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PAGE 1

Problem 17: Eigenvalue and Eigenvector Calculation

Given the matrix:

\[ \begin{bmatrix} 3 & -2 \\ 4 & -1 \end{bmatrix} \]

To find the eigenvalues, we solve the characteristic equation \( \det(A - \lambda I) = 0 \):

\[ \begin{vmatrix} 3 - \lambda & -2 \\ 4 & -1 - \lambda \end{vmatrix} = 0 \]
\[ (3 - \lambda)(-1 - \lambda) + 8 = 0 \]
\[ \lambda^2 - 2\lambda + 5 = 0 \]

Using the quadratic formula:

\[ \lambda = \frac{2 \pm \sqrt{4 - 20}}{2} = 1 \pm 2i \]

Finding Eigenvectors for \( \lambda = 1 + 2i \)

We solve \( (A - \lambda I)\vec{x} = \vec{0} \):

\[ \begin{bmatrix} 2 - 2i & -2 & \bigm| & 0 \\ 4 & -2 - 2i & \bigm| & 0 \end{bmatrix} \]

Multiply row 1 by \( 2 + 2i \):

\[ \begin{bmatrix} 8 & -4 - 4i & \bigm| & 0 \\ 4 & -2 - 2i & \bigm| & 0 \end{bmatrix} \]

Row reduction leads to:

\[ \begin{bmatrix} 2 & -1 - i & \bigm| & 0 \\ 0 & 0 & \bigm| & 0 \end{bmatrix} \]

Let \( x_2 = r \). Then:

\[ 2x_1 = (1 + i)x_2 \]

Choose \( x_2 = 1 \) or \( 1 - i \). The eigenvector \( \vec{x} \) is:

\[ \vec{x} = \begin{bmatrix} \frac{1+i}{2} \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 - i \end{bmatrix} \]
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7.5 Homogeneous Systems with Constant Coefficients

Solve \( \vec{x}' = A\vec{x} \) where \( A \) is a constant matrix.

If this were a scalar equation \( y' = ay \), then the solution is in the form of \( y = e^{rt} \) (because the exponential is related to its own derivative).

For the system \( \vec{x}' = A\vec{x} \), an exponential is expected, but it has to be a vector. Use the form:

\[ \vec{x} = \vec{v}e^{rt} \]

where \( \vec{v} \) is some constant vector.

Plug into the Differential Equation (DE):

\[ \vec{x}' = \vec{v}re^{rt} \]
\[ \vec{v}re^{rt} = A\vec{v}e^{rt} \]

Divide by \( e^{rt} \) (because \( e^{rt} \neq 0 \)):

\[ A\vec{v} = r\vec{v} \]
\[ (A - rI)\vec{v} = \vec{0} \]
\[ \det(A - rI) = 0 \]

This is the same as the equation we solve to find eigenvalues and eigenvectors. So, \( r \) must be an eigenvalue and \( \vec{v} \) must be an eigenvector.

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Linearly Independent Solutions and Systems

There are n linearly independent solutions \( \vec{x}^{(n)} = \vec{v}_n e^{\lambda_n t} \) because there are \( n \) eigenvalue/eigenvector pairs.

General Solution:

\[ \vec{x}(t) = c_1 \vec{v}_1 e^{\lambda_1 t} + c_2 \vec{v}_2 e^{\lambda_2 t} + \dots + c_n \vec{v}_n e^{\lambda_n t} \]

Example

\[ y'' + 5y' + 6y = 0 \]

Characteristic eq. \[ r^2 + 5r + 6 = 0 \implies r = -2, \, r = -3 \]

So \[ y = c_1 e^{-2t} + c_2 e^{-3t} \] \[ y' = c_1 \cdot -2e^{-2t} + c_2 \cdot -3e^{-3t} \]

Change to System:

\[ x_1 = y \]\[ x_2 = y' \]
\[ x_1' = x_2 \]\[ x_2' = -6x_1 - 5x_2 \]
\[ \vec{x}' = \begin{bmatrix} 0 & 1 \\ -6 & -5 \end{bmatrix} \vec{x} \]
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Eigenvalues:

\[ \begin{vmatrix} -\lambda & 1 \\ -6 & -5-\lambda \end{vmatrix} = 0 \]

\[ (-\lambda)(-5-\lambda) + 6 = 0 \]\[ \lambda^2 + 5\lambda + 6 = 0 \quad \text{same as characteristic eq.} \]\[ \lambda = -2, \, \lambda = -3 \]

Find Eigenvectors

\( \lambda_1 = -2 \)

\[ (A - \lambda I)\vec{v} = \vec{0} \quad \det(A - \lambda I) = 0 \]

\[ \left[ \begin{array}{cc|c} 2 & 1 & 0 \\ -6 & -3 & 0 \end{array} \right] \]
\[ \left[ \begin{array}{cc|c} 2 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right] \]

\[ v_2 = r \quad 2v_1 + v_2 = 0 \quad v_1 = -\frac{1}{2}v_2 = -\frac{1}{2}r \]

Choose \( r = -2 \)

\[ \vec{v}_1 = \begin{bmatrix} 1 \\ -2 \end{bmatrix} \]

Note: first element is often made to be 1 by choosing r

\( \lambda_2 = -3 \)

\[ \vec{v}_2 = \begin{bmatrix} 1 \\ -3 \end{bmatrix} \]
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General Solution and Fundamental Matrix

The general solution is given by:

\[ \vec{x} = c_1 \begin{bmatrix} 1 \\ -2 \end{bmatrix} e^{-2t} + c_2 \begin{bmatrix} 1 \\ -3 \end{bmatrix} e^{-3t} \]

Recall that:

\[ \vec{x} = \begin{bmatrix} y \\ y' \end{bmatrix} \]

Component Equations

  • 1st row: \( y = c_1 e^{-2t} + c_2 e^{-3t} \)
  • 2nd row: \( y' = c_1 \cdot -2e^{-2t} + c_2 \cdot -3e^{-3t} \)

If we collect the solutions and put them into a matrix, this matrix is called the fundamental matrix.

\[ \Phi(t) = \begin{bmatrix} e^{-2t} & e^{-3t} \\ -2e^{-2t} & -3e^{-3t} \end{bmatrix} \]
\[ \det[\Phi(t)] = \text{Wronskian} \]
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Example: Solving a System of Differential Equations

\[ \vec{x}' = \begin{bmatrix} 3 & -2 \\ 4 & -3 \end{bmatrix} \vec{x} \]
\[ \vec{x}(0) = \begin{bmatrix} 3 \\ 0 \end{bmatrix} \]

Finding Eigenvalues

\[ \begin{vmatrix} 3-\lambda & -2 \\ 4 & -3-\lambda \end{vmatrix} = 0 \]

\( (3-\lambda)(-3-\lambda) + 8 = 0 \)

\( \lambda^2 - 1 = 0 \)

\( \lambda = \pm 1 \)

Finding Eigenvectors

For \( \lambda_1 = 1 \):

\( (A - \lambda I)\vec{v} = \vec{0} \)

\[ \left[ \begin{array}{cc|c} 2 & -2 & 0 \\ 4 & -4 & 0 \end{array} \right] \rightarrow \left[ \begin{array}{cc|c} 1 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right] \]

Note: The matrix above represents \( (A - \lambda I) \).

\[ \vec{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

For \( \lambda_2 = -1 \):

\[ \left[ \begin{array}{cc|c} 4 & -2 & 0 \\ 4 & -2 & 0 \end{array} \right] \]
\[ \vec{v}_2 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]
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General Solution and Phase Plot

The general solution for the system is given by:

\[ \vec{x}(t) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^t + c_2 \begin{bmatrix} 1 \\ 2 \end{bmatrix} e^{-t} \]

Given the initial condition \( \vec{x}(0) = \begin{bmatrix} 3 \\ 0 \end{bmatrix} \), we set up the following system of equations:

\[ \vec{x}(0) = \begin{bmatrix} 3 \\ 0 \end{bmatrix} = \begin{bmatrix} c_1 + c_2 \\ c_1 + 2c_2 \end{bmatrix} \]

Solving for the constants:

\( 3 = -c_2 \implies c_2 = -3 \)
\( c_1 = 6 \)

Substituting these values back into the general solution:

\[ \vec{x}(t) = \begin{bmatrix} 6 \\ 6 \end{bmatrix} e^t - \begin{bmatrix} 3 \\ 6 \end{bmatrix} e^{-t} = \begin{bmatrix} 6e^t - 3e^{-t} \\ 6e^t - 6e^{-t} \end{bmatrix} \]

Phase Plot: \( x_1 \) vs \( x_2 \)

For the system \( \vec{x}' = A\vec{x} \), the origin \( \vec{x} = \vec{0} \) is an equilibrium point.

We analyze the behavior along the eigenvectors:

  • \( \vec{x}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^t \): Corresponds to a positive eigenvalue (\( \lambda = 1 \)). Solutions move AWAY from equilibrium.
  • \( \vec{x}_2 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} e^{-t} \): Corresponds to a negative eigenvalue. Solutions move TOWARD equilibrium.
Phase plot in the x1-x2 plane showing trajectories moving toward or away from the origin along eigenvectors.
PAGE 8

Vector Field Visualization

The following diagram illustrates the vector field for the system, showing the direction and magnitude of the flow at various points in the phase plane.

A detailed vector field plot on a Cartesian grid with axes ranging from -4.75 to 4.75.

The grid lines and arrows provide a numerical overview of the system's dynamics, highlighting the saddle-point behavior near the origin where trajectories diverge along one axis and converge along another.

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Page 9

Equilibrium Types: Saddle Points

This kind of equilibrium is called a saddle point (stable in some directions, unstable in others).

→

eigenvalues are real and of opposite signs

Example

\[ \vec{x}' = \begin{bmatrix} 0 & 1 \\ -6 & -5 \end{bmatrix} \vec{x} \quad \text{has e-values} \quad -2, -3 \]

Solution:

\[ \vec{x} = c_1 \begin{bmatrix} 1 \\ -2 \end{bmatrix} e^{-2t} + c_2 \begin{bmatrix} 1 \\ -3 \end{bmatrix} e^{-3t} \]
negative e-values
Phase portrait showing trajectories converging toward the origin along two distinct lines in the x1-x2 plane.
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Page 10

Vector Field and Trajectories

The following computer-generated plot illustrates the vector field and specific solution trajectories for the system discussed on the previous page.

Detailed vector field plot with red trajectory curves converging to the origin in a stable node pattern.
Note: The grid spans from -4.75 to 4.75 on both axes. The trajectories clearly show the influence of the two negative eigenvalues, with paths curving toward the origin.
PAGE 11

Equilibrium Stability Analysis

This equilibrium is an asymptotically stable node.

→ two real eigenvalues and both are negative

If both positive and real → unstable node

(opposite of the above)