From 28055b9c8a840b539f74dfad291ab1dc986486a7 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Fri, 26 May 2023 18:32:42 +0200
Subject: Clarified conditions for nontrivial isolated eigenvalue solution.

---
 2-point.tex | 12 +++++++++++-
 1 file changed, 11 insertions(+), 1 deletion(-)

(limited to '2-point.tex')

diff --git a/2-point.tex b/2-point.tex
index 3bacf89..af0af87 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -1173,7 +1173,17 @@ assuming the last equation is satisfied. The trivial solution, which gives the b
 \[
   \lambda_\mathrm{min}=\mu-\sqrt{4f''(1)}=\mu-\mu_\mathrm m
 \]
-as expected. We need to first the nontrivial solutions with nonzero $X$, but because the coefficients are so nasty this will be a numeric problem... Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is zero.
+as expected. We need to first the nontrivial solutions with nonzero $X$, but
+because the coefficients are so nasty this will be a numeric problem...
+Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is
+zero. In this case, if the associated normalized eigenvector is $\hat X$, its magnitude is set by
+\begin{equation}
+  \|X\|^2=\frac1{\hat X^TC\hat X}\left(1-\frac{f''(1)}{y^2}\right)
+\end{equation}
+In practice, we find that $\hat X^TC\hat X$ is positive. Therefore, for the
+solution to make sense we must have $y^2>f''(1)$. In practice, there is at most
+\emph{one} $y$ which produces a zero eigenvalue of $B-yC$ and satisfies this
+inequality, so the solution seems to be unique.
 
 \section{Conclusion}
 
-- 
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