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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. 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  
 +
<center><math>D(\lambda)=det\left(A-\lambda I\right)</math><center> 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\\
 
<center><math>A=\left[\begin{matrix}-5 & 2\\

Revision as of 12:36, 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) $<center> is used. Here is an example of determining eigenvectors and eigenvalues where the matrix A is given by <center>$ A=\left[\begin{matrix}-5 & 2\\ 2 & -2 \end{matrix}\right] $
.

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]

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