| Line 14: | Line 14: | ||
By solving the quadratic equation for <math>\lambda</math>, we will have two eigenvalues <math>\lambda_{1}=-1</math> and <math>\lambda_{2}=-6</math>. By substituting <math>\lambda's</math> into Eq [eq:1] | By solving the quadratic equation for <math>\lambda</math>, we will have two eigenvalues <math>\lambda_{1}=-1</math> and <math>\lambda_{2}=-6</math>. By substituting <math>\lambda's</math> into Eq [eq:1] | ||
| + | |||
| + | <math> | ||
| + | \[ | ||
| + | \left(A-\lambda_{1}I\right)\vec{x}=\left[\begin{matrix}-5-\lambda_{1} & 2\\ | ||
| + | 2 & -2-\lambda | ||
| + | \end{matrix}\right]\left[\begin{matrix}x_{1}\\ | ||
| + | x_{2} | ||
| + | \end{matrix}\right]=0 | ||
| + | \] | ||
| + | </math> | ||
Revision as of 12:43, 29 April 2014
Let define a n-by-n matrix A and a non-zero vector $ \vec{x}\in\mathbb{R}^{n} $. If there exists a scalar value $ \lambda $ which satisfies the vector equation $ A\vec{x}=\lambda\vec{x} $, we define $ \lambda $ as an eigenvalue of the matrix A, and the corresponding non-zero vector $ \vec{x} $ is called an eigenvector of the matrix A. To determine eigenvalues and eigenvectors a characteristic equation
is used. Here is an example of determining eigenvectors and eigenvalues where the matrix A is given by
Then the characteristic equation
By solving the quadratic equation for $ \lambda $, we will have two eigenvalues $ \lambda_{1}=-1 $ and $ \lambda_{2}=-6 $. By substituting $ \lambda's $ into Eq [eq:1]
$ \[ \left(A-\lambda_{1}I\right)\vec{x}=\left[\begin{matrix}-5-\lambda_{1} & 2\\ 2 & -2-\lambda \end{matrix}\right]\left[\begin{matrix}x_{1}\\ x_{2} \end{matrix}\right]=0 \] $
