A nonzero vector \({\mathbf v}\) is an eigenvector of \(A\) if \(A {\mathbf v} = \lambda {\mathbf v}\) for some \(\lambda \in {\mathbf R}\text{.}\) The constant \(\lambda\) is called an eigenvalue of \(A\text{.}\) Letting
\begin{equation*}
A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix} \quad \text{and} \quad
\mathbf v
=
\begin{pmatrix} x \\ y \end{pmatrix} \neq \mathbf 0,
\end{equation*}
we have \(A \mathbf x = \lambda \mathbf v\) or \(A \mathbf v - \lambda \mathbf v = \mathbf 0\text{.}\) In matrix form this is
\begin{align*}
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}
- \lambda \begin{pmatrix} x \\ y \end{pmatrix}
\amp =
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}
- \begin{pmatrix}
\lambda & 0 \\
0 & \lambda
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}\\
\amp =
\begin{pmatrix}
a- \lambda & b \\
c & d - \lambda
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}\\
\amp =
\begin{pmatrix} 0 \\ 0 \end{pmatrix}.
\end{align*}
This matrix equation is certainly true if \((x, y) = (0, 0)\text{.}\) However, we seek nonzero solutions to this system. This will occur exactly when the determinant of
\begin{equation*}
A - \lambda I = \begin{pmatrix}
a- \lambda & b \\
c & d - \lambda
\end{pmatrix}
\end{equation*}
is zero. In this case
\begin{equation*}
\det(A - \lambda I)
=
\det\begin{pmatrix}
a - \lambda & b \\
c & d - \lambda
\end{pmatrix}
=
\lambda^2 - (a + d) \lambda + (ad - bc).
\end{equation*}
We say that
\begin{equation*}
\det(A - \lambda I) = \lambda^2 - (a + d) \lambda + (ad - bc)
\end{equation*}
is the characteristic polynomial of \(A\text{.}\) We summarize the results of this discussion in the following theorem.