Prof. Alejandro Ucan-Puc
An automobile assembly line has 3 workstations (Body, Paint, Assembly) with production rates interacting via:
$$A = \begin{pmatrix} 1.1 & 0.2 & 0.1 \\ 0.3 & 0.9 & 0.2 \\ 0.1 & 0.1 & 1.0 \end{pmatrix}$$
By switching to an eigenvector basis $B = \{v_1, v_2, v_3\}$ representing decoupled production trade-offs, $A$ becomes diagonal!
Definition: Let $B = \{v_1, \dots, v_n\}$ and $C = \{w_1, \dots, w_n\}$ be two bases for $V$. The change of basis matrix from $B$ to $C$ is:
$$P_{C \leftarrow B} = \begin{pmatrix} [v_1]_C & [v_2]_C & \dots & [v_n]_C \end{pmatrix}$$Satisfying: $[v]_C = P_{C \leftarrow B} [v]_B$.
Definition: An $n \times n$ matrix $A$ is diagonalizable if it is similar to a diagonal matrix $D$, i.e., there exists an invertible matrix $P$ such that:
$$A = P D P^{-1} \quad \text{or} \quad D = P^{-1} A P$$Theorem: $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors.
To compute the state after $k$ production cycles $A^k$:
$$A^k = (P D P^{-1})^k = P D^k P^{-1}$$where $D^k = \begin{pmatrix} \lambda_1^k & & 0 \\ & \ddots & \\ 0 & & \lambda_n^k \end{pmatrix}$.
Diagonalization simplifies matrix operations by expressing operators in an optimal eigen-coordinate system.
Next Session: Orthogonality, Inner Products, and Least Squares.