For a square matrix \(A\), an eigenvector \(v\) is a non-zero vector such that:
\[ A \vec{v} = \lambda \vec{v} \]\( \lambda \) is the eigenvalue (scalar) — it tells how much \(v\) is stretched or shrunk.
Intuition: eigenvectors are special directions that the transformation only stretches/scales (no rotation).
Solve the characteristic equation:
\[ \det(A - \lambda I) = 0 \]Example for \(2×2, \, A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}\):
\[ A - \lambda I = \begin{pmatrix} 3-\lambda & 1 \\ 0 & 2-\lambda \end{pmatrix}, \quad \det(A) = (3-\lambda)(2-\lambda) = 0 \]\( \lambda = 3\) or \( \lambda = 2\)
For each eigenvalue \(\lambda\), solve \( (A - \lambda I)v = 0\).
Example for \(\lambda = 3\) in earlier matrix:
\[ (A - 3I)v = \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]Solution: \(y = 0,\, x\) free → eigenvector \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\)