PAGE 1

7.6 Repeated Eigenvalues (part 2)

\[ \vec{x}' = A\vec{x} \]

if \( A \) has repeated eigenvalues but has enough eigenvectors then solutions are formed normally: \( e^{\lambda t} \vec{v} \)

for example,

\[ \vec{x}' = \begin{bmatrix} 2 & 0 & 0 \\ -4 & 6 & 4 \\ 0 & 0 & 2 \end{bmatrix} \vec{x} \]

\[ \lambda = 2, 2, 6 \]

repeated

\[ \vec{v} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \text{ for } \lambda = 6 \]

for \( \lambda = 2 \), \( (A - \lambda I)\vec{v} = \vec{0} \)

\[ \begin{bmatrix} 0 & 0 & 0 & 0 \\ -4 & 4 & 4 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

two free variables \( \rightarrow \) two eigenvectors

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

PAGE 2

Solutions:

\[ e^{6t} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, e^{2t} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, e^{2t} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \]

if there are not enough eigenvectors, then we need to supplement with generalized eigenvectors.

example

\[ \vec{x}' = \begin{bmatrix} -11 & 0 & -4 \\ -1 & -9 & -1 \\ 1 & 0 & -7 \end{bmatrix} \vec{x} \]

\[ \lambda = -9, -9, -9 \]

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

\[ \begin{bmatrix} -2 & 0 & -4 & 0 \\ -1 & 0 & -1 & 0 \\ 1 & 0 & 2 & 0 \end{bmatrix} \]

\[ \rightarrow \dots \rightarrow \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

2 pivots

1 free variable

\( \rightarrow \) 1 eigenvector

missing Two

PAGE 3

Generalized Eigenvectors

The only ordinary eigenvector is \( \vec{v}_1 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \).

Missing \( \vec{v}_2, \vec{v}_3 \rightarrow \) generalized eigenvectors.

\[ (A - \lambda I) \vec{v}_1 = \vec{0} \quad \text{because } \vec{v}_1 \text{ is an ordinary eigenvector} \]\[ (A - \lambda I) \vec{v}_2 = \vec{v}_1 \]\[ (A - \lambda I) \vec{v}_3 = \vec{v}_2 \]

Calculating \( \vec{v}_2 \):

\[ A - \lambda I = \begin{bmatrix} -2 & 0 & -4 \\ -1 & 0 & -1 \\ 1 & 0 & 2 \end{bmatrix} \]

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

The augmented column represents \( \vec{v}_1 \). From the reduced row echelon form:

\[ \vec{v}_2 = \begin{bmatrix} -2 \\ r \\ 1 \end{bmatrix} \quad \begin{array}{l} \text{choose ANY } r \\ \text{let's choose } 0 \end{array} \]

\[ \vec{v}_2 = \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \]

PAGE 4

Calculating \( \vec{v}_3 \)

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

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

The augmented column represents \( \vec{v}_2 \). From the reduced row echelon form:

\[ \vec{v}_3 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \quad \text{2nd element is free, chose 0} \]

Forming the Three Solutions

Now we form the three solutions:

1st: \( e^{\lambda t} \vec{v}_1 = e^{-9t} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \)

2nd: \( e^{\lambda t} (t \vec{v}_1 + \vec{v}_2) = e^{-9t} \left( t \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \right) \)

3rd: \( e^{\lambda t} (\frac{1}{2} t^2 \vec{v}_1 + t \vec{v}_2 + \vec{v}_3) = e^{-9t} \left( \frac{t^2}{2} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} + \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \right) \)

Gen. solution is linear combo of them.

PAGE 5

The Second Way to Find These Generalized Eigenvectors

\[ (A - \lambda I) \vec{v}_1 = \vec{0} \]

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

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

\( \vec{v}_1 \) is true eigenvector

\( \vec{v}_2, \vec{v}_3 \): generalized eigenvectors

multiply by \( (A - \lambda I) \)

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

multiply by \( (A - \lambda I) \) again

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

\( (A - \lambda I)^{k+1} \) IS ALWAYS a matrix of zeros if there are \( k \) eigenvectors missing for this \( \lambda \)

\[ A - \lambda I = \begin{bmatrix} -2 & 0 & -4 \\ -1 & 0 & -1 \\ 1 & 0 & 2 \end{bmatrix} \]

\[ (A - \lambda I)^3 = \begin{bmatrix} -2 & 0 & -4 \\ -1 & 0 & -1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} -2 & 0 & -4 \\ -1 & 0 & -1 \\ 1 & 0 & 2 \end{bmatrix} \text{ (same)} \]

\[ = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

PAGE 6

all zeros

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

\[ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

\[ \vec{v}_3 = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \]

\( a, b, c \) are free

Choose ANY vector to be \( \vec{v}_3 \) EXCEPT zero vector or any multiple of the true eigenvector

\[ \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \]

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

Now find \( \vec{v}_2 \) from \( (A - \lambda I) \vec{v}_3 = \vec{v}_2 \)

\[ \begin{bmatrix} -2 & 0 & -4 \\ -1 & 0 & -1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix} \]

\[ \vec{v}_2 = \begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix} \]

PAGE 7

Finding Generalized Eigenvectors

Now, find \(\vec{v}_1\) from \((A - \lambda I)\vec{v}_2 = \vec{v}_1\) even though we have already found a true eigenvector.

\[ \begin{bmatrix} -2 & 0 & -4 \\ -1 & 0 & -1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} -2 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} 8 \\ 3 \\ -4 \end{bmatrix} \text{ (Note: Calculation in image shows result as } \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \text{ based on context)} \]

\[ \vec{v}_1 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \]

This happens to be the same as the true eigenvector we found, but this is NOT always the case.

Solutions

\[ e^{\lambda t} \vec{v}_1 = e^{-5t} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \] \[ e^{\lambda t} (t \vec{v}_1 + \vec{v}_2) = e^{-5t} \left( t \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} -2 \\ -1 \\ -1 \end{bmatrix} \right) \] \[ e^{\lambda t} \left( \frac{t^2}{2} \vec{v}_1 + t \vec{v}_2 + \vec{v}_3 \right) = e^{-5t} \left( \frac{t^2}{2} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} -2 \\ -1 \\ -1 \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right) \]

Solutions will look different depending on choice of \(\vec{v}_3\), but the general solution is the same (\(c_1, c_2, c_3\) will adjust for the different forms).

PAGE 8

Summary of Two Methods

First, find the true eigenvector.

Method 1

Use true eigenvector as \(\vec{v}_1\), then find the rest from:

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

Method 2

\[ (A - \lambda I)^{k+1} = \text{zero matrix} \quad (k \text{ eigenvectors missing}) \]

Choose the top one to be whatever EXCEPT zero vector or any multiple of true eigenvector.

Then:

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

← use this as \(\vec{v}_1\) even if it is not the true eigenvector