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PAGE 1

5.2 (Continued)

Eigenvalues are complex

Example

\[ \vec{x}' = \begin{bmatrix} -1 & 2 \\ -2 & -1 \end{bmatrix} \vec{x} \]

eigenvalues:

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

\[ -1 - \lambda = \pm 2i \]

complex \( \lambda \) are in complex conjugate pairs

eigenvectors:

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

\[ \lambda = -1 + 2i \]
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Eigenvector Calculation for Complex Eigenvalues

Starting with the augmented matrix for the first eigenvalue:

\[\begin{bmatrix}-2i & 2 & 0 \\ -2 & -2i & 0\end{bmatrix}\]
\(\rightarrow\)
\[\begin{bmatrix}-2 & -2i & 0 \\ -2i & 2 & 0\end{bmatrix}\]

Row Operations:

  • Multiply row 1 by \(-i\)
  • Add to row 2
\(\rightarrow\)
\[\begin{bmatrix}-2 & -2i & 0 \\ 0 & 0 & 0\end{bmatrix}\]
\(\rightarrow\)
\[\begin{bmatrix}1 & i & 0 \\ 0 & 0 & 0\end{bmatrix}\]

\(\vec{v} = \begin{bmatrix}-i \\ 1\end{bmatrix}\)

Second Eigenvalue: \(\lambda = -1 - 2i\)

\[\begin{bmatrix}2i & 2 & 0 \\ -2 & 2i & 0\end{bmatrix}\]
\(\rightarrow \dots \rightarrow\)
\[\begin{bmatrix}1 & -i & 0 \\ 0 & 0 & 0\end{bmatrix}\]
\(\vec{v} = \begin{bmatrix}i \\ 1\end{bmatrix}\)

Eigenvectors are also conjugate pairs

PAGE 3

Complex Eigenvalues and Real Solutions

Solutions are still \( e^{\lambda t} \vec{v} \) but usually we want it to not contain imaginary numbers.

Case 1:

\[ \lambda = -1 + 2i, \quad \vec{v} = \begin{bmatrix} -i \\ 1 \end{bmatrix} \]

Case 2:

\[ \lambda = -1 - 2i, \quad \vec{v} = \begin{bmatrix} i \\ 1 \end{bmatrix} \]

Starting with the first case:

\[ e^{\lambda t} \vec{v} = e^{(-1 + 2i)t} \begin{bmatrix} -i \\ 1 \end{bmatrix} = e^{-t} e^{i(2t)} \begin{bmatrix} -i \\ 1 \end{bmatrix} \]

Euler's Identity:

\[ e^{it} = \cos(t) + i \sin(t) \]

Applying Euler's identity to our expression:

\[ e^{-t} (\cos(2t) + i \sin(2t)) \begin{bmatrix} -i \\ 1 \end{bmatrix} \]
\[ = e^{-t} \begin{bmatrix} \sin(2t) - i \cos(2t) \\ \cos(2t) + i \sin(2t) \end{bmatrix} \]
\[ = e^{-t} \begin{bmatrix} \sin(2t) \\ \cos(2t) \end{bmatrix} + i e^{-t} \begin{bmatrix} -\cos(2t) \\ \sin(2t) \end{bmatrix} \]
PAGE 4

Solving Systems with Complex Eigenvalues

Repeat with \(\lambda = -1 - 2i\) and eigenvector \(\vec{v} = \begin{bmatrix} 1 \\ i \end{bmatrix}\):

\[ e^{\lambda t} \vec{v} = \dots = e^{-t} \begin{bmatrix} \sin(2t) \\ \cos(2t) \end{bmatrix} - i e^{-t} \begin{bmatrix} -\cos(2t) \\ \sin(2t) \end{bmatrix} \]

Note:

Conjugate pairs again.

The real and imaginary parts of either solution are themselves solutions to \(\vec{x}' = A\vec{x}\).

So, we use them to form the general solution:

  • Real part: \(e^{-t} \begin{bmatrix} \sin(2t) \\ \cos(2t) \end{bmatrix}\)
  • Imaginary part: \(e^{-t} \begin{bmatrix} -\cos(2t) \\ \sin(2t) \end{bmatrix}\)

General Solution

\[ \vec{x} = c_1 e^{-t} \begin{bmatrix} \sin(2t) \\ \cos(2t) \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -\cos(2t) \\ \sin(2t) \end{bmatrix} \]
PAGE 5

Phase Diagram Analysis for Oscillating Systems

\(\sin(2t)\) and \(\cos(2t)\) are periodic, so \(x_1\) and \(x_2\) are oscillating, so the phase diagram consists of spirals:

  • Into origin if the real part of \(\lambda\) is negative.
  • Away from origin if the real part of \(\lambda\) is positive.
  • If the real part is 0, the phase diagram consists of ovals.

Determining Spiral Direction

How to figure out if the spiral is in a clockwise direction or counterclockwise?

Easy Way:

\[ \vec{x}' = \begin{bmatrix} -1 & 2 \\ -2 & -1 \end{bmatrix} \vec{x} \]

Tangent vectors to spirals

Pick a convenient \(\vec{x}\) then see the direction of \(\vec{x}'\).

For example:

\[ \vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \rightarrow \vec{x}' = \begin{bmatrix} -1 \\ -2 \end{bmatrix} \]

left and down

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A phase portrait in a Cartesian coordinate system with axes labeled x_1 and x_2. The diagram shows a spiral sink or source where trajectories spiral outwards from the origin in a counter-clockwise direction. Arrows on the spiral curves indicate the direction of flow away from the center.
Figure 1: Phase portrait showing a spiral trajectory in the \(x_1-x_2\) plane.

5.5 Repeated Eigenvalues

\[ \vec{x}' = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \vec{x} \quad \lambda = 1, 1 \quad \text{algebraic multiplicity is Two} \]

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

\[ \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \vec{v} = \begin{bmatrix} a \\ b \end{bmatrix} = a \begin{bmatrix} 1 \\ 0 \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

linearly independent

\[ \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \quad \text{geometric multiplicity is Two} \]
PAGE 7

Solutions to Linear Systems

Solutions: \( e^{\lambda t} \vec{v} \), \( e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} \), \( e^t \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

General solution: \( \vec{x} = c_1 e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^t \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

Phase Diagram

A phase diagram in a Cartesian coordinate system with axes labeled x_1 and x_2. Numerous green trajectories (solution curves) radiate outward from the origin in all directions. Each trajectory has an arrow pointing away from the origin, indicating that the origin is an unstable node (source) where solutions grow exponentially over time.

Figure 1: Phase diagram showing an unstable node at the origin.

Now let's look at \( \vec{x}' = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \vec{x} \)

\[ \lambda = 2, 2 \]

Solving \( (A - \lambda I) \vec{v} = \vec{0} \) only produces one \( \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \)

Missing one eigenvector

→ matrix A is defective

(here, a defect of one)

PAGE 8

Solutions to Linear Systems with Generalized Eigenvectors

One solution: \( e^{\lambda t} \vec{v} \)

2nd solution: \( e^{\lambda t} (t \vec{v} + \vec{u}) \)

Note on \( \vec{u} \):

\( \vec{u} \) is a generalized eigenvector.

where \( (A - \lambda I) \vec{u} = \vec{v} \) and \( \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \)

here,

\[ \begin{bmatrix} 0 & 1 & | & 1 \\ 0 & 0 & | & 0 \end{bmatrix} \quad \vec{u} = \begin{bmatrix} a \\ 1 \end{bmatrix} \]

Choose any \( a \) so \( \vec{u} \) is linearly independent from \( \vec{v} \) and \( \vec{u} \neq \vec{0} \).

Let's use \( a = 0 \), so \( \vec{u} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

Final Solutions:

\[ e^{\lambda t} \vec{v} = e^{2t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]\[ e^{\lambda t} (t \vec{v} + \vec{u}) = e^{2t} \left( t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) \]