determin.ant — Linear Algebra from Scratch

Where you'll see this: computing matrix powers efficiently. To find A¹⁰⁰ directly would be 99 matrix multiplications. Via diagonalization, A¹⁰⁰ = P D¹⁰⁰ P⁻¹, and D¹⁰⁰ is trivial — just raise each diagonal entry to the 100th power. This powers Markov chains, population models, and graph analysis.

Diagonalization is the process of expressing a matrix in its "natural coordinate system" — aligned with the eigenvectors — where it becomes as simple as possible.

The decomposition

If A has n linearly independent eigenvectors $\mathbf{v}_1, \mathbf{v}_2$ , then:

A = P D P^{-1}

where P has the eigenvectors as columns, and D is diagonal with eigenvalues:

P = \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 \end{bmatrix}, \quad D = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}

How to read it

Think of $A\mathbf{x}$ as three steps:

P⁻¹x — re-express x in the eigenvector basis
Dx — scale each component (diagonal = just multiplication)
Px — convert back to the standard basis

Diagonalization is a change of basis into a coordinate system where the transformation is pure scaling. In that system, the matrix is just two independent scalar multiplications.

When it fails

Not every matrix is diagonalizable over ℝ:

Complex eigenvalues → no real eigenvectors
Repeated eigenvalue without enough independent eigenvectors (defective matrix)

Rotation matrices are the classic example: no real eigenvectors, can't be diagonalized over ℝ.

Try it

Edit the matrix and see the P, D, P⁻¹ decomposition. The product P·D·P⁻¹ should reconstruct A exactly. Try a rotation matrix — it will report that diagonalization fails.

Matrix powers via diagonalization:

import numpy as np
A = np.array([[3, 1], [1, 3]])
eigenvalues, P = np.linalg.eig(A)
D = np.diag(eigenvalues)
P_inv = np.linalg.inv(P)

# A^10 = P @ D^10 @ P_inv
A_power_10 = P @ np.diag(eigenvalues**10) @ P_inv
print(np.allclose(A_power_10, np.linalg.matrix_power(A, 10)))  # True