Given
A=
\begin{pmatrix} A & B \\
fractionReduce(A*D,B) & D \end{pmatrix}
.
Determine the entry {\color{red}X} such that
v=
\begin{pmatrix} {\color{red}X} \\
Y \end{pmatrix}
is an eigenvector for the eigenvalue \lambda = 0.
\color{red} X
=
- B/A * Y
Since \lambda = 0, we are looking for {\color{red}X} in v=
\begin{pmatrix} {\color{red}X} \\
Y \end{pmatrix},
such that the matrix-vector product A\cdot v = \lambda \cdot v = 0 is the zero vector.
This translates to the equation
\begin{pmatrix} A & B \\
fractionReduce(A*D,B) & D \end{pmatrix} \cdot
\begin{pmatrix} {\color{red}X} \\
Y \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \end{pmatrix}.
We calculate \begin{pmatrix} A & B \\
fractionReduce(A*D,B) & D \end{pmatrix} \cdot
\begin{pmatrix} {\color{red}X} \\
Y \end{pmatrix} =
\begin{pmatrix} negParens(A) \cdot {\color{red}X} + negParens(B) \cdot negParens(Y) \\
negParens(fractionReduce(A*D,B)) \cdot {\color{red}X} + negParens(D) \cdot negParens(Y)
\end{pmatrix} =
\begin{pmatrix} negParens(A) \cdot {\color{red}X} + B *Y \\
negParens(fractionReduce(A*D,B)) \cdot {\color{red}X} + D *Y \end{pmatrix}.
The two coordinates are zero for {\color{red}X} = fractionReduce(- B*Y,A).