Difference between revisions of "Sujin Test"

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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> 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

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~~<center><math>D(\lambda)=det\left(A-\lambda I\right)</math></center>~~

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~~is used. Here is an example of determining eigenvectors and eigenvalues where the matrix A is given by~~

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~~<center><math>A=\left[\begin{matrix}-5 & 2\\~~

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~~2 & -2~~

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~~\end{matrix}\right]</math></center>.~~

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~~Then the characteristic equation~~

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~~<center><math>D(\lambda)=\left(-5-\lambda\right)\left(-2-\lambda\right)-4=\lambda^{2}+7\lambda+6=0.</math></center>~~

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By solving the quadratic equation for <math>\lambda</math>, we will have two eigenvalues <math>\lambda_{1}=-1</math> and <math>\lambda_{2}=-6</math>. By substituting <math>\lambda's</math> into Eq [eq:1]

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~~<center><math>~~

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~~\left(A-\lambda_{1}I\right)\vec{x}=\left[\begin{matrix}-5-\lambda_{1} & 2\\~~

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~~2 & -2-\lambda~~

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~~\end{matrix}\right]\left[\begin{matrix}x_{1}\\~~

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~~x_{2}~~

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~~\end{matrix}\right]=0~~

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~~</math></center>~~

Difference between revisions of "Sujin Test" - Rhea

Latest revision as of 11:46, 13 May 2014

Alumni Liaison

@@ Line 1: / Line 1: @@
-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
-<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\\
-& -2
-\end{matrix}\right]</math></center>.
-Then the characteristic equation
-<center><math>D(\lambda)=\left(-5-\lambda\right)\left(-2-\lambda\right)-4=\lambda^{2}+7\lambda+6=0.</math></center>
-By solving the quadratic equation for <math>\lambda</math>, we will have two eigenvalues <math>\lambda_{1}=-1</math> and <math>\lambda_{2}=-6</math>. By substituting <math>\lambda's</math> into Eq [eq:1]
-<center><math>
-\left(A-\lambda_{1}I\right)\vec{x}=\left[\begin{matrix}-5-\lambda_{1} & 2\\
-& -2-\lambda
-\end{matrix}\right]\left[\begin{matrix}x_{1}\\
-x_{2}
-\end{matrix}\right]=0
-</math></center>