Download original notes
PAGE 1

5.5 (continued)

from last time:

\[ \vec{x}' = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \vec{x} \]
\[ \vec{x} = c_1 e^{2t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{2t} \left( t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) \]

phase diagram:

\( \vec{x} \to \vec{0} \) as \( t \to -\infty \) \( \to \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) more important (asymptote)

as \( t \) increases, move away

solutions go in the direction of \( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) while being nudged by \( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

A phase portrait on a Cartesian coordinate system with axes x_1 and x_2. The horizontal x_1 axis is labeled with the true eigenvector \begin{bmatrix} 1 \\ 0 \end{bmatrix}. The vertical x_2 axis is labeled with the generalized eigenvector \begin{bmatrix} 0 \\ 1 \end{bmatrix}. Blue trajectories emerge from the origin, initially following the direction of the generalized eigenvector before curving to become asymptotic to the true eigenvector axis. Red dashed lines and a small red vector (representing the sum of true and generalized eigenvectors) illustrate the directional 'nudge' that shapes the curves.
Phase portrait showing trajectories for a system with a repeated eigenvalue and a single eigenvector.

red vector: sum of true and generalized eigenvectors

\( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) (true eigenvector)

we do not see the generalized eigenvector as an asymptote

PAGE 2

3x3 System with \(\lambda\) Repeating 3 Times

Possibilities:

  1. Full set of eigenvectors
  2. Missing one eigenvector (defect of one)
  3. Missing two eigenvectors (defect of two)

Defect of Two is Actually Simpler

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

\(\lambda = 2, 2, 2\)

\[ \vec{v} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \]

missing two

Let \(\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\) (true)

\[ (A - \lambda I) \vec{v}_2 = \vec{v}_1 \quad \rightarrow \quad \left[ \begin{array}{ccc|c} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \]
\[ \vec{v}_2 = \begin{bmatrix} a \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \]
\[ (A - \lambda I) \vec{v}_3 = \vec{v}_2 \]

\(\vdots\)

\[ \vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \]
PAGE 3

Solutions to Linear Systems

  • Solution 1:\( e^{\lambda t} \vec{v}_1 \)
  • Solution 2:\( e^{\lambda t} (t \vec{v}_1 + \vec{v}_2) \)
  • Solution 3:\( e^{\lambda t} (\frac{1}{2} t^2 \vec{v}_1 + t \vec{v}_2 + \vec{v}_3) \)

General Solution

\[ \vec{x} = c_1 e^{2t} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + c_2 e^{2t} \left( t \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right) + c_3 e^{2t} \left( \frac{1}{2} t^2 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + t \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right) \]

Alternate Method

\[ \begin{aligned} (A - \lambda I) \vec{v}_1 &= \vec{0} \\ (A - \lambda I) \vec{v}_2 &= \vec{v}_1 \\ (A - \lambda I) \vec{v}_3 &= \vec{v}_2 \end{aligned} \]

Multiply last eq. by \( (A - \lambda I) \):

\[ (A - \lambda I)^2 \vec{v}_3 = (A - \lambda I) \vec{v}_2 = \vec{v}_1 \]

Again:

\[ (A - \lambda I)^3 \vec{v}_3 = (A - \lambda I) \vec{v}_1 = \vec{0} \quad \implies \quad (A - \lambda I)^3 \vec{v}_3 = \vec{0} \]
PAGE 4

Generalized Eigenvectors and Defect Properties

Not a coincidence: if defect is \(k\) then \((A - \lambda I)^{k+1} = \vec{0}\).

Also, \((A - \lambda I)^{k+1}\) is always a zero matrix.

Solving for Generalized Eigenvectors

So, \((A - \lambda I)^3 \vec{v}_3 = \vec{0}\)

\[ \text{becomes } \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

\(\vec{v}_3\) is arbitrary*

* \(\vec{v}_3 \neq \vec{0}\) and must be linearly independent from true eigenvectors.

\[ \vec{v}_3 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \text{ or } \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix} \text{ or } \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \text{ or } \begin{bmatrix} 0 \\ e \\ \pi \end{bmatrix} \]

Here, let's use \(\vec{v}_3 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}\).

Rebuilding the Chain

Now rebuild the entire chain, including \(\vec{v}_1\).

\[ (A - \lambda I) \vec{v}_3 = \vec{v}_2 = \dots = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \]
\[ (A - \lambda I) \vec{v}_2 = \vec{v}_1 = \dots = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \]
Matches the true eigenvector here, but not always.
PAGE 5

then form solutions as before

two methods: 1st is "stepping up"
2nd is "stepping down"

For the case w/ defect of one (missing one eigenvector)

"stepping down" is much easier

\[ \vec{x}' = \begin{bmatrix} 5 & -3 & -2 \\ 8 & -5 & -4 \\ -4 & 3 & 3 \end{bmatrix} \vec{x} \quad \lambda = 1, 1, 1 \]\[ \vec{v} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} \]
\[ \vec{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}, \vec{v}_2 = \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} \]

defect is 1 so

\[ (A - \lambda I)^{1+1} \vec{v}_3 = \vec{0} \]

zero matrix

\( \vec{v}_3 \) is arbitrary*

\( \vec{v}_3 \neq \vec{0} \) and linearly indp from true eigenvectors

PAGE 6

Generalized Eigenvectors and Solution Sets

Here, let's use ̑\(\vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\).

Step down: \((A - \lambda I)\vec{v}_3 = \vec{v}_2 = \dots = \begin{bmatrix} -2 \\ -4 \\ 2 \end{bmatrix}\)

\[ (A - \lambda I)\vec{v}_2 = \vec{v}_1 \]
\[ \begin{bmatrix} 4 & -3 & -2 \\ 8 & -6 & -4 \\ -4 & 3 & 2 \end{bmatrix} \begin{bmatrix} -2 \\ -4 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

Note on the Zero Vector Result:

Can't have \(\vec{0}\)

Because \(\vec{v}_2\) is a linear combination of the two true eigenvectors.

When this happens, pick either of the true eigenvectors to be \(\vec{v}_1\).

Here, \(\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}\).

General Solutions

  • Solution 1: \(e^{\lambda t} \vec{v}_1\)
  • Solution 2: \(e^{\lambda t} \vec{v}_2\)
  • Solution 3: \(e^{\lambda t} (t \vec{v}_2 + \vec{v}_3)\)