| Line 15: | Line 15: | ||
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> | + | <center><math> |
\left(A-\lambda_{1}I\right)\vec{x}=\left[\begin{matrix}-5-\lambda_{1} & 2\\ | \left(A-\lambda_{1}I\right)\vec{x}=\left[\begin{matrix}-5-\lambda_{1} & 2\\ | ||
2 & -2-\lambda | 2 & -2-\lambda | ||
| Line 21: | Line 21: | ||
x_{2} | x_{2} | ||
\end{matrix}\right]=0 | \end{matrix}\right]=0 | ||
| − | + | </math></center> | |
| − | </math> | + | |
Revision as of 11:44, 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]
