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•
A · x = 0 A n × n x
0 n × 1
•
A · x = λx λ
λ = 0
• T : V → V
V x T λ
• A n × n x A · x = λx A
λ A
◦
0
◦ λ S λ x A · x = λx
V S λ λ
∗ S λ
∗ S λ A · x1 = λx1 A · x2 = λx2 A · (x1 + x2) =λ(x1 + x2)
∗ S λ β A ·(βx) = β (A · x) = β (λx) =λ(βx)
• λx = (λI ) · x I
n × n A · x = λx = (λI ) · x (λI − A) · x = 0
x (λI
−A)
·x = 0 (λI
−A)
det(λI − A) = 0
• det(tI − A) n
t p(t) A
A
◦
A − tI tI − A tn
(−1)n
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◦
t2(t − 1)3 λ = 0, 0, 1, 1, 1
◦
A
•
◦
A tI − A
◦
1 i
√ 3
0 3 −80 0 π
π
2 0 1
0 3 20 0 2
• A
◦ tI − A
p(t)
◦ p(t) λ A
◦ λ x A · x = λx
(λI −A)·x = 0 x
λ
λ
• A =
1 00 1
◦ tI − A = t
− 1 00 t − 1
p(t) = det(tI − A) = (t − 1)2
◦ (t−1)2 = 0 t = 1 λ = 1, 1
◦
1 00 1
·
a
b
=
a
b
a
b
λ = 1
10
01
• A =
1 10 1
◦ tI − A =
t − 1 −10 t − 1
p(t) = det(tI − A) = (t − 1)2
◦ (t−1)2 = 0 t = 1 λ = 1, 1
◦
1 10 1
·
a
b
=
a
b
a
b
=
a + b
b
a b = 0
a
0
λ = 1
10
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• A =
1 0 1
−1 1 3−1 0 3
◦ tI −A =
t − 1 0 −1
1 t − 1 −31 0 t − 3
p(t) = (t−1)·
t − 1 −30 t − 3
+(−1)·
1 t − 11 0
=
(t −
1)2(t −
3) + (t −
1)
◦ p(λ) = (t − 1) · [(t − 1)(t − 3) + 1] = (t − 1)(t − 2)2 λ = 1, 2, 2
◦
∗ λ = 1
1 0 1
−1 1 3−1 0 3
·
a
b
c
= 1
a
b
c
a + c
−a + b + 3c
−a + 3c
=
a
b
c
c = 0 a = 0
0
b
0
0
10
.
∗ λ = 2
1 0 1−1 1 3
−1 0 3
·
a
b
c
= 2
a
b
c
a + c
−a + b + 3c
−a + 3c
=
2a
2b
2c
a = c b = 2c
c
2c
c
1
21
.
• λ A k
λ k
◦ λ
λ λ
λ
◦ (t − 1)3(t − 3)2 λ = 1
λ = 3
◦
λ
∗ k
(λI −
A) ·
x = 0
∗ λ k λI − A
B B k
∗ B λI −A
k B
∗
∗ k
• v1, v2, . . . ,vn A λ1, λ2, . . . , λn v1, v2, . . . ,vn
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◦ det(B) = det(P −1AP ) =det(P −1) det(A) det(P ) = det(A) det(P −1P ) = det(A)
◦ det(tI − B) = det(P −1 ·tI · P − P −1 · A · P ) = det(P −1(tI − A)P ) = det(tI − A)
∗
∗
◦ x A λ P −1 · x B λ
A · x = λx B · (P −1x) = P −1A(P P −1)x = P −1A · x = P −1 · (λx) = λ(P −1x)
∗ y B P · y A
∗ B A
• A B A
◦ A A
λ1, · · · , λn
λ1
λn
A
• A D
P D = P −1AP
◦ A =
−2 −63 7
P =
−1 21 −1
P −1 =
1 21 1
P −1AP =
4 00 1
= D
• A D = P −1AP A
◦ D Dk k D
◦ Dk = (P −1AP )k = P −1(Ak)P Ak = P · Dk · P −1
◦ A =
−2 −63 7
Dk =
4k 0
0 1
Ak =
−1 21 −1
·
4k 00 1
·
1 21 1
=
2 − 4k 2 − 2 · 4k−1 + 4k −1 + 2 · 4k
∗ k
k = −1
7
4
3
2
−3
4 −1
2
A−1
k = 1
2 B =
0 −21 3
B2 =
−2 −63 7
= A
• n × n A n
◦ A n v1, · · · , vn λ1, · · · , λn
P =
| | |
v1 · · · vn| | |
A
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∗ v1, · · · , vn A ·P =
| | |
Av1 · · · Avn| | |
=
| | |
λ1v1 · · · λnvn| | |
∗
| | |
λ1v1 · · · λnvn| | |
=
| | |
v1 · · · vn| | |
·
λ1
λn
= P · D
∗ A
· P = P
· D P
D = P −1AP
◦ D = P −1AP AP = P D
∗ P =
| | |
v1 · · · vn| | |
AP = P D
| | |
Av1 · · · Avn| | |
=
| | |
λ1v1 · · · λnvn| | |
Av1 = λ1v1 . . . Avn = λnvn v1, · · · , vn
P P
◦ n
n
•
◦
J 1J 2
J n
J 1, · · · , J n
J =
λ 1λ 1
1λ
λ
◦
2 1 0
0 2 00 0 3
J 1 =
2 10 2
J 2 = [3]
◦
x (λI − A)k · x = 0 k
k = 1
∗
• p(x) A p(A)
0
◦ A =
2 23 1
det(tI − A) =
t − 2 −2−3 t − 1
= (t − 1)(t − 2) − 6 =
t2 − 3t − 4 A2 =
10 6
9 7
A2 − 3A − 4I =
10 6
9 7
−
6 69 3
−
4 00 4
=
0 00 0
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◦ A A D = P −1AP D p(x)
A
∗ D λ1, · · · , λn A
A p(λ1) = · · · = p(λn) = 0
∗ D
p(D) = pλ1
λn =
p(λ1)
p(λn) =
0
0 = 0
∗ 0 = p(D) = p(P −1AP ) =P −1 [ p(A)] P p(A) = P · 0 · P −1 = 0
◦ A
J A D p(J ) = 0
• A
A = P −1DP
◦ A
◦ A
◦ A
◦ D
A P
A D
• A =
0 −23 5
D
P
D = P
−1
AP
◦ tI − A =
t 2−3 t − 5
det(tI − A) = t(t − 5) + 6 = t2 − 5t + 3 = (t − 2)(t − 3)
λ = 2, 3
◦
∗ λ = 2
0 −23 5
·
a
b
= 2
a
b
−2b
3a + 5b
=
2a
2b
a = −b
−b
b
λ = 2
−11
∗ λ = 3
0 −23 5
·
a
b
= 3
a
b
−2b
3a + 5b
=
3a
3b
a = −2
3b
−2
3 bb
λ = 3
−23
◦ A D =
2 00 3
P =
−1 −21 3
◦ P −1 =
−3 −21 1
P −1AP =
−3 −21 1
0 −23 5
−1 −21 3
=
2 00 3
=
D
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◦ D =
3 00 2
• A =
1 −1 0
0 2 00 2 1
D
P D = P −1AP
◦ tI −A =
t − 1 1 0
0 t − 2 00 −2 t − 1
det(tI −A) = (t−1)· t − 2 0
−2 t − 1
= (t−1)2(t−2)
λ = 1, 1, 2
◦
∗ λ = 1
1 −1 0
0 2 00 2 1
· a
b
c
=
a
b
c
a − b
2b
2b + c
=
a
b
c
b = 0
a
0c
λ = 1
100
001
∗ λ = 2
1 −1 00 2 00 2 1
· a
b
c
= 2
ab
c
a − b2b
2b + c
=
2a2b
2c
a = −b c = 2b
−b
b
2b
λ = 2
−1
12
◦ λ = 1 A D =
1 0 00 1 00 0 2
P =
1 0 −1
0 0 10 1 2
◦ P −1 =
1 1 0
0 −2 10 1 0
P −1AP =
1 1 0
0 −2 10 1 0
1 −1 0
0 2 00 2 1
1 0 −1
0 0 10 1 2
=
1 0 0
0 1 00 0 2
= D
•
A = 1 1 1
0 1 10 0 1
D
P D = P −1AP
◦ tI −A =
t − 1 −1 −1
0 t − 1 −10 0 t − 1
det(tI −A) = (t−1)3 tI −A
λ = 1, 1, 1
◦
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∗ λ = 1
1 1 1
0 1 10 0 1
·
a
b
c
=
a
b
c
a + b + c
b + c
c
=
a
b
c
a = b = 0
0
0c
λ = 1
0
01
◦ λ = 1
A
