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1、Linear AlgebraEigenvalues and Eigenvectors of a MatrixDefinition.Let A be an n-order square matrix.If the number 0 and n-dimensional non-zero column vector satisfy:A=0,then 0 is the eigenvalue of matrix A and is a eigenvector of A belonging to 0.1.Eigenvalue and Eigenvector1.Eigenvalue and Eigenvect
2、orNote:If A is a singular matrix(|A|=0),then the homogeneous linear system Ax=0 has non-zero solutions.If is the non-zero solution of Ax=0,then A=0=0.Therefore,it can be seen that 0=0 is the eigenvalue of the singular matrix A,and the non-zero solutions of the linear system Ax=0 are all the eigenvec
3、tors of A belonging to the eigenvalue 0=0.Generally,it follows from A=0 that(0E A)=0.It can be seen that is a non-zero solution to the homogeneous linear system(0 E A)x=0.So we have|0 E A|=0.Definition.Let A be an n-order square matrix and be a parameter,then we call the determinantas the characteri
4、stic polynomial of square matrix A.And det(E A)=0 is called the characteristic equation of A.2 2.Characteristic Polynomial.Characteristic PolynomialSteps for computing the eigenvalues and eigenvectors of matrix A:(1)The eigenvalues of A are the solutions of the characteristic equation,and the nth-or
5、der square matrix A has n eigenvalues.(2)The eigenvectors of A belonging to the eigenvalues i are all non-zero solutions of homogeneous linear system(iE E A A)x x=0 0.Example.Find the eigenvalues and eigenvectors of the matrix Solution.The characteristic polynomial of A is =(=(-1-1)()(-2)-2)2 2-1=-1
6、=(-1-1)2 2(-3)-3)So the eigenvalues are 1=2=1,3=3.For 1=2=1,we need to solve(EA)x=0.then we get the same solution equation:and a fundamental solution 1=(0,0,1)T.So k 1(k0)is all the eigenvectors belonging to 1=2=1.SinceFor 3=3,we need to solve(3EA)x=0.then we get the same solution equation:and its fundamental solution 2=(1,1,1)T.Since So k2(k0)is all the eigenvectors belonging to 3=3.