# Chapter2(1813) - 1 Ch2/MATH1813/YMC/2009 Chapter 2...

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Ch2/MATH1813/YMC/20091Chapter 2.Eigenvalues, Eigenvectors, Diagonalization2.1.Eigenvalues and EigenvectorsDefinition 2.1LetAMn(R). A nonzero vectorxMn×1(R)is called aneigenvectorofAifAxis a scalar multiple ofx, that is,Ax=λx(2.1)for someλR. The scalarλis called aneigenvalueofAandxis said to be an eigenvector ofAcorresponding toλ.In other words, the effect of multiplyingxby the matrixAis equivalent to multiplyingxby thescalarλ. The geometric interpretation is that an eigenvector of a given matrix is a vector whichis rescaled by a constant (eigenvalue), possibly reverse the direction, when we apply the matrix tothe vector. For example, an eigenvalue2means that the eigenvector is doubled in length, whilean eigenvalue-1means that the eigenvector stays the same in length but is reversed in direction.Example 2.2LetA=6-116-4!andx=18!. ThenAx=6-116-4!·18!=-2-16!=-218!=-2x.Hence-2is an eigenvalue ofAand the vector18!is an eigenvector ofAcorresponding to theeigenvalue-2.2RemarkNote that ifxis an eigenvector ofAcorresponding toλ, thenkx,k6= 0, is also aneigenvector ofAcorresponding toλ.Equation (2.1) can be written as(A-λIn)x=0.LetB=A-λIn. Ifλis an eigenvalue ofA, then there exists a nonzero vectorxsuch thatBx= 0. Proposition 1.46 then implies thatBisnotinvertible, that is,Bis singular, ordet(B) = 0.Therefore, in order to find eigenvalues ofA, we need to solvedet(A-λIn) = 0.(2.2)This equation is called thecharacteristic equationofA. The following gives the procedures offinding eigenvalues and their corresponding eigenvectors.
Ch2/MATH1813/YMC/20092Procedures to find eigenvalues and their corresponding eigenvectors1. Find all the eigenvalues ofAby solving the characteristic equation (2.2).2. Ifλ0is an eigenvalue ofA, then the eigenvectors ofAcorresponding toλ0are exactlythose NONTRIVIAL solutions of the system(A-λ0In)x=0.Example 2.3LetA=6-116-4!.Find all the eigenvalues ofAand their correspondingeigenvectors.Solution.The characteristic equation ofAis given by6-λ-116-4-λ= 0.Hence the eigenvalues ofAare the roots of the equationλ2-2λ-8=0, that is,λ1=-2andλ2= 4. Forλ1=-2, we haveA+ 2I2=8-116-2!.To find eigenvectors corresponding toλ1=-2, we solve8-116-2!·xy!=00!.Calculation shows that the solution to this system is given byxy!=α18!for anyαR\ {0}. Thus every vector of the formα18!is an eigenvector ofAcorrespondingto the eigenvalueλ1=-2. Similarly, forλ2= 4, we solve2-116-8!·xy!=00!,which givesxy!=β12!for anyβR\ {0}, and so we have obtained eigenvectors ofAcorresponding to the eigenvalueλ2= 4.2
Ch2/MATH1813/YMC/20093Example 2.4LetA=5-411!.Find all the eigenvalues ofAand their correspondingeigenvectors.