(New page: 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 de...) |
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Let define a n-by-n matrix A | Let define a n-by-n matrix A | ||
| − | and a non-zero vector \vec{x}\in\mathbb{R}^{n} | + | and a non-zero <math>vector \vec{x}\in\mathbb{R}^{n}</math> |
. If there exists a scalar value \lambda | . If there exists a scalar value \lambda | ||
which satisfies the vector equationA\vec{x}=\lambda\vec{x}, | which satisfies the vector equationA\vec{x}=\lambda\vec{x}, | ||
Revision as of 11:28, 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]
