Alumno
Department 1, Institute of Mathematics Chair for Numerical Mathematics an Scientific Computing Prof. Dr. G. Bader, Dr. A. Pawell
Problem Session to theCourse: Mathematics I
Environmental and Resource Management WS 2002/03 Solutions to Sheet No. 13 (Deadline: January, 27/28 2002)
Homework
H 13.1: Eigenvalues of A: λ1 = −1, λ2/3 = 1 Eigenvectors: λ1= −1: 2 0 0 0 1 1 x = 0, 0 1 1 λ2/3 = 1: 1 c2 = √ (0, 1, 1)T , 2 C= 0
1 √ 2 1 − √2
x = (0, 1, −1)T ,
1 c1 = √ (0, 1, −1)T 2
c3 = (1, 0, 0)T
1 √ 2 1 √ 2
0 1 0 . 0
q(x)= xT Ax = x2 + 2x2 x3 . 1 B = C, C T AC = diag(1, −1, −1)
2 2 2 q(Cy) = (Cy)T A(Cy) = y T C T ACy = y T diag(1, −1, −1)y = y1 − y2 − y3 .
H 13.2:
1−λ det(A − λE) = det 0 0
1 1−λ −1 1 = 5 −1 − λ
(1 − λ)[(1 − λ)(−1 − λ) + 5] = (1 − λ)[4 + λ2 ] = 0
⇒
λ1 = 1,
λ2/3 = ±2i
Eigenvectors: 1−λ 0 0 λ1 = 1: 1 1−λ −1 1 x = 0 5 −1 − λ
0 1 1 0 0 5 , 0 −1−2
x = t(1, 0, 0)T
λ1 = 2i:
1 − 2i 1 1 x = 0 0 1 − 2i 5 0 −1 −1 − 2i 2i , −1 − 2i, 1 1 − 2i
T
1 − 2i 1 1 0 −1 −1 − 2i x = 0, x = 0 0 0 λ1 = −2i: 1 + 2i 1 1 x = 00 1 + 2i 5 0 −1 −1 + 2i 1 + 2i 1 1 0 −1 −1 + 2i x = 0, x = 0 0 0 H 13.3: det(A − λE) = det 1−λ 1 1 3−λ
−2i , −1 + 2i, 1 1 + 2i
T
= (1 − λ)(3 − λ) − 1 = λ2 − 4λ + 2.
Eigenvalues:λ1/2 = 2 ± √ 2 √ 2 x = 0, x= √ (−1 + 4−2 2 1 √ 2, 1)T
Eigenvectors: λ = 2 + −1 − √ √
2 √1 1 1− 2
λ=2−
2 √ 2 √1 1 1+ 2 x = 0, x= √ (−1 − 4+2 2 1 √ 2, 1)T
−1 +
2
B
2 √ 4−2 2 √1 √ 4−2 2
−1+ √
√
−1− 2 √ √
√
√ 2+ 2 0
√
4+2 2 1 √ 4+2 2
B −1 = B T ,
B −1 AB =
0√ 2− 2
Additional Problems
P 13.1: 5−λ 2 det(A − λE) = det 0 2 0 5−λ 0 = 03−λ
(3 − λ) (5 − λ)2 − 4 = λ2 − 10λ + 21 = 0 ⇒ λ1 = 7, λ2,3 = 3. 1 1 1 x1 = √ 2 0 0 x3 = 0 1
−2 2 0 2 −2 0 x = 0, 0 0 −4 2 2 0 2 2 0 x = 0, 0 0 0
1 1 ...
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