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7.6 Multiple/Repeated Eigenvalues

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

solutions are \( e^{\lambda t} \vec{v} \)

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

\( \lambda, \vec{v} \) pairs

no issues if \( \lambda \)'s are distinct or complex
potential problems if \( \lambda \)'s are repeated

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

\( \lambda = 1, 1 \)

the eigenvalue of 1 is repeated twice (algebraic multiplicity of two)

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

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

here, both a and b are free

let \( a = r, b = s \)

then \( \vec{v} = \begin{bmatrix} r \\ s \end{bmatrix} = r \begin{bmatrix} 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

the eigenspace has a dimension of two

\( \rightarrow \) two basis vectors (eigenvectors)

\( \rightarrow \) geometric multiplicity of two

form solutions using these eigenvectors

general solution:

\[ \vec{x} = c_1 e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^t \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

if matrix is complete, enough eigenvectors to form linearly indp solutions \( \rightarrow \) solutions are formed the usual way.

if algebraic multiplicity = geo. multiplicity we say the matrix A is complete

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Defective Matrices and Repeated Eigenvalues

now look at

\[ \vec{x}' = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \vec{x} \quad \lambda = 1, 1 \]

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

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

only one free variable

\[ \vec{v} = \begin{bmatrix} a \\ b \end{bmatrix} \quad \begin{aligned} b &= 0 \\ a &= \text{free} = r \end{aligned} \]

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

eigenspace dimension is one

only one eigenvector

we are missing a vector

first solution: \( e^{\lambda_1 t} \vec{v}_1 = e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} \)

2nd one: \( e^{\lambda t} \vec{v}_2 \)

?

when we are missing eigenvectors, we say the matrix A is defective

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in scalar case, if the roots of characteristic eq. are repeated, we multiply the existing solution by \( t \) to form a new one

\[ y'' + 10y' + 25y = 0 \] \[ r^2 + 10r + 25 = 0 \rightarrow r = -5, -5 \] \[ y_1 = e^{-5t} \quad y_2 = t e^{-5t} \]

so, we might expect we form the 2nd solution in \( \vec{x}' = A \vec{x} \) by doing the same thing:

\( \vec{x}_1 = e^{\lambda t} \vec{v}_1 \)

\( \vec{x}_2 = e^{\lambda t} \vec{v}_1 t \)

BUT, this does NOT work

The Correct Form for Defective Matrices

instead, we need

\[ \vec{x}_2 = e^{\lambda t} (t \vec{v}_1 + \vec{v}_2) \]

↑

Ordinary eigenvector

↑

need to find (generalized eigenvector)

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How to find \(\vec{v}_2\)?

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

eigenvalue \(\lambda\) only one eigenvector \(\vec{v}_1\)

the second solution is \(\vec{x}_2 = e^{\lambda t} (t \vec{v}_1 + \vec{v}_2) = t e^{\lambda t} \vec{v}_1 + e^{\lambda t} \vec{v}_2\)

sub into \(\vec{x}' = A \vec{x}\)

\[ \vec{x}_2' = t \lambda e^{\lambda t} \vec{v}_1 + e^{\lambda t} \vec{v}_1 + \lambda e^{\lambda t} \vec{v}_2 \]

\[ t \lambda e^{\lambda t} \vec{v}_1 + e^{\lambda t} \vec{v}_1 + \lambda e^{\lambda t} \vec{v}_2 = A t e^{\lambda t} \vec{v}_1 + A e^{\lambda t} \vec{v}_2 \]

Compare like terms

\(t e^{\lambda t}\) terms:

\(\lambda \vec{v}_1 = A \vec{v}_1\) \(\rightarrow\) nothing new, definition of \(\lambda\) being an eigenvalue and \(\vec{v}_1\) an eigenvector

\(e^{\lambda t}\) terms:

\(\vec{v}_1 + \lambda \vec{v}_2 = A \vec{v}_2\)

gives us \(\vec{v}_2\)

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

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also, from \((A - \lambda I) \vec{v}_2 = \vec{v}_1\)

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

\[ (A - \lambda I)^2 \vec{v}_2 = (A - \lambda I) \vec{v}_1 = \vec{0} \]

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

so, there are two ways to find \(\vec{v}_2\)

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

or

\[ (A - \lambda I)^2 \vec{v}_2 = \vec{0} \]

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Revisiting Systems of Differential Equations

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

\( \lambda = 1, 1 \) \( \vec{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \)

1st solution: \( e^{\lambda t} \vec{v}_1 \)
2nd solution: \( e^{\lambda t} (t \vec{v}_1 + \vec{v}_2) \)

Finding the Generalized Eigenvector \( \vec{v}_2 \)

Let's find \( \vec{v}_2 \) using \( (A - \lambda I) \vec{v}_2 = \vec{v}_1 \)

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

\( b = 1 \) \( a = r \)

\[ \vec{v}_2 = \begin{bmatrix} r \\ 1 \end{bmatrix} = r \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

Choose ANY \( r \) here, choose \( r = 0 \)

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

1st solution: \( e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} \)
2nd solution: \( e^t \left( t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) \)

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General Solution

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

Second Way to Find \( \vec{v}_2 \)

Using the property: \( (A - \lambda I)^2 \vec{v}_2 = \vec{0} \)

\( A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \quad \lambda = 1 \)

\( A - \lambda I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \)

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

\( (A - \lambda I)^2 \vec{v}_2 = \vec{0} \)

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

\( \rightarrow \vec{v}_2 \) is almost arbitrary

Choose ANY \( \vec{v}_2 \) EXCEPT \( (A - \lambda I) \vec{v}_2 = \vec{0} \)

Basically, ANY \( \vec{v}_2 \) that is linearly independent from the true eigenvector \( \vec{v}_1 \)

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

choose ANY \( \vec{v}_2 \) indp from that

so, choose \( \vec{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \) but \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) is ok, too

For \( 3 \times 3 \) matrix (or beyond \( 3 \times 3 \)), the second way is better

in general, if we are missing \( n-1 \) eigenvectors, then

\[ (A - \lambda I)^n = \text{zero matrix} \]

so, choose the missing one from \( (A - \lambda I)^n \vec{v} = \vec{0} \)

then build the rest using \( (A - \lambda I) \vec{v}_n = \vec{v}_{n-1} \)

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Example

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

\( \lambda = 2, 2, 6 \) \( \vec{v} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \) for \( \lambda = 6 \)

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

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

two free variables → two eigenvectors (good, A is complete)

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

Solution is formed normally

\[ \vec{x} = c_1 e^{6t} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + c_2 e^{2t} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + c_3 e^{2t} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \]