update: auto commit

report.tex

@@ -919,7 +919,9 @@ $}
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   \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 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 have the same characteristic speed $\shift{^\xi}$
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