29. The Characteristic Equation
Dated: 12-06-2025
\[\det(A - \lambda I) = \vec 0\]
Here \(\lambda\) is the eigenvalue and \(I\) is the identity matrix. This equation is also called characteristic polynomial.
Example
Find eigen values of
\[
A
=
\begin{bmatrix}
2 & 3\\
3 & -6
\end{bmatrix}
\]
\[(A - \lambda I) \vec x = \vec 0\]
\[
A - \lambda I
=
\begin{bmatrix}
2 & 3 \\
3 & -6
\end{bmatrix}
-
\begin{bmatrix}
\lambda & 0 \\
0 & \lambda
\end{bmatrix}
=
\begin{bmatrix}
2-\lambda & 3 \\
3 & -6-\lambda
\end{bmatrix}
\]
By definition
\[
\det(A - \lambda I) =
\det
\left(
\begin{bmatrix}
2-\lambda & 3 \\
3 & -6-\lambda
\end{bmatrix}
\right)
= 0
\]
\[
\det
\left(
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\right)
= ad-bc
\]
\(\(\det(A-\lambda I) = (2-\lambda)(-6-\lambda)-(3)(3)\)\)
\(\(= -12+6\lambda-2\lambda+\lambda^2-9\)\)
\(\(= \lambda^2+4\lambda-21\)\)
\(\(\lambda^2+4\lambda-21=0\)\)
\(\((\lambda-3)(\lambda+7)=0\)\)
Hence, the eigen values are \(3\) and \(-7\).
Similarity
\(A_{n \times n}\) and \(B_{n \times n}\) are similar if there exists and invertible matrix \(P\) such that
\[P^{-1}AP = B\]
\[A = PBP^{-1}\]
The act of changing \(A\) into \(P^{-1} A P\) is called a similarity transformation.
Theorem
If \(A_{n \times n}\) and \(B_{n \times n}\) are similar then they have same characteristic polynomial and hence the same eigenvalues.
Row operations on a matrix usually change its eigenvalues.
Application to Dynamic Systems
Let
\[
A =
\begin{bmatrix}
.95 & .03 \\
.05 & .97
\end{bmatrix}
\]
Analyze the behavior of long term dynamical system defined by
\[\vec x_{k + 1} = A \vec x_k \quad k = 0, 1, 2, \ldots\]
with
\[
x_0
=
\begin{bmatrix}
0.6 \\
0.4
\end{bmatrix}
\]
First step is to find eigen values of \(A\) and a basis for each eigenspace.
\[\det(A - \lambda I) = 0\]
\[
0 =
\det
\left(
\begin{bmatrix}
0.95 - \lambda & 0.03 \\
0.05 & 0.97 - \lambda
\end{bmatrix}
\right)
=
(0.95-\lambda)(0.97-\lambda) - (0.03)(0.05)
\]
\(\(= \lambda^2 - 1.92\lambda + 0.92\)\)
Applying quadratic formula
\[\lambda = \frac{1.92 \pm \sqrt{(1.92)^2 - 4(0.92)}}{2} = \frac{1.92 \pm \sqrt{0.0064}}{2} = \frac{1.92 \pm 0.08}{2} = 1 \text{ or } 0.92\]
\[Ax = \lambda x\]
\(\((Ax - \lambda x) = 0\)\)
\(\((A-\lambda I)x = 0\)\)
For \(\lambda = 1\)
\[
\begin{bmatrix}
0.95 & 0.03 \\
0.05 & 0.97
\end{bmatrix}
-
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} = 0
\]
$$
\begin{bmatrix}
-0.05 & 0.03 \
0.05 & -0.03
\end{bmatrix}
\begin{bmatrix}
x_1 \
x_2
\end{bmatrix} = 0
$$
\[-0.05x_1 + 0.03 = 0\]
\[0.05x_1 - 0.03x_2 = 0 \implies x_1 = \frac{0.03}{0.05}x_2 \implies x_1 = \frac{3}{5}x_2\]
In parametric form, it becomes
\[x_1 = \frac{3}{5}t \text{ and } x_2 = t\]
For \(\lambda = 0.92\)
\[
\begin{bmatrix}
0.95 & 0.03 \\
0.05 & 0.97
\end{bmatrix}
-
\begin{bmatrix}
0.92 & 0 \\
0 & 0.92
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} = 0
\]
$$
\begin{bmatrix}
0.03 & 0.03 \
0.05 & 0.05
\end{bmatrix}
\begin{bmatrix}
x_1 \
x_2
\end{bmatrix} = 0
$$
\[0.03x_1 + 0.03x_2 = 0\]
\(\(0.05x_1 + 0.05x_2 = 0 \implies x_1 = -x_2\)\)
\[x_1 = t \text{ and } x_2 = -t\]
Thus, the eigen vectors corresponding to \(\lambda = 1\) and \(\lambda = .92\) are
\[
v_1
=
\begin{bmatrix}
3 \\
5
\end{bmatrix}
\]
\[
v_2
=
\begin{bmatrix}
1 \\
-1
\end{bmatrix}
\]
Next step is to write \(x_0\) in terms of \(v_1\) and \(v_2\).
\(\{v_1, v_2\}\) is basis for \(\mathbb R^2\).
So there exists weights \(c_1\) and \(c_2\) such that
\[
x_0 = c_1 \vec{v}_1 + c_2 \vec{v}_2 =
\begin{bmatrix}
\vec{v}_1 & \vec{v}_2
\end{bmatrix}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}
\]
$$
\begin{bmatrix}
c_1 \
c_2
\end{bmatrix}
=
\begin{bmatrix}
\vec{v}_1 & \vec{v}_2
\end{bmatrix}^{-1} x_0
=
\begin{bmatrix}
3 & 1 \
5 & -1
\end{bmatrix}^{-1}
\begin{bmatrix}
0.60 \
0.40
\end{bmatrix}
$$
\[
\begin{bmatrix}
3 & 1 \\
5 & -1
\end{bmatrix}^{-1}
=
\frac{1}{
\begin{vmatrix}
3 & 1 \\
5 & -1
\end{vmatrix}
}
\text{Adj}
\begin{bmatrix}
3 & 1 \\
5 & -1
\end{bmatrix}
=
\frac{1}{-8}
\begin{bmatrix}
-1 & -1 \\
-5 & 3
\end{bmatrix}
\]
\[
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}
=
\frac{1}{-8}
\begin{bmatrix}
-1 & -1 \\
-5 & 3
\end{bmatrix}
\begin{bmatrix}
0.60 \\
0.40
\end{bmatrix}
=
\begin{bmatrix}
0.125 \\
0.225
\end{bmatrix}
\]
Since \(v_1\) and \(v_2\) are eigen vectors of \(A\) then
\[A v_1 = (1)v_1\]
\[A v_2 = (0.92)v_2\]
\[\vec x_1 = A \vec x_0\]
\[= c_1 A v_1 + c_2 A v_2\]
\[= c_1 v_1 + c_2 (0.92) v_2\]
\[x_2 = A x_1 = c_1 A v_1 + c_2 (0.92) A v_2 = c_1 v_1 + c_2 (0.92)^2 v_2\]
\[\vec{x}_k = c_1 \vec{v}_1 + c_2 (0.92)^k \vec{v}_2 \quad k \ge 0\]
\[
\vec{x}_k
=
0.125
\begin{bmatrix}
3 \\
5
\end{bmatrix}
+
0.225 (0.92)^k
\begin{bmatrix}
1 \\
-1
\end{bmatrix}
\quad k \ge 0
\]
As
\[
k \to \infty \implies (0.92)^k \to 0 \implies
\vec x_k =
\begin{bmatrix}
.375 \\
.625
\end{bmatrix}
=
.125 \vec v_1
\]
References
Read more about notations and symbols.