From 2e09570e8fdbe696c38427f9b6c5451717d5ba4f Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Tue, 14 Apr 2020 13:50:40 -0400
Subject: I think I have a homogenous solution

---
 execsumm/document.tex | 21 +++++++++++++++++++--
 1 file changed, 19 insertions(+), 2 deletions(-)

diff --git a/execsumm/document.tex b/execsumm/document.tex
index f3cf093..97270f1 100644
--- a/execsumm/document.tex
+++ b/execsumm/document.tex
@@ -34,14 +34,31 @@ $$-\lambda{(\lambda-{-b+\sqrt{b^2-4c}\over2})
 %           - 2C_1C_2\rload(3R_1+\rload).$$
 Let $r_1$ and $r_2$ designate these two non-zero roots.
 
+The trivial zero eigenvalue corresponds to a unit x-direction vector
+by inspection. The two remaining roots, in the general case of a non-%
+degenerate system, which hasn't been explicitly ruled out, the middle
+row can be ``ignored'' because it is linearly independent in the
+following system:
+$$
+\pmatrix{-r&-C_2\rload &0\cr
+         * &*          &*\cr
+         0 &C_2R_1     &-r-C_1R_1}
+$$
+With $x = 1$, $\displaystyle y = -{r\over C_2\rload}$ and
+$\displaystyle z = y{C_2R_1\over r+C_1R_1}.$
+
 \def\bu{\par\leavevmode\llap{\hbox to \parindent{\hfil $\bullet$ \hfil}}}
 \noindent The eigenvalues and respective eigenvectors are:
 
 \bu $\lambda_1 = 0, v_1 = \pmatrix{1\cr0\cr0}$
 
-\bu $\lambda_2 = {r_1\over C_1C_2R_1\rload}, v_2 = ??$
+\def\num#1{\pmatrix{1\cr
+                    -{r_#1\over C_2\rload}\cr
+                    -{R_1r_#1\over \rload(r_#1+C_1R_1)}}}
+
+\bu $\displaystyle\lambda_2 = {r_1\over C_1C_2R_1\rload}, v_2 = \num1.$
 
-\bu $\lambda_3 = {r_2\over C_1C_2R_1\rload}, v_3 = ??$
+\bu $\displaystyle\lambda_3 = {r_2\over C_1C_2R_1\rload}, v_3 = \num2.$
 
 \section{Nonhomogenous System}
 
-- 
cgit