# Linear Algebra

Last updated: Sat, 17 Nov 2018

$A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$

$\lambda = 1, 3$ are the eigenvalues, which can be determined by solving $det(A - \lambda I) = 0$.

The Cayley-Hamilton theorem states that $(A - \lambda_1 I) ... (A - \lambda_n I) = 0$; hence the matrix can be factorized as $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ and $\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}$

Eigenvector $\mathbb{v}$ can be obtained by solving $A \mathbb{v} = \lambda \mathbb{v}$.

Vector spaces are defined over a field $\mathbb{F}$.