Eigenvalues and Eigenvectors

An eigenvector of a matrix $A$ is a nonzero vector $\vec{v}$ that satisfies:

$$A\vec{v} = \lambda\vec{v}$$

The scalar $\lambda$ is the corresponding eigenvalue. The transformation $A$ only stretches $\vec{v}$, without changing its direction.

Finding eigenvalues — Solve the characteristic equation:

$$\det(A - \lambda I) = 0$$

This yields a polynomial of degree $n$ whose roots are the eigenvalues.

Finding eigenvectors — For each $\lambda_i$, solve the homogeneous system:

$$(A - \lambda_i I)\vec{v} = \vec{0}$$

Example

$$A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} \implies \det(A-\lambda I) = \lambda^2 - 7\lambda + 10 = 0$$

So $\lambda_1 = 5$, $\lambda_2 = 2$.

Diagonalisation — If $A$ has $n$ linearly independent eigenvectors, then:

A = PDP^{-1}

where $D = \text{diag}(\lambda_1,\ldots,\lambda_n)$ and $P$ has eigenvectors as columns. This makes computing $A^k = PD^kP^{-1}$ trivial.

Physics link: Normal modes of coupled oscillators are eigenvectors of the system matrix. Each eigenvalue gives $\omega^{2}$, the squared angular frequency of that mode.

Common pitfall: Eigen-vectors keep their line, not their length or even sense: eigenvalues may be negative (flips), zero (collapse), or fractional (shrink). "Unchanged by the matrix" is only true when $\lambda = 1$.

An eigenvector $\vec{v}$ of a matrix $A$ satisfies:

$$A\vec{v} = \lambda\vec{v}$$

The scalar $\lambda$ is the corresponding eigenvalue. Under the transformation $A$, eigenvectors only scale — they do not rotate.

Eigenvalues are found from the characteristic equation $\det(A - \lambda I) = 0$.

Eigenvectors reveal the intrinsic axes of a transformation — diagonalisation, stability analysis, and PCA all depend on them.