Prof. Alejandro Ucan-Puc
An assembly line with three workstations (A, B, C) has dynamics modeled by:
$$T = \begin{pmatrix} 0.6 & 0.2 & 0.1 \\ 0.3 & 0.7 & 0.2 \\ 0.1 & 0.1 & 0.7 \end{pmatrix}$$
Is there a steady production vector $v \neq 0$ such that $T(v) = v$?
This corresponds to an eigenvector associated with eigenvalue $\lambda = 1$.
Definition: Let $A$ be an $n \times n$ matrix. A non-zero vector $v \in \mathbb{R}^n$ is an eigenvector of $A$ if there exists a scalar $\lambda \in \mathbb{R}$ such that:
$$A v = \lambda v$$The scalar $\lambda$ is called the eigenvalue associated with $v$.
The equation $Av = \lambda v \iff (A - \lambda I)v = 0$ has a non-trivial solution $v \neq 0$ if and only if:
$$\det(A - \lambda I) = 0$$This $n$-th degree polynomial $p(\lambda) = \det(A - \lambda I)$ is the characteristic polynomial of $A$.
Definition: For a given eigenvalue $\lambda$, the set of all eigenvectors plus the zero vector forms a subspace called the eigenspace of $\lambda$:
$$E_\lambda = \text{Ker}(A - \lambda I) = \{ v \in \mathbb{R}^n \mid (A - \lambda I)v = 0 \}$$Geometric Multiplicity: $\text{dim}(E_\lambda) = \text{dim}(\text{Ker}(A - \lambda I))$.
Eigenvalues dictate the long-term behavior, growth, or equilibrium states of continuous and discrete process models.
Next Session: Diagonalization and Change of Basis $A = P D P^{-1}$.