Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 189
which leads to this eigenspace and eigenvector.
⎛
V √
2+ 3
= { ⎝ (1/(2 + √ ⎞
3) 2 ) · b 3
(1/(2 + √ 3)) · b 3
⎠ ∣ b 3 ∈ C}
b 3
Finally, substituting x = λ 3 = 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
⎛
⎝ (1/(2 + √ ⎞
3) 2 )
(1/(2 + √ 3)) ⎠
1
(−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
((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
which gives this eigenspace and eigenvector.
⎛
⎞
⎛
(1/(−2 + √ ⎞
3) 2 )
(1/(2 + √ 3) 2 ) · b 3
V √
2− 3
= { ⎝ (1/(2 − √ 3)) · b 3
⎠ ∣ b 3 ∈ C}
b 3
⎝
(1/(−2 + √ 3))
1
Five.II.3.25 With respect to the natural basis B = 〈1, x, x 2 〉 the matrix representation is this.
⎛
⎞
Rep B,B (t) = ⎝ 5 6 2
0 −1 −8⎠
1 0 −2
Thus the characteristic
⎛
equation
0 = ⎝ 5 − x 6 2 ⎞
0 −1 − x −8 ⎠ = (5 − x)(−1 − x)(−2 − x) − 48 − 2 · (−1 − x)
1 0 −2 − x
is 0 = −x 3 + 2x 2 + 15x − 36 = −1 · (x + 4)(x − 3) 2 . To find the associated eigenvectors, consider this
system.
(5 − x) · b 1 + 6 · b 2 + 2 · b 3 = 0
(−1 − x) · b 2 − 8 · b 3 = 0
b 1 + (−2 − x) · b 3 = 0
Plugging in x = λ 1 = 4 gives
b 1 + 6 · b 2 + 2 · b 3 = 0 b 1 + 6 · b 2 + 2 · b 3 = 0 b 1 + 6 · b 2 + 2 · b 3 = 0
−ρ 1 +ρ 2
−ρ 1 +ρ 2
−5 · b 2 − 8 · b 3 = 0 −→ −5 · b 2 − 8 · b 3 = 0 −→ −5 · b 2 − 8 · b 3 = 0
b 1 − 6 · b 3 = 0
−6 · b 2 − 8 · b 3 = 0
−6 · b 2 − 8 · b 3 = 0
( ) ( ) ( ) ( )
0 0 2 3
−1 0
−2 1
Five.II.3.26 λ = 1, and , λ = −2, , λ = −1,
0 1 1 0
1 0
1 0
Five.II.3.27 Fix the natural basis B = 〈1, x, x 2 , x 3 〉. The map’s action is 1 ↦→ 0, x ↦→ 1, x 2 ↦→ 2x, and
x 3 ↦→ 3x 2 and its representation is easy to compute.
⎛ ⎞
0 1 0 0
T = Rep B,B (d/dx) = ⎜0 0 2 0
⎟
⎝0 0 0 3⎠
0 0 0 0
We find the eigenvalues with this computation.
−x 1 0 0
0 = |T − xI| =
0 −x 2 0
0 0 −x 3
= x 4
∣ 0 0 0 −x∣
Thus the map
⎛
has the single
⎞
eigenvalue
⎛ ⎞
λ = 0.
⎛
To
⎞
find the associated eigenvectors, we solve
0 1 0 0 b 1
b 1
⎜0 0 2 0
⎟ ⎜b 2
⎟
⎝0 0 0 3⎠
⎝b 3
⎠ = 0 · ⎜b 2
⎟
⎝b 3
⎠ =⇒ b 2 = 0, b 3 = 0, b 4 = 0
0 0 0 0 b 4 b 4
B,B
B
B
B,B
⎠