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 $P$ is the matrix whose columns are the corresponding eigenvectors.

Why this matters: once diagonalised, matrix powers become trivial:

$$A^n = PD^nP^{-1}$$

Since $D$ is diagonal, $D^n$ just raises each diagonal entry to the $n$-th power — no matrix multiplication needed. This is the key to computing long-run behaviour of linear dynamic systems.

When is $A$ diagonalisable? A sufficient condition: $A$ has $n$ distinct eigenvalues (for an $n \times n$ matrix). If eigenvalues repeat, diagonalisability depends on whether there are enough linearly independent eigenvectors.

The recipe:

1. Find eigenvalues by solving $\det(A - \lambda I) = 0$.
2. For each eigenvalue, find the eigenvector by solving $(A - \lambda I)\mathbf{v} = \mathbf{0}$.
3. Form $P$ (eigenvectors as columns) and $D$ (eigenvalues on diagonal).
4. Verify: $A = PDP^{-1}$.

Economic application: in a Leontief model, the economy's state after $n$ periods is $\mathbf{x}_n = A^n \mathbf{x}_0$. Diagonalising $A$ makes this computation practical — and reveals which sectors dominate in the long run (the eigenvector of the largest eigenvalue).

Why diagonalise

Common pitfall: Assuming every matrix diagonalises. Distinct eigenvalues guarantee it; repeated eigenvalues may not deliver enough independent eigenvectors — check before you decompose.

Matrix Powers and Steady States

Diagonalisation turns the intimidating problem of computing $A^{100}$ into something a pocket calculator can handle.

Example: let $A = \begin{pmatrix} 0.6 & 0.3 \\ 0.4 & 0.7 \end{pmatrix}$ — a Markov transition matrix (columns sum to 1). The eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 0.3$.

$$A^n = PD^nP^{-1} = P \begin{pmatrix} 1^n & 0 \\ 0 & 0.3^n \end{pmatrix} P^{-1}$$

As $n \to \infty$, $0.3^n \to 0$, so $A^n$ converges to a matrix determined entirely by the $\lambda = 1$ eigenvector — the stationary distribution.

$$\lim_{n\to\infty} A^n = P \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} P^{-1}$$

Key insight: the long-run behaviour of a linear system is controlled by the dominant eigenvalue. If $|\lambda_{\text{max}}| < 1$, the system decays to zero. If $|\lambda_{\text{max}}| = 1$ (Markov), it converges to a steady state. If $|\lambda_{\text{max}}| > 1$, it explodes.

Practical use: population dynamics, market-share evolution, economic growth models, and any system described by a linear recurrence $\mathbf{x}_{n+1} = A\mathbf{x}_n$ — diagonalisation gives you the closed-form solution in one step.

$$\mathbf{x}_n = c_1 \lambda_1^n \mathbf{v}_1 + c_2 \lambda_2^n \mathbf{v}_2$$

Each term is an eigenvalue raised to a power times an eigenvector — the system decomposes into independent exponential modes.

Tip: In any Markov matrix the dominant eigenvalue is exactly 1, and its eigenvector — normalised to sum to one — is the long-run market share. Everything else decays at the second eigenvalue's rate: $|\lambda_2|$ close to 1 means slow convergence, close to 0 means the past is forgotten fast.

Common pitfall: Believing the starting distribution matters in the long run. It doesn't — the stationary distribution is a property of the transition matrix alone. Initial conditions only decide how long the transient lasts.