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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, 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 | ||
| − | < | + | <center><math>A=\left[\begin{matrix}-5 & 2\\ |
2 & -2 | 2 & -2 | ||
| − | \end{matrix}\right]< | + | \end{matrix}\right]</math></center>. |
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. | ||
Revision as of 12:33, 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 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]
