\xi_1,\xi_2,\xi_3)^T.
+      \end{equation}
+    }
+    
   \end{frame}
 
   \begin{frame}{Existing first order formulations}