| Line 1: | Line 1: | ||
| − | Let define a n-by-n matrix A | + | 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 equationA <math>\vec{x}=\lambda\vec{x}</math>, |
| − | + | ||
| − | + | ||
| − | + | ||
we define \lambda | we define \lambda | ||
as an eigenvalue of the matrix A | as an eigenvalue of the matrix A | ||
Revision as of 11:30, 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 equationA $ \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 A=\left[\begin{matrix}-5 & 2\\
2 & -2 \end{matrix}\right].
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]
