Linear Algebra Exercises-n-Answers.pdf

188 Linear Algebra, by Hefferon
Five.II.3.24 (a) The characteristic equation is
3 − x −2 0
0 =
−2 3 − x 0
∣ 0 0 5 − x∣ − 11x 2 + 35x − 25 = (x − 1)(x − 5) 2
and so the eigenvalues are λ 1 = 1 and also the repeated eigenvalue λ 2 = 5. To find eigenvectors,
consider this system.
(3 − x) · b 1 − 2 · b 2 = 0
−2 · b 1 + (3 − x) · b 2 = 0
(5 − x) · b 3 = 0
For λ 1 = 1 we get
2 · b 1 − 2 · b 2 = 0
−2 · b 1 + 2 · b 2 = 0
4 · b 3 = 0
leading to this eigenspace and eigenvector.
⎛ ⎞
⎛ ⎞
b 2
{ ⎝b 2
⎠ ∣ b 2 ∈ C} ⎝ 1 1⎠
0
0
For λ 2 = 1 the system is
−2 · b 1 − 2 · b 2 = 0
−2 · b 1 − 2 · b 2 = 0
0 · b 3 = 0
leading to this. ⎛
{ ⎝ −b ⎞ ⎛
2
b 2
⎠ + ⎝ 0 ⎞
⎛
0 ⎠ ∣ b 2 , b 3 ∈ C} ⎝ −1
⎞ ⎛
1 ⎠ , ⎝ 0 ⎞
0⎠
0 b 3 0 1
(b) The characteristic equation is
−x 1 0
0 =
0 −x 1
∣ 4 −17 8 − x∣ = −x3 + 8x 2 − 17x + 4 = −1 · (x − 4)(x 2 − 4x + 1)
and the eigenvalues are λ 1 = 4 and (by using the quadratic equation) λ 2 = 2 + √ 3 and λ 3 = 2 − √ 3.
To find eigenvectors, consider this system.
−x · b 1 + b 2 = 0
−x · b 2 + b 3 = 0
4 · b 1 − 17 · b 2 + (8 − x) · b 3 = 0
Substituting x = λ 1 = 4 gives the system
−4 · b 1 + b 2 = 0
−4 · b 2 + b 3 = 0
4 · b 1 − 17 · b 2 + 4 · b 3 = 0
−4 · b 1 + b 2 = 0
ρ 1+ρ 3
−→ −4 · b 2 + b 3 = 0
−16 · b 2 + 4 · b 3 = 0
leading to this eigenspace and eigenvector.
⎛
V 4 = { ⎝ (1/16) · b ⎞
3
(1/4) · b 3
⎠ ∣ b 2 ∈ C}
Substituting x = λ 2 = 2 + √ 3 gives the system
(−2 − √ 3) · b 1 + b 2 = 0
(−2 − √ 3) · b 2 + b 3 = 0
4 · b 1 − 17 · b 2 + (6 − √ 3) · b 3 = 0
b 3
−4 · b 1 + b 2 = 0
−4ρ 2+ρ 3
−→ −4 · b 2 + b 3 = 0
0 = 0
⎛
⎞
⎝ 1 4 ⎠
16
(−4/(−2− √ 3))ρ 1 +ρ 3
−→
(−2 − √ 3) · b 1 + b 2 = 0
(−2 − √ 3) · b 2 + b 3 = 0
+ (−9 − 4 √ 3) · b 2 + (6 − √ 3) · b 3 = 0
(the middle coefficient in the third equation equals the number (−4/(−2− √ 3))−17; find a common
denominator of −2 − √ 3 and then rationalize the denominator by multiplying the top and bottom
of the frsction by −2 + √ 3)
((9+4 √ 3)/(−2− √ (−2 − √ 3) · b 1 + b 2 = 0
3))ρ 2+ρ 3
−→ (−2 − √ 3) · b 2 + b 3 = 0
0 = 0