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7.6 Repeated Eigenvalues (part 3)

last time: \( \lambda \) repeated 3 times, missing 2 eigenvectors

today: \( \lambda \) repeated 3 times, missing 1 eigenvector

\[ \vec{x}' = \begin{bmatrix} 5 & -3 & -2 \\ 8 & -5 & -4 \\ -4 & 3 & 3 \end{bmatrix} \vec{x} \]

\( \lambda = 1, 1, 1 \)

ordinary eigenvectors

\[ \vec{v} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} \]

missing one

Two methods

  1. using ordinary eigenvector and use \( (A - \lambda I) \vec{v}_2 = \vec{v}_1 \) etc to build other solutions "up"
  2. choose the "top" one arbitrarily and use \( (A - \lambda I) \vec{v}_3 = \vec{v}_2 \) etc to build others "down"

1st method

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two solutions: \( \vec{x}_1 = e^{\lambda t} \vec{v}_1 = e^t \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} \)

\( \vec{x}_2 = e^{\lambda t} \vec{v}_2 = e^t \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} \)

missing \( \vec{x}_3 \)

third one is: \( \vec{x}_3 = e^{\lambda t} (t \vec{u} + \vec{v}_3) \)

some linear combo of the two ordinary eigenvectors

\( \vec{u} = c_1 \vec{v}_1 + c_2 \vec{v}_2 \)

and \( (A - \lambda I) \vec{v}_3 = \vec{u} = c_1 \vec{v}_1 + c_2 \vec{v}_2 \)

first, find what \( c_1 \) and \( c_2 \) are:

\[ (A - \lambda I) \vec{v}_3 = \vec{u} = c_1 \vec{v}_1 + c_2 \vec{v}_2 \]
\[ c_1 \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} = \begin{bmatrix} c_1 \\ 2c_2 \\ 2c_1 - 3c_2 \end{bmatrix} \begin{bmatrix} 4 & -3 & -2 & c_1 \\ 8 & -6 & -4 & 2c_2 \\ -4 & 3 & 2 & 2c_1 - 3c_2 \end{bmatrix} \]
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look at row 1 and row 2

system is consistent only if \( 2c_1 = 2c_2 \rightarrow c_1 = c_2 \)

choose any \( c_1, c_2 \) such that \( c_1 = c_2 \)

let's choose \( c_1 = 2 \) for convenience (\( c_1 = c_2 \) so \( c_2 = 2 \))

\( (A - \lambda I) \vec{v}_3 = \vec{u} \) becomes

\[ \left[ \begin{array}{ccc|c} 4 & -3 & -2 & 2 \\ 8 & -6 & -4 & 4 \\ -4 & 3 & 2 & -2 \end{array} \right] \rightarrow \dots \rightarrow \left[ \begin{array}{ccc|c} 1 & -3/4 & -1/2 & 1/2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \]

2nd, 3rd elements are free

\[ \vec{v} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} \frac{1}{2} + \frac{3}{4}a + \frac{1}{2}b \\ a \\ b \end{bmatrix} \]

choose \( a = 0, b = -1 \)

\[ \vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} \]
\[ \vec{x}_3 = e^{\lambda t} (t \vec{u} + \vec{v}_3) \]

\( \downarrow \)

\[ c_1 \vec{v}_1 + c_2 \vec{v}_2 = \begin{bmatrix} 2 \\ 4 \\ -2 \end{bmatrix} \]
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\[ \vec{x}_3 = e^t \left( t \begin{bmatrix} 2 \\ 4 \\ -2 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} \right) \]

General solution:

\[ \vec{x} = c_1 e^t \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} + c_2 e^t \begin{bmatrix} 0 \\ 1 \\ -3 \end{bmatrix} + c_3 e^t \left( t \begin{bmatrix} 2 \\ 4 \\ -2 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} \right) \]

2nd method:

\( \vec{v}_1, \vec{v}_2 \) are ordinary eigenvectors

\[ \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ -3 \end{bmatrix} \]

but we don't necessarily use them in our solutions

\( (A - \lambda I) \vec{v}_3 = \vec{u} = c_1 \vec{v}_1 + c_2 \vec{v}_2 \)

multiply by \( (A - \lambda I) \) on both sides

\[ (A - \lambda I)^2 \vec{v}_3 = c_1 \underbrace{(A - \lambda I) \vec{v}_1}_{\vec{0}} + c_2 \underbrace{(A - \lambda I) \vec{v}_2}_{\vec{0}} \]
\[ (A - \lambda I)^2 \vec{v}_3 = \vec{0} \]
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Generalized Eigenvectors: Top-Down Method

Starting with the matrix calculation for \( A - \lambda I \):

\[ A - \lambda I = \begin{bmatrix} 4 & -3 & -2 \\ 8 & -6 & -4 \\ -4 & 3 & 2 \end{bmatrix} \]

Calculating the square of the matrix:

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

Solving for the generalized eigenvector \( \vec{v}_3 \) such that \( (A - \lambda I)^2 \vec{v}_3 = \vec{0} \):

\[ (A - \lambda I)^2 \vec{v}_3 = \vec{0} \rightarrow \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

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

Let's choose \( \vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \)

but \( \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} \) is ok, too

(and many other choices)

* : not \( \vec{0} \) and must be linearly independent from existing eigenvectors:

\[ \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} \]
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Using this "top down" method, we don't necessarily reuse the ordinary eigenvectors, we MUST find them this way:

\[ (A - \lambda I) \vec{v}_3 = \vec{v}_2 \] \[ \begin{bmatrix} 4 & -3 & -2 \\ 8 & -6 & -4 \\ -4 & 3 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -2 \\ -4 \\ 2 \end{bmatrix} = \vec{v}_2 \]

Usually we then do \( (A - \lambda I) \vec{v}_2 = \vec{v}_1 \)

\[ \text{but here, } \begin{bmatrix} 4 & -3 & -2 \\ 8 & -6 & -4 \\ -4 & 3 & 2 \end{bmatrix} \begin{bmatrix} -2 \\ -4 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
because \( \vec{v}_2 \) is a linear combo of the two ordinary eigenvectors

do NOT use \( \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \) to be \( \vec{v}_1 \), choose either of the two ordinary eigenvectors to be \( \vec{v}_1 \)

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here, let's choose \( \vec{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} \)

solutions:

\[ \vec{x}_1 = e^{\lambda t} \vec{v}_1 = e^t \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} \]

the one we found on last page

\[ \vec{x}_2 = e^{\lambda t} \vec{v}_2 = e^t \begin{bmatrix} -2 \\ -4 \\ 2 \end{bmatrix} \]
\[ \vec{x}_3 = e^{\lambda t} (\vec{v}_2 t + \vec{v}_3) = e^t \left( t \begin{bmatrix} -2 \\ -4 \\ 2 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} \right) \]

general solution:

\[ \vec{x} = c_1 e^t \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} + c_2 e^t \begin{bmatrix} -2 \\ -4 \\ 2 \end{bmatrix} + c_3 e^t \left( t \begin{bmatrix} -2 \\ -4 \\ 2 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} \right) \]