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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]

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood