Given
A=
\begin{pmatrix} A & {\color{red}b} \\
{\color{teal}c} & D \end{pmatrix}
.
Find the entries {\color{red}b} and {\color{teal}c} such that
v=
\begin{pmatrix} X \\
Y \end{pmatrix}
is an eigenvector for the eigenvalue \lambda = 0.
\color{red} b
=
B
\color{teal} c
=
C
We are looking for {\color{red}b} and {\color{teal}c}
with A = \begin{pmatrix} A & {\color{red}b} \\
{\color{teal}c} & D \end{pmatrix} and
v= \begin{pmatrix} X \\
Y \end{pmatrix},
such that the matrix-vector product A\cdot v = \lambda \cdot v = {\color{blue}0} equals the
zero vector, as \lambda = 0.
We calculate die left-hand side and get
\begin{pmatrix} A & {\color{red}b} \\
{\color{teal}c} & D \end{pmatrix} \cdot
\begin{pmatrix} X \\
Y \end{pmatrix} =
\begin{pmatrix} negParens(A) \cdot negParens(X) + Y {\color{red}b} \\
{\color{teal}c} \cdot negParens(X) + negParens(D) \cdot negParens(Y)
\end{pmatrix} =
{\color{blue}
\begin{pmatrix} A*X + Y {\color{red}b} \\
X {\color{teal}c} + D *Y \end{pmatrix}
}.
Both equations
\color{blue}A*X + Y {\color{red}b} = 0 and
\color{blue}X {\color{teal}c} + D *Y =0
deliver
{\color{red}b} =
fractionReduce(- A*X,Y) und {\color{teal}c} = fractionReduce(- D*Y,X).