DETERMINANTS AND EIGENVALUES 1. Introduction Gauss ...

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92 II. DETERMINANTS AND EIGENVALUESFor any specific λ, this is a homogeneous system of two equations in two unknowns.By the theory developed in the previous sections, we know that it will have nonzerosolutions precisely in the case that the rank is smaller than two. A simplecriterion for that to be the case is that the determinant of the coefficient matrixshould vanish, i.e.,[ ]2 − λ 1det=(2−λ) 2 −1=01 2−λor 4 − 4λ + λ 2 − 1=λ 2 −4λ+3=(λ−3)(λ − 1) = 0.The roots of this equation are λ = 3 and λ = 1. Thus, these and only these scalarsλ can be eigenvalues for appropriate eigenvectors.First consider λ = 3. Putting this in (2) yields(3)[2 − 3 11 2−3]v =Gauss-Jordan reduction yields[ ]−1 1→1 −1[−1 11 −1[ ]1 −1.0 0]v = 0.(As is usual for homogeneous systems, we don’t need to explicitly write downthe augmented matrix, because there are zeroes to the right of the ‘bar’.) Thecorresponding system is v 1 − v 2 = 0, and the general solution is v 1 = v 2 with v 2free. A general solution vector has the form[ ] [ ] [ ]v1 v2 1v = = = vv 2 v 2 .2 1Put v 2 = 1 to obtain[ ]1v 1 =1which will form a basis for the solution space of (3). Any other eigenvector forλ = 3 will be a non-zero multiple of the basis vector v 1 .Consider next the eigenvalue λ = 1. Put this in (2) to obtain(4)[2 − 1 11 2−1]v =[1 11 1]v = 0.In this case, Gauss-Jordan reduction—which we omit—yields the general solutionv 1 = −v 2 with v 2 free. The general solution vector is[ ] [ ]v1 −1v = = vv 2 .2 1Putting v 2 = 1 yields the basic eigenvector[ ]−1v 2 = .1
4. EIGENVALUES AND EIGENVECTORS FOR n × n MATRICES 93The general caseWe redo the above algebra for an arbitrary n × n matrix.eigenvector condition as followsA − λI =⎢⎣Av = λvAv − λv =0Av−λIv =0(A−λI)v =0.First, rewrite theThe last equation is the homogeneous n × n system with n × n coefficient matrix⎡a 11 − λ a 12 ... a 1n⎤a 21 a 22 − λ ... a 2n. . ... .a n1 a n2 ... a nn − λIt has a non-zero solution vector v if and only if the coefficient matrix has rankless than n, i.e., if and only if it is singular. By Theorem 2.1, this will be true ifand only if λ satisfies the characteristic equation⎡⎤a 11 − λ a 12 ... a 1na(5) det(A − λI) = det ⎢ 21 a 22 − λ ... a 2n⎣ .... ... .a n1 a n2 ... a nn − λ⎥⎦ .⎥⎦ =0.As in the example, the strategy for finding eigenvalues and eigenvectors is as follows.First find the roots of the characteristic equation. These are the eigenvalues.Then for each root λ, find a general solution for the system(6) (A − λI)v =0.This gives us all the eigenvectors for that eigenvalue. The solution space of thesystem (6), i.e., the null space of the matrix A − λI, is called the eigenspace correspondingto the eigenvalue λ.Example 3. Consider the matrix⎡A = ⎣ 1 4 3⎤4 1 0⎦.3 0 1The characteristic equation is⎡det(A − λI) = det ⎣ 1 − λ 4 4 1−λ 30⎤⎦3 0 1−λ=(1−λ)((1 − λ) 2 − 0) − 4(4(1 − λ) − 0) + 3(0 − 3(1 − λ)=(1−λ) 3 −25(1 − λ) =(1−λ)((1 − λ) 2 − 25)=(1−λ)(λ 2 − 2λ − 24) = (1 − λ)(λ − 6)(λ +4)=0.
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