{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ch5.1 - 5.1 Eigenvalues and Eigenvectors 163 CHAPTER 5...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5.1 Eigenvalues and Eigenvectors 163 CHAPTER 5 EIGEN VAL UES AND EIGENVECTORS 5.1 Eigenvalues and Eigenvectors 2 — —1 1 0 AX = [—1 2 —1] [I] = [0] = 0X therefore X is an eigenvector of A corresponding to the eigenvalue 0 —1 — 2 1 0 (a) det(AI — A) = A2 — 2A — 3 = (A — 3)(A + 1); the eigenvalues areA = 3 and A = —1 (b) det(AI — A) = A2 — 8A + 16 = (A — 4)2; the eigenvalue isA = 4 (c) det(AI — A) = A2 — 12; the eigenvalues are A = x/fi = 2V3 and A = — 12 = —2\/3 (d) det(AI — A) = A2 + 3; no real eigenvalues (e) det(AI — A) = A2; the eigenvalue is A = 0 (f) det(AI — A) = A2 — 2A + 1 = (A — 1)2,' the eigenvalue isA = 1 A — 4 0 —1 (a) Cofactor expansion along the second column yields det(AI — A) = 2 A — 1 0 2 0 A — 1 = (A — 1) A g 4 [fl = (A — 1)[(A — 4)(A — 1) — (—1)(2)] = (A — 1)(A2 — 5A + 6) =(A — 1)(A — 2)(A — 3); the characteristic equation is (A — 1)(A — 2)(A — 3) = 0 A — 3 0 5 (b) We use the arrow technique to evaluate the determinant: det(AI — A) = —% A + 1 0 = —1 —1 A + 2 [(A—3)(A+1)(A+2)+0+1]—[—5(A+1)+0+0]=A3—2A; the characteristic equation is A3 — 2A = 0 A + 2 0 —1 (c) We use the arrow technique to evaluate the determinant: det(AI — A) = 6 A + 2 0 = —19 —5 A+4 [(A+2)2(A+4)+0+30]—[19(A+2)+0+0]=A3+8A2+A+8; the characteristic equation is A3 + 8A2 + A + 8 = 0 164 Chapter 5: Eigenvalues and Eigenvectors A + 1 0 —1 (d) We use the arrow technique to evaluate the determinant: det(AI — A) = 1 A — 3 0 = 4 —13 A + 1 [(A+1)2(A—3)+0+13]—[—4(A—3)+0+0]=A3—A2—A—2; the characteristic equation is A3 — A2 — A — 2 A — 5 0 —1 (e) We use the arrow technique to evaluate the determinant: det(AI — A) = —1 A — 1 0 = 7 —1 A [(A—5)(A—1)A+0—1]—[—7(A—1)+0+0] =A3—6A2+12A—8; the characteristic equation is A3 — 6A2 + 12A — 8 = 0 A — 5 —6 —2 0 A+1 8 —1 0 A+2 [(A—5)(A+1)(A+2)+48+0]—[2(A+1)+0+0]=A3—2A2—15A+36; (1‘) We use the arrow technique to evaluate the determinant: det(AI — A) = the characteristic equation is A3 — 2A2 — 15A + 36 = 0 (a) The reduced row echelon form of I — A = —3 0 —1 1 0 0 2 0 0 is 0 0 1 .The general solution of 2 0 0 0 0 0 x1 0 0 (I — A)X = 0 is x1 = 0, x2 = if, x3 = 0. In vector form, [362] = [if] = t [1]. A basis for the eigenspace x3 0 0 corresponding to A = 1 is {(010)}. ‘l —2 0 —1 1 0 The reduced row echelon form of 21 — A = l 2 1 0] is 0 1 _21 .The general solution of 2 0 1 0 0 0 X1 _3t _1 (21 — A)X = 0 is x1 = —lt, x2 = if, x3 = t. In vector form, x2 = 2 = t 2 .A basis for the 2 t 1 x3 t 1 eigenspace corresponding to A = 2 is {(—1,2,2)} (scaled by a factor of 2 for convenience). —1 0 —1 1 0 1 The reduced row echelon form of 31 — A = l 2 2 0] is [0 1 —1l. The general solution of 2 0 2 0 0 0 951 —t —1 (31 — A)X = 0 is 361 = —t, 362 = if, x3 = t. In vector form, [362] = [ if] = t[ 1]. A basis for the eigenspace x3 15 1 corresponding to A = 3 is {(—1,1,1)}. 5.1 Eigenvalues and Eigenvectors 165 _3 _ 5 3 (b) The reduced row echelon form of 0] — A = —% is _ l . The general solution of —1 —1 2 3 5t 3 (01—A)X=0 isx1 =—t, x2=—t, x3 =t. |nvectorform,l: l—l 3 E 3 — —t 1 . A basis for the eigenspace 3 1 corresponding to A = 0 is {(513)} (scaled by a factor 3 for convenience). —20—15\/§ The reduced row echelon form of x/EI — A = _E \E + 1 0 is 0 1 —2—v§ 6+5vE 20+15v§ 2+\/§ The general solution of (WI — A)X = 0 is x1 = 20+15fit20+15fi x1 6+5x/7t 6+5\F Invectorform, X2 = 2+v2 2+v2 _ x = x =t. 6+5\/§ I 2 6+5“? ’ 3 6+5\/—: 6+5\F A basis for the eigenspace corresponding to A = x/E is {(20 + 15%, 2 + x/Z 6 + 5%)} (scaled by a factor 6 + SW for convenience). —20+15\/E The reduced row echelon form of —\/§I — A = _E —\/E + 1 0 is 0 1 72+x/E 675V? —1 —1 —\/E + 2 0 0 0 . . 20—15% 2—\/§ The general solution of (—x/EI — A)X = 0 IS 361 = 645%? , x2 = _5\/§ t, 363 — t 2045/31f 20—15% x1 675/? 6—5/3 |n vector form, X2 27v? 2 t Z—fi x3 6_5\/§ 6—5\/§ t 1 A basis for the eigenspace corresponding to A = — 2 is {(20 — 15x5, 2 — x/Z 6 — EVE} (scaled by a factor 6 — SW for convenience). (c) The reduced row echelon form of —81 — A = The general solution of (—81 — A)X = 0 is x1 = —%t, x2 = —%t, x3 = t. 166 Chapter 5: Eigenvalues and Eigenvectors 1 1 X1 — E l: —E In vector form, X2 = _l t = t _1__ x3 6 6 t A basis for the eigenspace corresponding to A = —8 is {(—1, —1,6)} (scaled by a factor 6 for convenience). 1 3 0 —1 1 0 — 3 (d) The reduced row echelon form of 21 —A = 1 —1 0 is 0 1 _l . 4 13 3 3 0 0 The general solution of (21 — A)X = 0 is x1 = gt, x2 = £15,363 = 1 1 3t 3 |nvectorform, :2 — =1? 3. 3: 3 A basis for the eigenspace corresponding to A = 2 is {(1,1,3)} (scaled by a factor 3 for convenience). —3 0 —1 —1 1 0 is 7 —1 2 The general solution of (2] — A)X— — 01s x1: —lt, x2 = —%t, x3 = . 1 1 x1 — 3 ‘5 3 In vector form, 362 = _ l t = t _1. x3 3 3 t A basis for the eigenspace corresponding to A = 2 is {(—1, —1,3)} (scaled by a factor 3 for convenience). (e) The reduced row echelon form of 21 — A = C) O H C) H O Owll—‘WIH (f) The reduced row echelon form of 31 — A = 0 4 8] is [0 1 2]. The general solution of (31— A)X = 0 is x1 = 5t, x2 = —2t, x3 = t. In vector form, [x2] = [—215] = t [—2]. A basis for the eigenspace corresponding to A. = 3 is {(5, —2,1)}. 2 8 .The general solution of —9 —6 The reduced row echelon form of —4-1 — A = l_— 0 0:]' 15E: — 3 1 —2t —2 (—41 —A)X = 0 is 361 = —2t, x2 = St, x3 = t. In vector form, [X2] =l: :0]: till 3. A basis for the eigenspace corresponding to A. = —4 is {(—6,8,3)} (scaled by a factor 3 for convenience). 10. 12. 14. 5.1 Eigenvalues and Eigenvectors 167 In each part we employ the procedure described in Example 3 in Section 5.1. (a) det(AI —A) = A4 + A3 — 3A2 — A + 2. The only integer solutions of the characteristic equation are $1 and $2. Since det(11 — A) = 0, A — 1 must be a factor of the characteristic polynomial. Dividing A — 1 into A4 + A3 — 3A2 — A + 2 leads to det(AI — A) = (A — 1)(A3 + 2A2 — A — 2). The cubic polynomial can be found to be zero at A = 1 as well. Dividing A — 1 into A3 + 2A2 — A — 2 yields det(AI —A) = (A — 1)2(A2 + 3A + 2) = (A — 1)2(A + 1)(A + 2). We conclude that the eigenvalues are 1, —1, and —2. (b) det(AI — A) = A4 — 8A3 + 19A2 — 24A + 48. The only integer solutions of the characteristic equation are $1, $2, $3, $4, $6, $8, $12, $16, $24, and $48. Successively substituting these into the characteristic polynomial, we find det(41 — A) = 0 so that A — 4 must be a factor of the polynomial. Dividing A — 4 into A4 — 8A3 + 19A2 — 24A + 48 we obtain det(AI — A) = (A — 4)(A3 — 4A2 + 3A — 12). The cubic polynomial can have integer zeros $1, $2, $3, $4, $6, and $12. Substituting A = 3 results in a zero value, therefore dividing A — 4 into A3 — 4A2 + 3A — 12 leads to det(AI — A) = (A — 4)2(A2 + 3). The only real eigenvalue is 4. (a) The matrix is upper triangular, therefore by Theorem 5.1.2 its eigenvalues are —1 and 5, the entries on the main diagonal. (b) The matrix is lower triangular, therefore by Theorem 5.1.2 its eigenvalues are 3, 7, and 1, the entries on the main diagonal. (c) The matrix is diagonal, therefore by Theorem 5.1.2 its eigenvalues are — g, 1, and g, the entries on the main diagonal. We begin by finding eigenvalues and eigenvectors of the matrix A. A+1 2 2 —1 A—2 —1 1 1 A =A3—A2—A+1=A2(A—1)—1(A—1)=(A2—1)(A—1)=(A+1)(A—1)2 det(AI —A) = = [(A + 1)(A 2)A 2 2] [2(A 2) (A + 1) — 2A] From the characteristic equation (A + 1)(A — 1)2 = 0 it follows that A has eigenvalues 1 and —1. 168 16. 18. 26. Chapter 5: Eigenvalues and Eigenvectors 2 2 2 1 1 1 —1 —1 —1 is 0 0 0 . 1 1 1 0 0 0 The general solution of (I — A)X = 0 is x1 = —S — t, x2 = S, x3 = t. The reduced row echelon form of I — A = A basis for the eigenspace of the matrix A corresponding to A = 1 is {(—1,1,0), (—1,0,1)}. 0 2 2 1 0 —2 The reduced row echelon form of —11 — A = —1 — —1 is 0 1 1 . 1 1 —1 0 0 0 The general solution of (I — A)X = 0 is 361 = 215, 362 = —t, x3 = t. A basis for the eigenspace of the matrix A corresponding to A. = 1 is {(2, —1,1)}. Using Theorem 5.1.4, we conclude that the eigenvalues of the matrix A25 are 125 = 1 and (—1)25 = —1. Furthermore, the bases for eigenspaces corresponding to these eigenvalues are {(—1,1,0), (—1,0,1)} and {(2, —1,1)}, respectively. Since pol) = det (/11 — A), it follows that p(0) = det(—A) = (—1)"det (A). (a) n = 3 and p(0) = 5 therefore det(A) = —5 (b) n = 4- and p(0) = 7 therefore det(A) = 7 . a a A — a —a Denoting A = L1: {1:}, we have det(/11 — A) = l —a2111 A _ 21222 = (/1 — a11)(l — (122) — (112(121 = 12 — (“11 ‘l' “22)ll + 011022 — “12022 h—x —,_/ tr(A) det (A) A + 2 —2 —3 detW—A): 2 ,1—3 —2 =A3—6/12+11/1—6,- 4 —2 A — 5 The only integer solutions of the characteristic equation are i1, i2, i3, i6. Since det(11 — A) = 0, we divide A — 1 into A3 — 6/12 + 11/1 — 6 to obtain detUJ —A) = (/1 — 1)(/12 — 5/1 + 6) = (/1 — 1)(/1 — 2)(/1 — 3). 3 — —3 1 0 —1 2 — —2 is 0 1 0 . 4 — —4 0 0 0 The general solution of (I — A)X = 0 is x1 = t, x2 = 0, x3 = t. The reduced row echelon form of I — A = A basis for the eigenspace of the matrix A corresponding to A = 1 is {(1,0,1)}. 4 — —3 1 —% 0 The reduced row echelonform of 21—A= 2 — —2 is 0 0 1. 4 — ‘3 0 0 0 The general solution of (21 — A)X = 0 is x1 = it, x2 = t, x3 = 0. 5.1 Eigenvalues and Eigenvectors 169 A basis for the eigenspace of the matrix A corresponding to A = 2 is {(1,2,0)}. 5 — —3 1 0 —1 The reduced row echelon form of 31 — A = 2 0 —2 is 0 1 —1 . 4 — —2 0 0 0 The general solution of (31 — A)X = 0 is x1 = t, x2 = t, x3 = t. A basis for the eigenspace of the matrix A corresponding to A = 3 is {(111)}. (a) From the result of Exercise 23, the matrix A‘1 has 0 eigenvalue 1 with a basis for the corresponding eigenspace {(1,0,1)}, 0 eigenvalue % with a basis for the corresponding eigenspace {(1,2,0)}, 0 eigenvalue 2 with a basis for the corresponding eigenspace {(1,1,1)}. (b) From the result of Exercise 24, the matrix A — 31 has 0 eigenvalue —2 with a basis for the corresponding eigenspace {(1,0,1)}, 0 eigenvalue —1 with a basis for the corresponding eigenspace {(1,2,0)}, 0 eigenvalue 0 with a basis for the corresponding eigenspace {(1,1,1)}. (c) From the result of Exercise 24, the matrix A + 2] has 0 eigenvalue 3 with a basis for the corresponding eigenspace {(1,0,1)}, 0 eigenvalue 4 with a basis for the corresponding eigenspace {(1,2,0)}, 0 eigenvalue 5 with a basis for the corresponding eigenspace {(1,1,1)}. . (a) Since the degree of pol) is 6, A is a 6 X 6 matrix (see Exercise 17) (b) 29(0) qt 0, therefore 0 is not an eigenvalue of A. From parts (a) and (t) of Theorem 5.1.6, A is invertible. (c) A has three eigenspaces since it has three distinct eigenvalues, each corresponding to an eigenspace. 5.2 Diagonalization l4 —1 2 4 matrices. I = 18 does not equal I: 1| = 14 therefore, by Table 1 in Section 5.2, A and B are not similar ...
View Full Document

{[ snackBarMessage ]}