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| − | Let define a n-by-n matrix A and a non-zero vector <math>\vec{x}\in\mathbb{R}^{n}</math>. If there exists a scalar value <math>\lambda</math> which satisfies the vector | + | Let define a n-by-n matrix A and a non-zero vector <math>\vec{x}\in\mathbb{R}^{n}</math>. If there exists a scalar value <math>\lambda</math> which satisfies the vector equation <math>A\vec{x}=\lambda\vec{x}</math>, we define <math>\lambda</math> |
| − | + | as an eigenvalue of the matrix A, and the corresponding non-zero vector <math>\vec{x}</math> is called an eigenvector of the matrix A. To determine eigenvalues and eigenvectors a characteristic equation <math>D(\lambda)=det\left(A-\lambda I\right)</math> is used. Here is an example of determining eigenvectors and eigenvalues where the matrix A is given by | |
| − | as an eigenvalue of the matrix A | + | |
| − | + | <math><center>A=\left[\begin{matrix}-5 & 2\\ | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
2 & -2 | 2 & -2 | ||
| − | \end{matrix}\right]. | + | \end{matrix}\right]<center></math>. |
| + | |||
Then the characteristic equation D(\lambda)=\left(-5-\lambda\right)\left(-2-\lambda\right)-4=\lambda^{2}+7\lambda+6=0. | Then the characteristic equation D(\lambda)=\left(-5-\lambda\right)\left(-2-\lambda\right)-4=\lambda^{2}+7\lambda+6=0. | ||
By solving the quadratic equation for \lambda | By solving the quadratic equation for \lambda | ||
Revision as of 11:32, 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
$ <center>A=\left[\begin{matrix}-5 & 2\\ 2 & -2 \end{matrix}\right]<center> $.
Then the characteristic equation D(\lambda)=\left(-5-\lambda\right)\left(-2-\lambda\right)-4=\lambda^{2}+7\lambda+6=0.
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]
