update: auto commit

report.tex

@@ -289,16 +289,15 @@
     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
-    {
-      % \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.
+        \begin{split}
+          \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{split}
       \end{equation}
-    }
     
   \end{frame}