Difference between revisions of "Sujin Test" - Rhea

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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 equationA <math>\vec{x}=\lambda\vec{x}</math>,
- 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
-\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]

Latest revision as of 11:46, 13 May 2014