{\color{blue}\lambda_1} =
{\color{red}X} =
Given
A =
\begin{pmatrix} A & B\\
C & D \end{pmatrix}.
Determine the eigenvalues {\color{blue}\lambda_{1,2}}
and the missing entry {\color{red}X}, such that
\begin{pmatrix} {\color{red}X} \\
Y \end{pmatrix}
is an eigenvector for the eigenvalue {\color{blue}\lambda_{1,2}} is.
{\color{blue}\lambda_1} =
{\color{red}X} =
{\color{blue}\lambda_2} =
{\color{red}X}=
We are looking for {\color{blue}\lambda} and {\color{red}X}
satisfying A \cdot \begin{pmatrix} {\color{red}X} \\ Y \end{pmatrix} =
{\color{blue}\lambda} \cdot \begin{pmatrix} {\color{red}X} \\ Y \end{pmatrix} .
At first, we determine {\color{blue}\lambda} as roots of the characteristic
polynomial
\lambda^2 - L1+L2 \cdot \lambda + L1 *L2 =
(\lambda - L1) \cdot (\lambda - L2). This shows that we have the two eigenvalues {\color{blue}\lambda} = L1 and {\color{blue}\lambda} = L2.
We calculate \begin{pmatrix} A & B\\
C & 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(C) \cdot {\color{red}X} + negParens(D) \cdot negParens(Y)
\end{pmatrix} =
\begin{pmatrix}
negParens(A) \cdot {\color{red}X} + B*Y \\
negParens(C) \cdot {\color{red}X} + D*Y
\end{pmatrix}
and consider first {\color{blue}\lambda} = L1.
Thus, we are looking for {\color{red}X} with
\begin{pmatrix}
A {\color{red}X} + B*Y \\
C {\color{red}X} + D*Y
\end{pmatrix} = {\color{blue}L1} \cdot
\begin{pmatrix} {\color{red}X} \\ Y \end{pmatrix}.
With the 2nd coordinate
C {\color{red}X} + D*Y
= L1*Y we get
{\color{red}X} = fractionReduce((L1-D)*Y,C).
These values {\color{red}X} = fractionReduce((L1-D)*Y,C)
and {\color{blue}\lambda} = L1
guarantee that the 1st equation fractionReduce(A*(L1-D)*Y + B*Y*C,C) =
fractionReduce(L1*(L1-D)*Y,C)
holds.
Similarly, we get {\color{red}X} = fractionReduce((L2-D)*Y,C)
for the eigenvector corresponding to the eigenvalue {\color{blue}\lambda} = L2.