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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 equation <math>A\vec{x}=\lambda\vec{x}</math>, we define <math>\lambda</math> | + | 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. |
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To determine eigenvalues and eigenvectors a characteristic equation | To determine eigenvalues and eigenvectors a characteristic equation | ||
Revision as of 11:37, 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]
