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\end{matrix}\right]</math></center>. | \end{matrix}\right]</math></center>. | ||
| − | + | Then the characteristic equation | |
| − | + | ||
| − | + | <center><math>D(\lambda)=\left(-5-\lambda\right)\left(-2-\lambda\right)-4=\lambda^{2}+7\lambda+6=0.</math></center> | |
| − | + | ||
| − | + | 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] | |
| − | + | ||
Revision as of 11:35, 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 $ D(\lambda)=det\left(A-\lambda I\right) $ 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]
