Matrix Diagonalisation

Decomposing a Matrix

A matrix $A$ is diagonalisable if it can be written as $A = PDP^{-1}$, where $D$ is a diagonal matrix of eigenvalues and $P$ has eigenvectors as columns.

Why this matters: matrix powers become trivial via $A^n = PD^nP^{-1}$. Raising $D$ to the $n$-th power requires zero matrix multiplications.

The Recipe:

1. Solve $\det(A - \lambda I) = 0$ for eigenvalues.
2. Solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$ for eigenvectors.
3. Form $P$ and $D$.

Pitfall: Distinct eigenvalues guarantee diagonalisability, but repeated ones may lack independent eigenvectors.

Matrix Powers and Steady States

Diagonalisation yields the closed-form solution: $\mathbf{x}_n = c_1 \lambda_1^n \mathbf{v}_1 + c_2 \lambda_2^n \mathbf{v}_2$.

Dominant eigenvalue controls long-run behaviour:

Example: For Markov matrices, $\lambda_1 = 1$ gives the stationary distribution. As $n \to \infty$, $0.3^n \to 0$, leaving only the dominant eigenvector.

Tip: In Markov matrices, the dominant eigenvalue is 1. Its normalised eigenvector is the long-run market share.