文档介绍：第五章特征值的估计 1
第五章特征值的估计
§ 特征值估计
一、特征值的上界
1
1 1 2
Th1 n×n
A ∈ R , M = ⋅ max{}aij − a ji ⇒ Im(λA ) ≤ M n(n − 1)
2 i, j 2
证明略去.
例1 An×n 是实对称矩阵⇒ Im(λA ) = 0, 即λA ∈ R .
η1
H
引理 1 ∀B = ()bij , ∀y = 满足 y = 1 ⇒ y By ≤ B .
n×n M 2 m∞
ηn
H
证 y By = bijηiη j ≤ max bij ⋅ηi η j
∑ i, j ∑
i, j i, j
1 2 2 1
≤ max bij ⋅η i + η j = max bij ⋅()n + n = B
i, j ∑ i, j m∞
2 i, j 2
1 1
Th2 A , λ≤ A , Re λ≤ A + AH , Im λ≤ A − AH .
n×n A m ()A ()A
∞ 2 m∞ 2 m∞
证设 A 的特征值为λ,特征向量 x 满足 x = 1, 则
2
Ax = λx ⇒λ= x H Ax, λ= x H AH x
(1) λ= x H Ax ≤ A
m∞
1 1
(2) λ+ λ= x H ()A + AH x ⇒ Re()λ= λ+ λ≤ A + AH
2 2 m∞
1 1
(3) λ−λ= x H ()A − AH x ⇒ Im()λ= λ−λ≤ A − AH
2 2 m∞
H
例2 A = A ⇒ Im()λ A = 0, A ∈ R
H
A = −A ⇒ Re()λ A = 0, 即λ A 是纯虚数或零.
Th4 A = a ,a , ,a ⇒λλλ≤ a a a .
n×n ( 1 2 L n ) 1 2 L n 1 2 2 2 L n 2
第五章特征值的估计 2
证 det A = 0 时, 结论成立.
det A ≠ 0 ⇒ a1 ,a2 ,L,an
b1 = a1 a1 = b1
b2 = a2 − k21b1 a2 = k21b1 + b2
LL LL
bn = an − kn,n−1bn−1 −L− kn1b1 an = kn1b1 +L+ kn,n−1bn−1 + bn
1 k21 L kn1
1 k
L n2
()a1 ,a2 ,L,an = (b1 ,b2 ,L,bn )
O M
1
∆
:
B =()b1 ,b2 ,L,bn det A = det B, bi ⊥b j (i ≠ j)
2 2
a = (k b + + k b ) + b
i 2 i1 1 L i,i−1 i−1 i 2
2 2 2
= k b + + k b + b ≥ b
i1 1 L i,i−1 i−1 2 i 2 i 2
2
det B = det B ⋅det B = det B ⋅ det B = det B H ⋅det B = det()B H B
2
b1
2 2
= det = b ⋅⋅ b
O ()1 2 L n 2
2
b
n 2
λλ= det A = det B = b b ≤ a a
1 L n 1 2 L n 2 1 2 L n 2
2 2 2
Th5 设 A 的特征值为λ, ,λ, 则λ+ + λ≤ A .
n×n 1 L n 1 L n F
λ1 *
L ∆
证对 A ,存在酉矩阵U , st. U H AU = =T
n×n O M
λn
2 2 2 2
由此可得λ+ + λ≤ T = A ().
1 L n F F
例3 设的特征值为, 则
An×n λ1 ,L,λn
2
AH A = AAH ⇔λ 2 + + λ= A 2
1 L n F
证必要性. AH A = AAH :存在酉矩阵 P, 使得
第五章特征值的估计 3
λ1
∆
H
P AP = O = Λ
λn
2 2 2 2