update: auto commit

report.tex

@@ -916,7 +916,10 @@ $}
       \mqty(1 & a \\ 0 & 1),
     \end{equation*}
     if we ask Mathematica to compute the eigenvectors, it will give us $\myvec{v}_1 = \mqty(1 \\ 0)$ and $\myvec{v}_2 = \mqty(0 \\ 0)$, because unless $a=0$, this matrix have a Jordan block. So the geometric multiplicity is less than the algebraic multiplicity except on a zero-measure set, that's why Mathematica thinks there is no second linearly independent eigenvector. If we simply check if the eigenvectors are linearly independent, we will miss the condition $a=0$ and get false.
+  \end{frame}
 
+  \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 $i$.
   \end{frame}
 
 \end{document}