7.4 Solution Curves of Linear Systems (continued)
\( \lambda = 1, 1 \)
\( \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) only need a generalized eigenvector
let \( \vec{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \)
Solution:
as \( t \to \infty \), both \( e^t \) and \( t e^t \) are important
both \( e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) and \( t e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) follow the same vector: \( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) → ordinary eigenvector
the generalized one: \( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \) affects orientation but is not really visible