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Linear Algebra, Optimisation and Numerical Methods / Eigenvalues, Diagonalisation, and Dynamical Applications

Eigenvalues and Eigenvectors: Ax = λx

This note opens Stage 3 of the unit with the question every earlier topic was quietly building toward: given a linear map, are there directions it merely stretches, rather than rotates or shears? Those special directions — eigenvectors — and their stretch factors — eigenvalues — are found by solving Ax=λxA\mathbf{x}=\lambda\mathbf{x}, an equation this note reduces to the determinant condition det⁡(A−λI)=0\det(A-\lambda I)=0: the characteristic polynomial, whose roots are exactly the eigenvalues.
Learning objectives. By the end of this note you should be able to:
  • state the eigenvalue equation and explain what it means geometrically;
  • derive the characteristic equation det⁡(A−λI)=0\det(A-\lambda I)=0 from Ax=λxA\mathbf{x}=\lambda\mathbf{x};
  • compute eigenvalues and eigenvectors by hand for 2×22\times2 and 3×33\times3 matrices;
  • verify an eigenpair by direct substitution.

In this lesson

  1. The eigenvalue equation
  2. The characteristic equation
  3. Computing eigenvalues and eigenvectors by hand

Before you start

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