Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 187
leading to this eigenspace and eigenvector.
( )
−2b2 ∣∣
{ b
b 2 ∈ C}
2
For λ 2 = 1 the system is
leading to this.
Five.II.3.22
2 · b 1 + 2 · b 2 = 0
−1 · b 1 − 1 · b 2 = 0
( )
−b2 ∣∣
{ b
b 2 ∈ C}
2
( )
−2
1
( )
−1
1
The characteristic equation
0 =
∣ −2 − x −1
5 2 − x∣ = x2 + 1
has the complex roots λ 1 = i and λ 2 = −i. This system
For λ 1 = i Gauss’ method gives this reduction.
(−2 − x) · b 1 − 1 · b 2 = 0
5 · b 1 (2 − x) · b 2 = 0
(−2 − i) · b 1 − 1 · b 2 = 0 (−5/(−2−i))ρ 1+ρ 2 (−2 − i) · b −→ 1 − 1 · b 2 = 0
5 · b 1 − (2 − i) · b 2 = 0
0 = 0
(For the calculation in the lower right get a common denominator
5
−2 − i − (2 − i) = 5
−2 − i − −2 − i
5 − (−5)
· (2 − i) =
−2 − i −2 − i
to see that it gives a 0 = 0 equation.) These are the resulting eigenspace and eigenvector.
( ) ( )
(1/(−2 − i))b2 ∣∣ 1/(−2 − i)
{
b
b 2 ∈ C}
2 1
For λ 2 = −i the system
leads to this.
Five.II.3.23
(−2 + i) · b 1 − 1 · b 2 = 0 (−5/(−2+i))ρ 1 +ρ 2 (−2 + i) · b −→ 1 − 1 · b 2 = 0
5 · b 1 − (2 + i) · b 2 = 0
0 = 0
( )
(1/(−2 + i))b2 ∣∣
{
b2 ∈ C}
b 2
( )
1/(−2 + i)
The characteristic equation is
1 − x 1 1
0 =
0 −x 1
∣ 0 0 1 − x∣ = (1 − x)2 (−x)
and so the eigenvalues are λ 1 = 1 (this is a repeated root of the equation) and λ 2 = 0. For the rest,
consider this system.
(1 − x) · b 1 + b 2 + b 3 = 0
−x · b 2 + b 3 = 0
(1 − x) · b 3 = 0
When x = λ 1 = 1 then the solution set is this eigenspace.
⎛ ⎞
1
{ ⎝b
0 ⎠ ∣ b 1 ∈ C}
0
When x = λ 2 = 0 then the solution set is this eigenspace.
⎛
{ ⎝ −b ⎞
2
b 2
⎠ ∣ b 2 ∈ C}
0
So these are eigenvectors associated with λ 1 = 1 and λ 2 = 0.
⎛
⎝ 1 ⎞ ⎛
0⎠
⎝ −1
⎞
1 ⎠
0 0
1