update: auto commit

report.tex

@@ -915,7 +915,7 @@ $}
     \begin{equation*}
       \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 output of Mathematica is correct except on a zero-measure set. If we simply check if the eigenvectors are linearly independent, we will miss the condition $a=0$ and get false.
+    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}