一、线性空间与线性变换
1. 线性空间
线性空间 ：设V F V ⋅ 加法交换律、加法结合律、数乘结合律、两个分配律，0元存在，1元存在，负元存在 这八条运算律，则称V
（1）加法交换律：α + β = β + α ( α + β ) + γ = α + ( β + γ ) k ( l α ) = ( k l ) α ( k + l ) α = k α + l α k ( α + β ) = k α + k β α + 0 = α 1 ⋅ α = α α + β = 0
例如：
函数空间：全体实函数集合构成 实数域R
矩阵空间：复数域C m × n C m × n C
多项式空间：实数域R n R [ x ] n + 1 R
2. 基、维数
基、坐标、维数
若设线性空间V n α 1 , α 2 , ⋯ , α n V α α 1 , α 2 , ⋯ , α n α = k 1 α 1 + k 2 α 2 + ⋯ + k n α n α 1 , α 2 , ⋯ , α n V 基 ，称k 1 , k 2 , ⋯ , k n α α 1 , α 2 , ⋯ , α n 坐标 ，称V n 维 线性空间，记作dim V = n
维数是矩阵中自由量的个数，比如2 × 2 2 × 2
线性空间的基并不唯一，证明一组向量是线性空间的基，两步走：
证明这组向量线性无关.
证明线性空间任意向量可由这组向量表示.
基变换
设( α 1 , α 2 , ⋯ , α n ) ( β 1 , β 2 , ⋯ , β n ) n V n ( F ) β i = c 1 i α 1 + c 2 i α 2 + ⋯ + c ni a i = [ α 1 , α 2 , ⋯ , α n ] c 1 i c 2 i ⋮ c ni i = 1 , 2 , ⋯ , n
写成矩阵的形式为[ β 1 β 2 ⋯ β n ] = [ α 1 α 2 ⋯ α n ] c 11 c 21 ⋯ c n 1 c 12 c 22 ⋯ c n 2 ⋯ ⋯ ⋯ ⋯ c 1 n c 2 n ⋯ c nn = [ α 1 α 2 ⋯ α n ] C
称矩阵C ( α 1 , α 2 , ⋯ , α n ) ( β 1 , β 2 , ⋯ , β n ) 过渡矩阵 。
过渡矩阵是可逆的，若B = A C A = B C − 1
坐标变换
任取ξ ∈ V ξ ( α 1 , α 2 , ⋯ , α n ) [ a 1 a 2 ⋯ a n ] T ( β 1 , β 2 , ⋯ , β n ) [ b 1 b 2 ⋯ b n ] T
b 1 b 2 ⋮ b n = C − 1 a 1 a 2 ⋮ a n
与基变换B = A C b = C − 1 a
例：在4维线性空间R 2 × 2 ε 1 = [ 0 1 1 1 ] ε 2 = [ 1 1 0 1 ] ε 3 = [ 1 0 1 1 ] ε 4 = [ 1 1 1 0 ] μ 1 = [ 1 0 0 0 ] μ 2 = [ 1 0 1 0 ] μ 3 = [ 1 1 1 0 ] μ 4 = [ 1 1 1 1 ] ε μ A = [ 1 3 2 4 ]
解：4维空间的过渡矩阵是4 × 4 μ 1 = c 11 ε 1 + c 21 ε 2 + c 31 ε 3 + c 41 ε 4 μ 2 = ⋯
算出[ μ 1 μ 2 μ 3 μ 4 ] = [ ε 1 ε 2 ε 3 ε 4 ] − 3 2 3 1 3 1 3 1 − 3 1 − 3 1 3 2 3 2 0 0 0 1 3 1 3 1 3 1 3 1
设向量A ( x 1 x 2 x 3 x 4 ) T [ 1 3 2 4 ] = x 1 [ 0 1 1 1 ] + x 2 [ 1 1 0 1 ] + x 3 [ 1 0 1 1 ] + x 4 [ 1 1 1 0 ]
算出x 1 = 3 7 x 2 = 3 4 x 3 = 3 1 x 4 = − 3 2
向量A y 1 y 2 y 3 y 4 = − 3 2 3 1 3 1 3 1 − 3 1 − 3 1 3 2 3 2 0 0 0 1 3 1 3 1 3 1 3 1 − 1 x 1 x 2 x 3 x 4 = − 1 − 1 − 1 4
3. 线性子空间、维数公式
设V F n W V α , β ∈ W k α + lβ ∈ W W V 对加法和乘法封闭 即可）
比如
列空间R(A) ：由矩阵的列组成的列向量的生成空间；列向量的线性组合可以表示为A x R ( A ) = { A x ∣ x 属于 n 维空间 }
核空间N(A) ：核空间为解空间，比如N ( A ) = { x ∣ A x = 0 } A x = 0 dim N ( A ) A
子空间的交、和 ：设V 1 V 2 V
V 1 ⋂ V 2 = { α ∣ α ∈ V 1 且 α ∈ V 2 } V 1 ⋂ V 2 V V 1 + V 2 = { α = α 1 + α 2 ∣ α 1 ∈ V 1 且 α 2 ∈ V 2 } V 1 + V 2 V
直和子空间 ：如果W = V 1 + V 2 V 1 ⋂ V 2 = { 0 } W V 1 V 2 W = V 1 ⊕ V 2
直和补子空间 ：对于V V 1 V V 2 V = V 1 ⊕ V 2 V 2 V 1
维数公式 ：
dim ( V 1 + V 2 ) = dim V 1 + dim V 2 − dim ( V 1 ⋂ V 2 ) dim ( W 1 ∩ W 2 ) ⩽ dim W i ⩽ dim ( W 1 + W 2 ) ⩽ dim V
4. 线性映射
线性映射 ：若存在映射ϕ : V 1 → V 2 V 1 α 1 , α 2 ϕ ( α 1 + α 2 ) = ϕ ( α 1 ) + ϕ ( α 2 ) ϕ ( λ α 1 ) = λ ϕ ( α 1 ) ϕ V 1 V 2 线性映射 ，称α 1 ϕ ( α 1 ) 原像 ，ϕ ( α 1 ) α 1 像 。
若α 1 , α 2 , ⋯ , α s ϕ ( α 1 ) , ϕ ( α 2 ) , ⋯ , ϕ ( α s ) ϕ ( ∑ k i α i ) = ∑ k i ϕ ( α i )
核空间可以用映射来定义：
设ϕ V 1 V 2 ϕ ( V 1 ) = { β = ϕ ( α ) ∈ V 2 , ∀ α ∈ V 1 } ϕ ( V 1 ) V 2 ϕ 值域 ，记作R ( ϕ ) N ( ϕ ) = ϕ − 1 ( 0 ) = { α ∈ V 1 ∣ ϕ ( α ) = 0 } N ( ϕ ) V 1 ϕ 核子空间 ，dim N ( ϕ ) ϕ 零度 。
秩与零度定理 ：设ϕ n V 1 m V 2 dim R ( ϕ ) + dim N ( ϕ ) = n
线性变换 ：设ϕ V V ϕ V
线性变换的矩阵表示 ：设ϕ V α 1 , α 2 , ⋯ , α n V ϕ ( V ) ϕ ( α j ) = ∑ i = 1 m c ij α i j = 1 , 2 , ⋯ , n ϕ ( α 1 , α 2 , ⋯ , α n ) = ( α 1 , α 2 , ⋯ , α n ) c 11 c 21 ⋮ c n 1 c 12 c 22 ⋮ c n 2 ⋯ ⋯ ⋮ ⋯ c 1 n c 2 n ⋮ c nn = ( α 1 , α 2 , ⋯ , α n ) C n C ϕ α 1 , α 2 , ⋯ , α n 矩阵表示 。（简称为在某基下的矩阵）
例：在R [ x ] 4 p 1 = x 3 p 2 = x 2 p 3 = x p 4 = 1 D
解：微分运算是一种线性变换，这里线性映射是微分运算D
⎩ ⎨ ⎧ D ( p 1 ) = 3 x 2 = 0 p 1 + 3 p 2 + 0 p 3 + 0 p 4 D ( p 2 ) = 2 x = 0 p 1 + 0 p 2 + 2 p 3 + 0 p 4 D ( p 3 ) = 1 = 0 p 1 + 0 p 2 + 0 p 3 + 1 p 4 D ( p 4 ) = 0 = 0 p 1 + 0 p 2 + 0 p 3 + 0 p 4 0 3 0 0 0 0 2 0 0 0 0 1 0 0 0 0
同一线性变换在不同基下的矩阵之间的关系 ：设ϕ V α 1 , α 2 , ⋯ , α n β 1 , β 2 , ⋯ , β n V { α i } { β i } P ϕ α 1 , α 2 , ⋯ , α n A β 1 , β 2 , ⋯ , β n B = P − 1 A P
证明：∵ ( β 1 , β 2 , ⋯ , β n ) = ( α 1 , α 2 , ⋯ , α n ) P ϕ ( α 1 , α 2 , ⋯ , α n ) = ( α 1 , α 2 , ⋯ , α n ) A ϕ ( β 1 , β 2 , ⋯ , β n ) = ( β 1 , β 2 , ⋯ , β n ) B
∴ ( β 1 , β 2 , ⋯ , β n ) B = ϕ ( β 1 , β 2 , ⋯ , β n ) = ϕ [( α 1 , α 2 , ⋯ , α n ) P ] = ϕ ( α 1 , α 2 , ⋯ , α n ) P = ( α 1 , α 2 , ⋯ , α n ) A P = ( β 1 , β 2 , ⋯ , β n ) P − 1 A P
由于β 1 , β 2 , ⋯ , β n B = P − 1 A P
5. 特征值、特征向量
设A n λ n η A η = λ η λ A η A λ
相关定义：
λ E − A A 特征矩阵 .行列式∣ λ E − A ∣ A 特征多项式.
n ∣ λ E − A ∣ = 0 A A 特征值 （或特征根）.矩阵A A 谱 ，记作σ ( A )
( λ E − A ) X = 0 A 特征方程组 .
性质1：
∣ A ∣ = λ 1 λ 2 ⋯ λ n A ∑ i = 1 n a ij = ∑ i = 1 n λ i = t r ( A )
性质2：
k A kλ A m λ m m 若A A − 1 λ − 1
更一般地，f ( A ) f ( λ ) f ( x ) x
（此时η A kλ λ m λ − 1 f ( λ )
特征子空间 ：n A λ 0 R n A λ 0 V λ 0 V λ 0 ( λ 0 E − A ) X = 0
代数重复度 ：设λ 1 , λ 2 , ⋯ , λ r A r p 1 , p 2 , ⋯ , p r p i λ i ∣ λ E − A ∣ = ( λ − λ 1 ) p 1 ( λ − λ 2 ) p 2 ⋯ ( λ − λ r ) p r
几何重复度 ：特征子空间V λ 0 q i λ i q i = n − rank ( λ i E − A )
性质3：
属于不同特征值的特征向量是线性无关的.
设λ 1 , λ 2 , ⋯ , λ r A r q i λ i η i 1 , η i 2 , ⋯ , η i q i λ i q i A η 11 , η 12 , ⋯ , η 1 q 1 ; η 21 , η 22 , ⋯ , η 2 q 2 ; ⋯ ; η r 1 , η r 2 , ⋯ , η r q r
一个特征向量不能属于不同的特征值.
矩阵A λ i q i p i
6. 相似、对角化
相似对角化的条件 ：A A 单纯矩阵 。n A n 每一个特征值的几何重复度等于代数重复度 ）（推论：无重根则一定可以相似对角化）。
引理：对于分块矩阵C = [ A 0 0 B ] C A B
两个相似的矩阵它们的行列式的值相同、秩相同、特征值相同且对应的特征向量也相同。
同时对角化的条件 ：若存在可逆矩阵P P − 1 A P P − 1 BP A B A B = B A A B = B A A B
例：判断A = 3 2 1 − 1 0 − 1 1 1 2
解：∣ λ E − A ∣ = ( λ − 1 ) ( λ − 2 ) 2 λ 1 = 1 , λ 2 = 2 λ 1 λ 2
λ 2 E − A = − 1 − 2 − 1 1 2 1 − 1 − 1 0 → 行 1 0 0 − 1 0 0 1 1 0 rank ( λ 2 E − A ) = 2 q 2 = n − rank ( λ 2 E − A ) = 1
几何重复度不等于代数重复度，故A
二、Jordan标准形
1. λ
λ a ij ( λ ) i = 1 , 2 , ⋯ , m j = 1 , 2 , ⋯ , n F A ( λ ) = a 11 ( λ ) a 21 ( λ ) ⋯ a m 1 ( λ ) a 12 ( λ ) a 22 ( λ ) ⋯ a m 2 ( λ ) ⋯ ⋯ ⋯ ⋯ a 1 n ( λ ) a 2 n ( λ ) ⋯ a mn ( λ ) 多项式矩阵 ，或λ a ij ( λ ) A ( λ )
比如特征矩阵λ E − A λ
λ 初等变换 ：
矩阵的任意两行（列）互换位置.
非零常数c
矩阵的某一行（列）的φ ( λ ) φ ( λ ) λ
（和数字矩阵类似，每种初等变换都有对应的可逆的初等矩阵，且一个矩阵左乘初等矩阵相当于作初等行变换，右乘相当于作初等列变换）
λ 等价 ：如果A ( λ ) B ( λ ) A ( λ ) B ( λ ) A ( λ ) ⋍ B ( λ )
（和数字矩阵类似，等价具有自反性A ( λ ) ⋍ A ( λ ) A ( λ ) ⋍ B ( λ ) 则 B ( λ ) ⋍ A ( λ ) A ( λ ) ⋍ B ( λ ) 且 B ( λ ) ⋍ C ( λ ) 则 A ( λ ) ⋍ C ( λ )
2.
不变因子、行列式因子、初等因子
λ m × n λ A ( λ ) = d 1 ( λ ) ⋱ d r ( λ ) 0 ⋱ 0 d i ( λ ) d i ( λ ) ∣ d i + 1 ( λ ) d i + 1 ( λ ) d i ( λ ) Simth标准型 ，d 1 ( λ ) , d 2 ( λ ) , ⋯ , d r ( λ ) A ( λ ) 不变因子 。
比如[ λ 0 0 λ + 1 ] λ + 1 λ [ 1 0 0 λ ( λ + 1 ) ]
行列式因子 ：设rank ( A ( λ )) = r A ( λ ) k 1 ≤ k ≤ r D k ( λ ) A ( λ ) k D k ( λ ) r λ r
例：求A ( λ ) = 1 − λ λ 1 + λ 2 λ 2 λ λ 2 λ − λ − λ 2
解：每个位置上的元素都构成一个一阶行列式因子，比如1 − λ λ D 1 = 1
3 × 3 λ 1 − λ λ λ 2 λ = λ ( − λ 2 − λ + 1 ) 1 − λ λ 2 + 1 λ 2 λ 2 = λ 3 ( − λ − 1 ) λ D 2 = λ
三阶子式就是这个矩阵的行列式，∣ A ( λ ) ∣ = − λ 3 − λ 2 D 3 = λ 3 + λ 2
观察可知，D 1 ( λ ) D 2 ( λ ) D 2 ( λ ) D 3 ( λ ) ⋯ λ
初等因子 ：对于λ d 1 ( λ ) , d 2 ( λ ) , ⋯ , d r ( λ ) d 1 ( λ ) = ( λ − λ 1 ) e 11 ( λ − λ 2 ) e 12 ⋯ ( λ − λ s ) e 1 s ⋯ d r ( λ ) = ( λ − λ 1 ) e r 1 ( λ − λ 2 ) e r 2 ⋯ ( λ − λ s ) e rs λ 1 , ⋯ , λ s e ij ( λ − λ j ) e ij λ
由于d i ( λ ) ∣ d i + 1 ( λ ) d i ( λ ) 0 ≤ e 11 ≤ e 21 ≤ ⋯ ≤ e r 1 ⋯ 0 ≤ e 1 s ≤ e 2 s ≤ ⋯ ≤ e rs
例：若λ ⎩ ⎨ ⎧ d 1 ( λ ) = 1 d 2 ( λ ) = λ ( λ − 1 ) d 3 ( λ ) = λ ( λ − 1 ) 2 ( λ + 1 ) 2 d 4 ( λ ) = λ 2 ( λ − 1 ) 3 ( λ + 1 ) 3 ( λ − 2 ) λ λ λ 2 λ − 1 ( λ − 1 ) 2 ( λ − 1 ) 3 ( λ + 1 ) 2 ( λ + 1 ) 3 ( λ − 2 )
例：若已知5 × 6 λ λ λ λ 2 ( λ − 1 ) ( λ − 1 ) 2 ( λ − 1 ) 2 ( λ + i ) 3
解：d 4 ( λ ) = λ 2 ( λ − 1 ) 2 ( λ + i ) 3 d 3 ( λ ) = λ ( λ − 1 ) 2 d 2 ( λ ) = λ ( λ − 1 ) d 1 ( λ ) = 1
故其Simth标准型为： 1 0 0 0 0 0 λ ( λ − 1 ) 0 0 0 0 0 λ ( λ − 1 ) 2 0 0 0 0 0 λ 2 ( λ − 1 ) 2 ( λ + i ) 3 0 0 0 0 0 0 0 0 0 0 0
特别地，对于分块对角λ A ( λ ) = A 1 ( λ ) ⋱ A t ( λ ) A 1 ( λ ) , ⋯ , A t ( λ ) A ( λ )
对角型矩阵可以很快地写成Simth标准型，例如[ λ 0 0 λ + 1 ] λ λ λ + 1 [ 1 0 0 λ ( λ + 1 ) ]
定理：等价的λ
根据Simth标准型的性质d i ( λ ) ∣ d i + 1 ( λ ) λ d 1 ( λ ) = D 1 ( λ ) d 2 ( λ ) = D 1 ( λ ) D 2 ( λ ) ⋯ d r ( λ ) = D r − 1 ( λ ) D r ( λ )
或者写成D 1 ( λ ) = d 1 ( λ ) D 2 ( λ ) = d 1 ( λ ) d 2 ( λ ) D 3 ( λ ) = d 1 ( λ ) d 2 ( λ ) d 3 ( λ ) ⋯ D r ( λ ) = d 1 ( λ ) d 2 ( λ ) ⋯ d r ( λ )
定理：λ
定理：A ( λ ) B ( λ ) 等价的充要条件 是它们有相同的各阶行列式因子 。
定理：A ( λ ) B ( λ ) 等价的充要条件 是它们有相同的不变因子 。
定理：A ( λ ) B ( λ ) 等价的充要条件 是它们有相同的秩 和相同的初等因子 。
有时利用行列式因子可以方便地求出Simth标准型，而不需要作复杂的初等变换
例：求 λ − a c 1 λ − a c 2 ⋱ ⋱ λ − a c n − 1 λ − a c 1 , c 2 , ⋯ , c n − 1 = 0
解：它是一个对角阵，D n ( λ ) = ( λ − a ) n
观察矩阵，有一个特殊的n − 1 c 1 c 2 ⋯ c n − 1 = 0 D n − 1 = 1 D n − 2 = ⋯ = D 1 = 1 d 1 ( λ ) = d 2 ( λ ) = ⋯ = d n − 1 ( λ ) = 1 d n ( λ ) = ( λ − a ) n
数字矩阵的相似与λ ：设A B n A B λ E − A λ E − B
数字矩阵的不变因子、初等因子 ：对于数字矩阵A λ E − A A λ E − A A
定理：两个同阶方阵A B 相似的充要条件 是它们有相同的各阶行列式因子 。
定理：两个同阶方阵A B 相似的充要条件 是是它们有相同的不变因子 。
定理：两个同阶方阵A B 相似的充要条件 是它们有相同的初等因子 。（由于特征矩阵行列式的值不为0，故rank ( λ E − A ) = rank ( λ E − B ) = n
3. 方阵的Jordan标准型
如果一个方阵A Λ = P − 1 A P Λ A A J A = P − 1 A P J A
Jordan标准型 ：准对角矩阵J = J 1 J 2 ⋱ J s n i J i = a i 1 a i 1 ⋱ ⋱ ⋱ 1 a i ( λ E − J i ) D n i ( λ ) = ( λ − a i ) n i D n − 1 ( λ ) = ⋯ = D 1 ( λ ) = 1 J i ( λ − a i ) n i J ( λ − a 1 ) n 1 ( λ − a 2 ) n 2 ⋯ ( λ − a s ) n s
矩阵相似于Jordan标准型的条件 ：若A ∈ C n × n A ( λ − a 1 ) n 1 ( λ − a 2 ) n 2 ⋯ ( λ − a s ) n s A ∼ J = diag ( J 1 , J 2 , ⋯ , J s ) J i
若不考虑Jordan块的排列顺序，则Jordan标准型是唯一的。
推论：n A A
Jordan标准型的求法 ：
（1）求出初等因子，然后根据初等因子与Jordan标准型的关系直接写出Jordan标准型.
（2）Jordan标准型对角线上的元素是矩阵的特征值，先算出特征值，然后讨论如何分块.
例：求A = − 1 − 4 1 1 3 0 0 0 2
解：先求出A λ E − A = λ + 1 − 4 1 − 1 λ − 3 0 0 0 λ − 2 ⋍ 1 1 ( λ − 1 ) 2 ( λ − 2 ) A ( λ − 1 ) 2 ( λ − 2 ) A 1 0 0 1 1 0 0 0 2 2 0 0 0 1 0 0 1 1
例：求A = 2 1 − 2 3 8 − 14 2 2 − 3
解：先求出A f ( λ ) = ∣ λ E − A ∣ = λ − 2 − 1 2 − 3 λ − 8 14 − 2 − 2 λ + 3 = ( λ − 1 ) ( λ − 3 ) 2 A λ 1 = 1 λ 2 = 3
对于λ 1 = 1 f ( λ )
对于λ 2 = 3 A rank ( 3 E − A ) = 2 q 2 = n − rank ( λ 2 E − A ) = 1 p 2 = 2 A
故Jordan标准型为 1 0 0 0 3 0 0 1 3 3 0 0 1 3 0 0 0 1
Jordan标准型的相似变换矩阵 ：若n A J P P − 1 A P = J P P
例：求A = 3 3 − 2 0 − 1 0 8 6 − 5 P
解：λ E − A = λ − 3 3 2 0 λ + 1 0 8 − 6 λ + 5 ⋍ 1 0 0 0 λ + 1 0 0 0 ( λ + 1 ) 2 ( λ + 1 ) ( λ + 1 ) 2 − 1 0 0 0 − 1 0 0 1 − 1
设相似变换矩阵按列分块写成P = [ X 1 X 2 X 3 ] P − 1 A P = J A P = P J A P = [ A X 1 A X 2 A X 3 ] P J = [ − X 1 − X 2 X 2 − X 3 ] A X 1 = − X 1 A X 2 = − X 2 A X 3 = X 2 − X 3 ⎩ ⎨ ⎧ ( E + A ) X 1 = 0 ( E + A ) X 2 = 0 ( E + A ) X 3 = X 2 α 1 = 0 1 0 α 2 = − 2 0 1 X 1 = α 1 X 2 = α 2 X 2 X 2 X 2 = k 1 α 1 + k 2 α 2 rank ( E + A ) = rank ( E + A ∣ X 2 ) [ E + A ∣ X 2 ] = 4 3 − 2 0 0 0 8 6 − 4 − 2 k 2 k 1 k 2 rank ( E + A ) = 1 k 1 = 3 k 2 = − 2 rank ( E + A ∣ X 2 ) = 1 X 2 = 4 3 − 2 X 3 = 1 0 0 P 0 1 0 4 3 − 2 1 0 0
先了解Jordan标准型的几个性质 ：
比如P − 1 A P = J ⟶ A [ P 1 P 2 P 3 P 4 P 4 P 5 ] = [ P 1 P 2 P 3 P 4 P 5 ] 2 1 2 1 2 3 1 3 A P 1 = 2 P 1 A P 4 = 3 P 4 A P 2 = P 1 + 2 P 2 A P 3 = P 2 + 2 P 3 A P 5 = P 4 + 3 P 5
对于给定的λ i λ i
证明：设A λ i s rank ( λ i E − J ) = n − s λ i λ i q i = n − rank ( λ i E − A ) ∵ rank ( λ i E − A ) = rank ( P − 1 ( λ i E − A ) P ) = rank ( λ i E − J ) ∴ rank ( λ i E − J ) = n − q i s = q i
特征值λ i λ i
设n A J P − 1 A P = J rank ( λ i E − A ) l = rank ( P − 1 ( λ i E − A ) l P ) = rank [ p − 1 ( λ i E − A ) P ] l = rank ( λ i E − J ) l = 1 , 2 , ⋯
对应n i J i ( J i − λ i E ) l J i = λ i 1 λ i 1 ⋱ ⋱ ⋱ 1 λ i n i × n i ⎩ ⎨ ⎧ rank ( J i − λ i E ) = n i − 1 rank ( J i − λ i E ) 2 = n i − 2 ⋮ rank ( J i − λ i E ) n i − 1 = 1 rank ( J i − λ i E ) h = 0 , h ≥ n i j = i rank ( J j − λ i E ) l = n j
根据以上性质，可以得出求解Jordan标准型的另一种方法 ：
（1）计算rank ( A − λ i E ) l rank ( J − λ i E ) l = rank ( A − λ i E ) l l = 1 , 2 , ⋯
（2）通过分析rank ( J − λ i E ) l λ i
例：已知10阶矩阵A ∣ λ E − A ∣ = ( λ − 2 ) 7 ( λ − 3 ) 3
解：只知道特征多项式是无法得知初等因子的，因此无法得知如何分块，要作以下讨论
当λ = 2 rank ( 2 E − A ) = 7 rank ( 2 E − A ) 2 = 4 rank ( 2 E − A ) 3 = 3 rank ( 2 E − A ) 4 = 3 λ = 2 10 − 7 = 3 λ ≥ 2 7 − 4 = 3 λ = 2 ≥ 3 4 − 3 = 1 λ = 2 λ = 2
当λ = 3 rank ( 3 E − A ) = 8 rank ( 3 E − A ) 2 = 7 rank ( 3 E − A ) 3 = 7 λ = 3 10 − 8 − 2 λ = 3 ≥ 2 8 − 7 = 1 λ = 3 λ = 3
4. Jordan标准型的应用
解微分方程组
若有微分方程组⎩ ⎨ ⎧ d t d x 1 = a 11 x 1 + a 12 x 2 + ⋯ + a 1 n x n d t d x 2 = a 21 x 1 + a 22 x 2 + ⋯ + a 2 n x n ⋮ d t d x n = a n 1 x 1 + a n 2 x 2 + ⋯ + a nn x n d t d X = A X X = x 1 ( t ) x 2 ( t ) ⋮ x n ( t ) d t d X = d t x 1 d t x 2 ⋮ d t x n J A P − 1 A P = J X = P Y Y = y 1 ( t ) y 2 ( t ) ⋮ y n ( t ) P d t d Y = A P Y d t d Y = P − 1 A P Y = J Y Y X = P Y X
例：求下面微分方程组的解⎩ ⎨ ⎧ d t d x 1 = − x 1 − 2 x 2 + 6 x 3 d t d x 2 = − x 1 + 3 x 3 d t d x 3 = − x 1 − x 2 + 4 x 3
解：写成d t d X = A X X = x 1 ( t ) x 2 ( t ) x 3 ( t ) A = − 1 − 1 − 1 − 2 0 − 1 6 3 4 A P − 1 A P = J = 1 0 0 0 1 0 0 1 1 P = − 1 1 0 2 1 1 2 0 1 P − 1 = − 1 1 − 1 0 1 − 1 2 − 2 3
令X = P Y d t d Y = J Y ⎩ ⎨ ⎧ d t d y 1 = y 1 d t d y 2 = y 2 + y 3 d t d y 3 = y 3 ⎩ ⎨ ⎧ y 1 = k 1 e t y 2 = ( k 3 t + k 2 ) e t y 3 = k 3 e t X = P Y ⎩ ⎨ ⎧ x 1 = − k 1 e t + 2 k 3 e t + 2 ( k 3 t + k 2 ) e t x 2 = k 1 e t + ( k 3 t + k 2 ) e t x 3 = k 3 e t + ( k 3 t + k 2 ) e t
A m = E
设A F n m A m = E A m
证明：若存在可逆矩阵P P − 1 A P = J A m = E J m = ( P − 1 A P ) m = P − 1 A m P = E J i m = λ i m m λ i m − 1 λ i m ⋯ ⋱ ⋱ ∗ ⋮ m λ i m − 1 λ i m J i m = E J i A λ i m
例：若n A A 2 = A J = 1 ⋱ 1 0 ⋱ 0
证：由于J 2 = J J i λ i 2 = λ J λ i J = diag ( 1 , ⋯ , 1 , 0 ⋯ , 0 ) ∼ A
三、内积空间、正规矩阵、Hermite矩阵
1. 内积空间
内积（欧氏空间） ：设V R n V α β α β ( α , β )
( α , β ) = ( β , α ) ( k α , β ) = k ( α , β ) ( α + β , γ ) = ( α , γ ) + ( β , γ ) ( α , α ) ≥ 0 α = 0 ( α , α ) = 0
这里α β γ V k n V 欧氏空间 。
欧氏空间不是唯一的，只要满足上面的运算条件的映射都是欧氏空间，比如
（1）设α = ( x 1 , x 2 , ⋯ , x n ) T β = ( y 1 , y 2 , ⋯ , y n ) T ( α , β ) = α T β ( ⋅ , ⋅ ) R n R n
（2）若规定( α , β ) = x 1 y 1 + 2 x 2 y 2 + ⋯ + n x n y n ( ⋅ , ⋅ ) R n R n
（3）容易验证，( A , B ) := tr ( A B T )
（4）类似的，( f , g ) := ∫ a b f ( x ) g ( x ) d x
其中符号“:=
内积（酉空间） ：设V C n V α β α β ( α , β )
( α , β ) = ( β , α ) ( k α , β ) = k ( α , β ) ( α , k β ) = k ( α , β ) ( α + β , γ ) = ( α , γ ) + ( β , γ ) ( α , α ) ≥ 0 α = 0 ( α , α ) = 0
这里α β γ V k n V 酉空间 。
酉空间也不是唯一的，比如
（1）设α = ( a 1 , a 2 , ⋯ , a n ) T β = ( b 1 , b 2 , ⋯ , b n ) T ( α , β ) := ( β ) T α = a 1 b 1 + a 2 b 2 + ⋯ + a n b n ( ⋅ , ⋅ ) C n C n
（2）在n 2 C n × n ( A , B ) := tr ( A B H ) B H B C n × n
（3）设C [ a , b ] ( f , g ) := ∫ a b f ( x ) g ( x ) d x C [ a , b ]
其中B H
一般地，我们这样定义内积 ：设V n { α i } V α = ∑ i = 1 n x i α i β = ∑ j = 1 n y i α i α β ( α , β ) = ( ∑ i = 1 n x i α i , ∑ j = 1 n y i α i ) = ∑ i , j = 1 n x i y j ( α i , α j ) g ij = ( α i , α j ) g ij G = g 11 g 21 ⋯ g n 1 g 12 g 22 ⋯ g n 2 ⋯ ⋯ ⋯ ⋯ g 1 n g 2 n ⋯ g nn G { α i } 度量矩阵 ，根据定义可知，g ij = g ji ( G ) T = G ( α , β ) = X T G Y
2. Hermite矩阵
复共轭转置矩阵 ：设A ∈ C n × n A A A H = ( A ) T A H A A H
A H = ( A T ) ( A + B ) H = A H + B H ( k A ) H = k A H ( A B ) H = B H A H ( A H ) H = A ∣ A ∣ = ∣ A ∣ ( A k ) H = ( A H ) k ( A H ) − 1 = ( A − 1 ) H
Hermite矩阵 ：设A ∈ C n × n A H = A A Hermite矩阵 （简称H-阵）；若A H = − A A 反Hermite矩阵 （简称反H-阵）。
这个定义和实对称矩阵和反实对称矩阵是对应的。若不考虑对角线上的元素，H-阵a ij a ji a ij a ji
实对称矩阵可以看作是特殊的H-阵，欧式空间和酉空间的度量矩阵也是H-阵。
任意矩阵都可以表示为A = 2 A + A T + 2 A − A T A = 2 A + A H + 2 A − A H
3. 向量的长度、Schmidt正交化
向量的长度 ：设V α ∈ V ∣∣ α ∣∣ = ( α , α )
∣∣ α ∣∣ ≥ 0 α = 0 ∣∣ α ∣∣ = 0 ∣∣ k α ∣∣ = ∣ k ∣ ∣∣ α ∣∣ k ∈ C 三角不等式：∣∣ α + β ∣∣ ≤ ∣∣ α ∣∣ + ∣∣ β ∣∣
柯西-施瓦茨（Cauchy-Schwarz）不等式：∣∣ ( α , β ) ∣∣ ≤ ∣∣ α ∣∣ ∣∣ β ∣∣
一般地，我们将向量α = ( a 1 , a 2 , ⋯ , a n ) ∣∣ α ∣∣ = ∑ i = 1 n ∣ a i ∣ 2 ∣ a i ∣ a i
向量的夹角 ：设V α β < α , β >= arccos ∣∣ α ∣∣ ∣∣ β ∣∣ ( α , β ) 0 ≤< α , β >≤ π
向量的正交 ：在酉空间V ( α , β ) = 0 α β α ⊥ β
单位化 ：长度为1的向量称为单位向量，对于任意一个非零的向量α ∣∣ α ∣∣ α
正交向量组：设{ α i } { α i }
正交向量组是一个线性无关的向量组。
标准正交向量组：如果一个正交向量组中任何一个向量都是单位向量，则称此向量组为标准正交向量组。
标准正交基：在n n n
Schmidt正交化与单位化 ：设{ α 1 , α 2 , ⋯ , α r } n V r r s p an { α 1 , α 2 , ⋯ , α r }
第一步：正交化
β 1 = α 1
β 2 = α 2 − ( β 1 , β 1 ) ( α 2 , β 1 ) β 1
⋮
β r = α r − ( β 1 , β 1 ) ( α 2 , β 1 ) β 1 − ⋯ − ( β r − 1 , β r − 1 ) ( α r , β r − 1 ) β r − 1
得到的{ β i }
第二步：单位化
η 1 = ∣∣ β 1 ∣∣ β 1 η 2 = ∣∣ β 2 ∣∣ β 2 ⋯ η r = β r β r
得到的{ η i }
4. 酉矩阵
酉矩阵 ：酉矩阵是复数域上的正交矩阵，设A n A H A = A A H = E A A ∈ U n × n
酉矩阵的充要条件 ：设A ∈ C n × n A A n
酉矩阵的性质（设A , B ∈ U n × n
A − 1 = A H ∈ U n × n 酉矩阵的特征值都是模长为1的复数.（即都分布在复平面的单位圆上）
∣ det ( A ) ∣ = 1 ∣ ⋅ ∣ A B , B A ∈ U n × n
Householder矩阵 ：设α ∈ C n × n α H α = 1 G = E − 2 α α H G
对比正交矩阵：设A n A T A = A A T = E A A ∈ E n × n
正交矩阵的行、列向量组都是标准正交向量组。
正交矩阵的性质（设A , B ∈ E n × n
A − 1 = A T ∈ E n × n 正交矩阵的实特征值一定为1或-1.
det ( A ) = ± 1 A B , B A ∈ E n × n
（2）的证明：A η = λ η η T A T = λ η T A η η T A T A η = λ η T A η A T A = E A η = λ η η T η = λ 2 η T η ( λ 2 − 1 ) η T η = 0 η T η η = 0 λ 2 − 1 = 0 λ = ± 1
正交矩阵分为两类：设A ∈ E 2 × 2 ∣ A ∣ = 1 ⇒ Q − 1 A Q = [ cos θ sin θ − sin θ cos θ ] ∣ A ∣ = − 1 ⇒ P − 1 A P = [ − 1 0 0 1 ] P Q A ∈ E 3 × 3 T − 1 A T = a 0 0 0 cos θ sin θ 0 − sin θ cos θ ∣ A ∣ = 1 a = 1 ∣ A ∣ = − 1 a = − 1
酉矩阵（正交矩阵）的几何意义：保持向量内积、长度、夹角不变。
比如，设A ∈ U n × n α , β ∈ C n ( α , β ) = α H β ( A α , A β ) = ( A α ) H ( A β ) = α H A H A β = α H β = ( α , β )
5. 幂等矩阵、投影变换
幂等矩阵 ：设A ∈ C n × n A 2 = A A
幂等矩阵的性质：
A T A H E − A E − A T E − A H A ( E − A ) = ( E − A ) A = 0 N ( A ) = R ( E − A ) N ( E − A ) = R ( A ) A x = x x ∈ R ( A ) C n = R ( A ) ⊕ N ( A ) R ( A ) ⋂ N ( A ) = 0
性质（3）的证明：设x ∈ N ( A ) A x = 0 x − A x = x − 0 = x ⇒ ( E − A ) x = x x ∈ R ( E − A ) N ( A ) ⊆ R ( E − A ) y ∈ R ( E − A ) x ∈ C n y = ( E − A ) x A y = A ( E − A ) x = ( A − A 2 ) x = ( A − A ) x = 0 y ∈ N ( A ) R ( E − a ) ⊆ N ( A ) N ( A ) = R ( E − A )
性质（5）的证明：设z ∈ R ( A ) ⋂ N ( A ) ∃ x ∈ C n z = A x A z = 0 A z = A 2 x = A x = z = 0 z ∈ R ( A ) ⋂ N ( A ) = { 0 } x ∈ C n x = A x + ( x − A x ) A x ∈ R ( A ) ( x − A x ) ∈ R ( E − A ) = N ( A )
幂等矩阵的结构定理 ：设A r n A P ∈ C n n × n P − 1 A P = [ E r 0 0 0 ]
推论：设A n tr ( A ) = rank ( A )
投影变换 ：设S T n V V = S ⊕ T V α α = x + y x ∈ S y ∈ T x α T S y α S T 投影 。由上式确定的投影变换τ : V → S ⊆ V τ ( α ) = x V T S 投影变换 。
幂等矩阵与投影变换的关系 ：设A n τ ( α ) = A α ∀ α ∈ C n C n N ( A ) R ( A )
证明：C n = R ( A ) ⊕ N ( A ) α = A α + ( α − A α ) A α ∈ R ( A ) ( α − A α ) ∈ R ( E − A ) = N ( A )
定理：则下列命题等价（设τ n V
τ V τ 2 = τ τ A A 2 = A
从1到2的证明：设τ v T S ∀ α ∈ V α = x + y x ∈ S , y ∈ T τ ( α ) = x ∵ τ 2 ( α ) = τ ( τ ( α )) = τ ( x ) = x = τ ( α )
从2到1的证明：反过来证，∀ α ∈ V α = τ ( α ) + α − τ ( α ) ∵ τ 2 = τ ∴ τ ( α − τ ( α )) = τ ( α ) − τ 2 ( α ) = 0 α − τ ( α ) ∈ N ( τ ) V = R ( τ ) + N ( τ ) ∀ z ∈ R ( τ ) ⋂ N ( τ ) ∃ x ∈ V z = τ ( x ) τ ( z ) = τ 2 ( x ) = τ ( x ) = z z ∈ N ( τ ) z = τ ( z ) = 0 R ( τ ) ⋂ N ( τ ) = { 0 }
从2到3的证明：设α 1 , ⋯ , α n V A τ τ ( α 1 , ⋯ , α n ) = ( α 1 , ⋯ , α n ) A τ 2 ( α 1 , ⋯ , α n ) = τ [( α 1 , ⋯ , α n ) A ] = τ ( α 1 , ⋯ , α n ) A = ( α 1 , ⋯ , α n ) A 2 ∵ τ 2 = τ τ 2 ( α 1 , ⋯ , α n ) = ( α 1 , ⋯ , α n ) A ( α 1 , ⋯ , α n ) A 2 = A
空间的正交 ：设S T n V x ∈ S y ∈ T ( x , y ) = 0 S T
正交和 ：设S T n V S T S + T S T S + T
正交投影 ：设n V S T α ∈ V α = x + y x ∈ S y ∈ T σ : V → S ⊆ V σ ( α ) = x V S V T S
幂等矩阵与正交投影变换的关系 ：设A n σ ( α ) = A α ∀ α = C n C n R ( A )
证明：前面已经证明过A R ( A ) N ( A ) ∀ x ∈ R ( A ) y ∈ N ( A ) = R ( E − A ) z 1 , z 2 ∈ C n x = A z 1 y = ( E − A ) z 2 ( x , y ) = ( A z 1 , ( E − A ) z 2 ) = z 2 H ( E − A ) H A z 1 = z 2 H ( E − A ) A z 1 = z 2 H ( A − A 2 ) z 1 = 0
6. Schur引理与不等式
正交相似 ：设A , B ∈ R n × n U ∈ E n × n U T A U = U − 1 A U = B A B
酉相似 ：设A , B ∈ C n × n U ∈ U n × n U H A U = U − 1 A U = B A B
Schur引理 ：任何一个n A
Schur不等式 ：设A ∈ C n × n λ 1 , λ 2 , ⋯ , λ n A ∑ i = 1 n ∣ λ i ∣ 2 ≤ ∑ i , j ∣ a ij ∣ 2 A
7. 正规矩阵
正规矩阵 ：设A ∈ C n × n A A H = A H A A
实正规矩阵 ：设A ∈ R n × n A A T = A T A A
关于正规矩阵的一些结论：
H-阵、反H-阵、正交矩阵、酉矩阵、对角矩阵都是正规矩阵。
设A A
设A A
正规矩阵的结构定理 ：设A ∈ C n × n A U U H A U = λ 1 λ 2 ⋱ λ n λ 1 , λ 2 , ⋯ , λ n A
证明：根据Schur引理，A A
推论1：n n
推论2：正规矩阵属于不同特征值的特征向量彼此正交 。
定理：设A
A A A A A A
（如果A
证明：
可知A = U diag ( λ 1 , λ 2 , ⋯ , λ n ) U H
（1）对于H-阵，A H = U diag ( λ 1 , λ 2 , ⋯ , λ n ) H U H = U diag ( λ 1 , λ 2 , ⋯ , λ n ) U H = A = U diag ( λ 1 , λ 2 , ⋯ , λ n ) U H λ i = λ i λ i
8.
Hermite矩阵的结构定理、次酉矩阵
性质：设A ∈ C n × n
A + A H A A H A H A A − A H 如果A A k k
如果A A − 1
如果A i A i A i A
如果A B k A + lB k l
如果A B A B B A A B = B A
定理：
设A ∈ C n × n A X ∈ C n X H A X
设A ∈ C n × n A n B B H A B A B H A B
H-阵的结构定理 ：设A ∈ C n × n A U ∈ U n × n U H A U = λ 1 λ 2 ⋱ λ n λ 1 , λ 2 , ⋯ , λ n ∈ R A
推论：实对称矩阵正交相似于实对角矩阵。
次酉矩阵 ：设{ α 1 , α 2 , ⋯ , α r } n n × r U 1 = [ α 1 α 2 ⋯ α r ] U 1 ∈ U n × r
次酉矩阵的充要条件 ：U 1 ∈ U n × r U 1 H U 1 = E r × r
证明：U 1 H U 1 = α 1 H α 2 ⋮ α r H [ α 1 α 2 ⋯ α r ] = α 1 H α 1 α 2 H α 1 ⋯ α r H α 1 α 1 H α 2 α 2 H α 2 ⋯ α r H α 2 ⋯ ⋯ ⋯ ⋯ α 1 H α r α 2 H α r ⋯ α r H α r i = j α i α j = 0 i = j α i α j = 1 r × r
同时是H-矩阵和幂等矩阵的充要条件 ：设A n A = A H = A 2 n × r U 1 ∈ U n × r A = U 1 U 1 H rank ( A ) = r
证明：根据幂等矩阵的性质知，幂等矩阵一定酉相似于一个分块对角阵，U H A U = [ E r 0 0 0 ] A = U [ E r 0 ] [ E r 0 ] U H = U 1 U 1 H
9. Hermite二次型
定义：有n x 1 , x 2 , ⋯ , x n f ( x 1 , x 2 , ⋯ , x n ) = ∑ i = 1 n ∑ j = 1 n a ij x i x j Hermite二次型 ，其中规定a ij = a ji X = x 1 x 2 ⋮ x n ∈ C n A = a 11 a 21 ⋯ a n 1 a 12 a 22 ⋯ a n 2 ⋯ ⋯ ⋯ ⋯ a 1 n a 2 n ⋯ a nn f ( x 1 , x 2 , ⋯ , x n ) = X H A X A Hermite二次型对应的矩阵 ，并称A Hermite二次型的秩 。
写矩阵表示，比如：f ( x 1 , x 2 , x 3 ) = i x 1 x 2 + x 1 x 3 − i x 1 x 2 + x 1 x 3 = [ x 1 x 2 x 3 ] 0 − i 1 i 0 0 1 0 0 x 1 x 2 x 3
对应Hermite二次型，可作可逆的线性替换X = C Y f ( x 1 , x 2 , ⋯ , x n ) = X H A X = Y H ( C H A C ) Y = Y H B Y B = C H A C B H = B B A 复合同 （复相合）。
标准型 ：Hermite二次型中只含有纯平方项无交叉项的二次型f ( y 1 , y 2 , ⋯ , y n ) = λ 1 y 1 y 1 + λ 2 y 2 y 2 + ⋯ + λ n y n y n
规范型 ：Hermite二次型中只含有纯平方项无交叉项而且系数只有1、-1、0的二次型f ( y 1 , y 2 , ⋯ , y n ) = y 1 y 1 + ⋯ + y s y s − y s + 1 y s + 1 − ⋯ − y r y r
定理1：对于任意一个Hermite二次型f ( x 1 , x 2 , ⋯ , x n ) = X H A X 酉线性替换 X = U Y f ( X ) f ( X ) = λ 1 y 1 y 1 + λ 2 y 2 y 2 + ⋯ + λ n y n y n λ 1 , λ 2 , ⋯ , λ n A
定理2：对于任意一个Hermite二次型f ( x 1 , x 2 , ⋯ , x n ) = X H A X 可逆的线性替换 X = P Y f ( X ) f ( X ) = y 1 y 1 + ⋯ + y s y s − y s + 1 y s + 1 − ⋯ − y r y r rank ( A ) = r
正定Hermite二次型 ：对于f ( x 1 , x 2 , ⋯ , x n ) = ∑ i = 1 n ∑ j = 1 n a ij x i x j = X H A X x 1 , x 2 , ⋯ , x n f ( X ) > 0 A f ( X ) ≥ 0 A
如果一个矩阵是正定的，那它一定是一个H-阵，且可逆的线性替换不改变二次型的定性 。
比如：f ( y 1 , y 2 , y 3 ) = 4 y 1 y 1 + 8 y 2 y 2 + 3 y 3 y 3 f ( y 1 , y 2 , y 3 ) = 12 y 2 y 2 + 9 y 3 y 3 y 1 = 1 y 2 = y 3 = 0 f ( y 1 , y 2 , y 3 ) = 0
定理：下列命题等价（f ( X ) = X H A X
f ( X ) 对于任意n P P H A P
A n 存在n P P H A P = E
存在n Q A = Q H Q
*存在正线上三角矩阵R A = R H R
证明：（2）可逆的线性替换不改变二次型的定性（3）可以通过酉线性替换为标准型，由于是正定的，系数为特征值一定大于零（4）通过可逆的线性替换化为规范型，系数只能是1、-1、0，再由于正定，系数只能为1，所以是单位阵（5）由4可得
正定的充要条件 ：n A = ( a ij ) A n a 11 a 21 ⋮ a k 1 a 12 a 22 ⋮ a k 2 ⋯ ⋯ ⋯ a 1 k a 2 k ⋮ a kk > 0 k = 1 , 2 , ⋯ , n
例1：证明A A = E
解：因为A U ∈ U n × n A = U λ 1 ⋱ λ n U H λ i λ i > 0 A ∣ λ i ∣ = 1 λ i diag ( λ i ) = E U U U H = E A = E
例2：若A B A B B A
解：由于A Q A = Q H Q A B = Q H QB = Q H QB Q H ( Q H ) − 1 ∼ QB Q H A B QB Q H B QB Q H B A
例3：设A B A + B
解：证明可逆就是要证明矩阵的行列式的值不为零。由于A Q A = Q H Q ∣ A + B ∣ = ∣ Q H Q + B ∣ = ∣ Q H ∣∣ E + ( Q H ) − 1 B Q − 1 ∣∣ Q ∣ B ( Q H ) − 1 B Q − 1 = ( Q − 1 ) H B ( Q − 1 ) ∣ E + ( Q H ) − 1 B Q − 1 ∣ = 0 A + B
和正定对应，半正定Hermite二次型 的性质：下面命题等价（f ( X ) = X H A X
f ( X ) 对于任意n P P H A P
A n 存在n P P H A P = [ E r 0 0 0 ] r = rank ( A )
存在秩为r n Q A = Q H Q
定理：设A G A = G 2
证明：存在酉矩阵U A = U diag ( λ 1 , ⋯ , λ n ) U H λ i > 0 λ i ≥ 0 G = U diag ( λ 1 , ⋯ , λ n ) U H
10.
Hermite矩阵偶在复合同下的标准型
定理 ：设A B n B P ∈ C n n × n P H A P = λ 1 λ 2 ⋱ λ n P H BP = E n × n λ 1 , λ 2 , ⋯ , λ n P A B
证明：由于B P 1 ∈ C n n × n P 1 H B P 1 = E n × n P 1 H A P 1 P 2 ∈ U n × n P 2 H P 1 H A P 1 P 2 = λ 1 ⋱ λ n λ 1 , ⋯ , λ n P 1 H A P 1 n P 2 H P 1 H B P 1 P 2 = E n × n P = P 1 P 2
由于∣ λ E − P 1 H A P 1 ∣ = ∣ λ P 1 H B P 1 − P 1 H A P 1 ∣ = ∣ P 1 H ∣∣ λ B − A ∣∣ P 1 ∣ P 1 H A P 1 λ 1 , λ 2 , ⋯ , λ n ∣ λ B A ∣ A B P
Hermite矩阵偶在复合同下的标准型 （即用二次型的语言来描述前面的定理）：对于给定的两个Hermite二次型f 1 ( X ) = X H A X = ∑ i , j = 1 n a ij x i x j f 2 ( X ) = X H BX = ∑ i , j = 1 n b ij x i x j f 1 ( X ) 非退化的线性替换 X = P Y f 1 ( X ) f 2 ( X ) f 1 = λ 1 y 1 y 1 + λ 2 y 2 y 2 + ⋯ + λ n y n y n f 2 = y 1 y 1 + y 2 y 2 + ⋯ + y n y n λ 1 , λ 2 , ⋯ , λ n ∣ λ B − A ∣ = 0
定义：设A B n B A X = λ BX λ n ∣ λ B − A ∣ = 0 A B λ 1 , λ 2 , ⋯ , λ n A B 广义特征值 ，将λ i A X = λ BX X λ i 广义特征向量 。
广义特征值于广义特征向量的性质：
有n
有n X 1 , X 2 , ⋯ , X n A X i = λ i B X i i = 1 , 2 , ⋯ , n
这n X i H B X j = δ ij X i H A X j = λ j δ ij δ ij = { 1 , i = j 0 , i = j
性质（3）的理解：P 1 H A P 1 P 1 H A P 1 n n n
四、矩阵的分解
1. 矩阵的满秩分解
满秩分解定理 ：设A ∈ C r m × n r m × n B ∈ C r m × r C ∈ C r r × n A = BC B 列满秩 矩阵，C 行满秩 矩阵，因此我们称此分解为矩阵的满秩分解 。
证明：假设A r A [ E r 0 D 0 ] P A = [ E r 0 D 0 ] A = P − 1 [ E r 0 ] [ E r D ] = BC
如果A r P A Q = [ E r 0 D 0 ]
例1：求A = 1 1 2 4 2 2 4 8 1 2 3 6 0 1 1 2 1 3 4 8 2 3 5 10
解：作初等行变换得A → 行 1 0 0 0 2 0 0 0 0 1 0 0 − 1 1 0 0 − 1 2 0 0 1 1 0 0 = M rank ( A ) = 2 B = 1 1 2 4 1 2 3 6 ∈ C 2 4 × 2 C = [ 1 0 2 0 0 1 − 1 1 − 1 2 1 1 ] ∈ C 2 2 × 6
实际上这种分解并不唯一，也可以把M B = 1 1 2 4 0 1 1 2 C = [ 1 0 2 0 1 1 0 1 1 2 2 1 ]
例2：求A = [ 0 0 0 0 1 2 2 4 3 6 ]
解：A → 行 [ 0 0 0 0 1 0 2 0 3 0 ] B = [ 1 2 ] C = [ 0 0 1 2 3 ] B = [ 2 4 ] C = [ 0 0 2 1 1 2 3 ]
矩阵的满秩分解并不唯一 ，一般地，我们选取简化阶梯型主元所在的列对应的列向量构成列满秩矩阵，将阶梯型矩阵全为零的行去掉后即可构成行满秩矩阵。
定理：如果A = B 1 C 1 = B 2 C 2 A
存在矩阵θ ∈ C r r × r B 1 = B 2 θ C 1 = θ − 1 C 2
C 1 H ( C 1 C 1 H ) − 1 ( B 1 H B 1 ) − 1 B 1 H = C 2 H ( C 2 C 2 H ) − 1 ( B 2 H B 2 ) − 1 B 2 H
2. 矩阵的正交三角分解（UR分解）
正交三角分解定理 ：A ∈ C n n × n A 唯一 地分解为A = U 1 R 1 A = R 2 U 2 U 1 , U 2 ∈ U n × n 酉矩阵 ，R 1 正线上三角矩阵 ，R 2 正线下三角矩阵 。（正线上三角矩阵：即主对角线上的元素都是正数的上三角矩阵）
证明：首先证明分解的存在性 。将矩阵A A = [ α 1 α 2 ⋯ α n ] A α 1 , α 2 , ⋯ , α n β 1 , β 2 , ⋯ , β n η 1 . η 2 , ⋯ , η n α 1 , α 2 , ⋯ , α n η 1 . η 2 , ⋯ , η n ⎩ ⎨ ⎧ α 1 = c 11 η 1 α 2 = c 21 η 1 + c 22 η 2 ⋯ α n = c n 1 η 1 + c n 2 η 2 + ⋯ + c nn η n c 11 β 1 c 22 β 2 ⋯ c nn β n A = [ α 1 α 2 ⋯ α n ] = [ η 1 η 2 ⋯ η n ] c 11 c 21 c 22 ⋯ ⋯ ⋱ c n 1 c n 2 ⋮ c nn = U R
再证明唯一性 。设A A = U R = U R U − 1 U = R R − 1 U − 1 U R R − 1 U − 1 U = E R R − 1 = E U = U R = R
对于另一种分解R 2 U 2 B T ∈ C n n × n B T = U 1 R 1 B = R 1 T U 1 T = R 2 U 2 R 1
推论1：如果A ∈ C r m × r A A A = U R U R r
推论2：如果A ∈ C r r × n A A A = R U U R r
推论3：如果A ∈ C r m × n A A = U 1 R 1 R 2 U 2 U 1 ∈ U r m × r U 2 ∈ U r r × n R 1 r R 2
例：求A = 1 1 0 0 1 0 1 0 − 1 0 0 1
解：容易看出A ∈ C 3 4 × 3 A = [ α 1 α 2 α 3 ] ⎩ ⎨ ⎧ β 1 = α 1 = [ 1 1 0 0 ] T β 2 = α 2 − ( β 1 , β 1 ) ( α 2 , β 1 ) β 1 = [ 2 1 − 2 1 1 0 ] T β 3 = α 3 − ( β 1 , β 1 ) ( α 3 , β 1 ) β 1 − ( β 2 , β 2 ) ( α 3 , β 2 ) β 2 = [ − 3 1 3 1 3 1 1 ] T ⎩ ⎨ ⎧ η 1 = [ 2 2 2 2 0 0 ] T η 2 = [ 6 6 − 6 6 3 6 0 ] T η 3 = [ − 6 3 6 3 6 3 2 3 ] T α i η i ⎩ ⎨ ⎧ α 1 = 2 η 1 α 2 = 2 6 η 1 + 2 2 η 2 α 3 = − 2 2 η 1 − 6 6 η 2 + 3 2 3 η 3 β i β i A = [ η 1 η 2 η 3 ] 2 0 0 2 2 2 6 0 − 2 2 − 6 6 3 2 3
3. 矩阵的奇异值分解
引理1：对于任意矩阵A rank ( A A H ) = rank ( A H A ) = rank ( A )
证明：A A H A H A A A H A A x = 0 A H A x = 0 A x = 0 A H A H A x = 0 x H x H A H A x = 0 A x x H A H = ( A x ) H
引理2：对于任意矩阵A A A H A H A
证明：用A A H x H A A H x = ( A H x ) H A H x ≥ 0 x ∈ C m A H x m A H A
定理：设A ∈ C r m × n λ i A A H m × m μ i A H A n × n λ 1 ≥ λ 2 ≥ ⋯ ≥ λ r > λ r + 1 = λ r + 2 = ⋯ = λ m = 0 μ 1 ≥ μ 2 ≥ ⋯ ≥ μ r > μ r + 1 = μ r + 2 = ⋯ = μ n = 0 λ i = μ i > 0 i = 1 , 2 , ⋯ , r α i = λ i = μ i > 0 i = 1 , 2 , ⋯ , r A 正奇异值 ，简称奇异值 。
上面定理简单地说就是，A A H A H A A A H λ i x A H A λ i A H x
证明：设A ∈ C r m × n A A H A H A
A A H x = λ i x ⟶ A H A A H x = λ i A H x A H x λ i A H A A H x x A A H A H x A H x = 0 A A H x = λ i x = 0 λ i x A H x
A H A A A H
定理：正规矩阵的奇异值 为其非零特征值的模长。
推论：
H-阵的奇异值是其非零特征值的绝对值.
正定H-阵的奇异值是其非零特征值.
酉矩阵的奇异值全部为1.
证明：根据结构定理知，正规矩阵A U H A U = diag ( λ 1 , ⋯ , λ n ) U H A H U = diag ( λ 1 , ⋯ , λ n ) U H A A H U = diag ( ∣ λ 1 ∣ 2 , ⋯ , ∣ λ n ∣ 2 )
例1：求A = 1 0 0 2 0 0
解：A A H = 5 0 0 0 0 0 0 0 0 A A H A 5
例2：求A = [ 0 2 − 1 0 1 0 ]
解：A A H = [ 2 0 0 4 ] 2
奇异值分解定理 ：设A ∈ C r m × n α 1 ≥ α 2 ≥ ⋯ ≥ α r A r m 酉矩阵 U n 酉矩阵 V U H A V = [ Δ 0 0 0 ] m × n Δ = α 1 ⋱ α r r × r α 1 ≥ α 2 ≥ ⋯ ≥ α r > 0
证明：由于rank ( A A H ) = rank ( A ) = r A A H α 1 2 ≥ α 2 2 ≥ ⋯ ≥ α r 2 > 0 α r + 1 2 = α r + 2 2 = α m 2 = 0 A A H m U H A A H U = [ Δ 2 0 0 0 ] U U = [ U 1 U 2 ] U 1 ∈ U r m × r U 2 ∈ U m − r m × ( m − r ) [ U 1 H U 2 H ] A A H [ U 1 U 2 ] = [ Δ 2 0 0 0 ] { U 1 H A A H U 1 = Δ 2 U 2 H A A H U 2 = 0 A A H Δ 2 U 1 Δ − 1 = ( Δ − 1 ) H ( Δ − 1 ) H U 1 H A A H U 1 Δ − 1 = E A H U 2 A H U 2 = 0
令V 1 = A H U 1 Δ − 1 V 1 ∈ U r n × r V 1 H V 1 = E V 2 V = [ V 1 V 2 ] V 1 V 2 0 = V 1 H V 2 = Δ − 1 U 1 H A V 2 ⟶ U 1 H A V 2 = 0
由前面得U H A V = [ U 1 H U 2 H ] A [ V 1 V 2 ] = [ U 1 H A V 1 U 2 H A V 1 U 1 H A V 2 U 2 H A V 2 ] = [ Δ 0 0 0 ]
奇异值分解式(SVD) ：把A = U [ Δ 0 0 0 ] V H A
留意到A A H = U [ Δ 2 0 0 0 ] U H ⟶ A A H U = U [ Δ 2 0 0 0 ] m × m U [ Δ 2 0 0 0 ] A H A = V [ Δ 2 0 0 0 ] V H ⟶ A H A V = V [ Δ 2 0 0 0 ] n × n
由此可知U A A H V A H A V 1 = A H U 1 Δ − 1
由此可知A
推论 ：设A ∈ C r m × n α 1 ≥ α 2 ≥ ⋯ ≥ α r A r 次酉矩阵 U r ∈ U r m × r V r ∈ U r n × r A = U r Δ V r H A U V
例1：求A = 2 0 0 i 0 0
解：A A H A α = 5 u 1 = 1 0 0 u 2 = 0 1 0 u 3 = 0 0 1 U = 1 0 0 0 1 0 0 0 1 V 1 = A H U 1 Δ − 1 = [ 2 − i 0 0 0 0 ] 1 0 0 5 1 = [ 5 2 − 5 i ] V 2 = [ − 5 i 5 2 ] V 1 V = [ 5 2 − 5 i − 5 i 5 2 ] A = U 5 0 0 0 0 0 V H
例2：求A = 0 − 1 0 1 1 0 2 0
解：由于A A H 4 × 4 B = A H B
B = A H = U [ 5 0 0 2 0 0 0 0 ] V H U = [ 0 1 1 0 ] V = [ V 1 V 2 ] = 5 1 0 5 2 0 0 − 2 1 0 2 1 5 2 0 − 5 1 0 0 2 1 0 2 1 A = B H = V 5 0 0 0 0 2 0 0 U H
4. 矩阵的极分解
极分解定理 （A A ∈ C n n × n 酉矩阵 U ∈ U n × n 正定H-阵 H 1 H 2 A = H 1 U = U H 2 唯一 的，同时有A A H = H 1 2 A H A = H 2 2 A = H 1 U = U H 2 极分解表达式 。
证明：设A ∈ C n n × n A H A H 2 A H A = H 2 2 ( H 2 H ) − 1 A H A H 2 − 1 = E ( A H 2 − 1 ) H ( A H 2 − 1 ) = E A H 2 − 1 A H 2 − 1 = U A = U H 2 A = U H 2 = U H 2 U H U = H 1 U U H 2 U H = H 1 A A H = H 1 2
极分解定理 （A A ∈ C n × n 酉矩阵 U ∈ U n × n 半正定H-阵 H 1 H 2 A = H 1 U = U H 2 A A H = H 1 2 A H A = H 2 2
证明：根据奇异值分解定理，存在酉矩阵U 1 U 2 A = U 1 α 1 ⋱ α n U 2 α 1 ≥ ⋯ ≥ α r > α r + 1 = ⋯ α n = 0 α 1 , ⋯ , α r A r A = ( U 1 diag ( α 1 , ⋯ , α n ) U 1 H ) ( U 1 U 2 ) = ( U 1 U 2 ) ( U 2 H diag ( α 1 , ⋯ , α n ) U 2 ) H 1 = U 1 diag ( α 1 , ⋯ , α n ) U 1 H H 2 = U 2 H diag ( α 1 , ⋯ , α n ) U 2 U = U 1 U 2 A = H 1 U = U H 2
5. 谱分解(正规矩阵)
正规矩阵的谱分解 ：设A U ∈ U n × n A = U diag ( λ 1 , λ 2 , ⋯ , λ n ) U H U A = [ α 1 α 2 ⋯ α n ] λ 1 λ 2 ⋱ λ n α 1 H α 2 H ⋮ α n H = λ 1 α 1 α 1 H + λ 2 α 2 α 2 H + ⋯ + λ n α n α n H α i A λ i A 谱分解表达式 。
设正规矩阵A r λ 1 , ⋯ , λ r λ i n i λ i n i α i 1 , ⋯ , α i n i A A = ∑ i = 1 r λ i ∑ j = 1 n i α ij α ij H = ∑ i = 1 r λ i G i G i = ∑ j = 1 n i α ij α ij H G i H = G i = G i 2 G i G k = 0 i = k
定理 ：设A n r λ 1 , ⋯ , λ r λ i n i A r n G 1 , G 2 , ⋯ , G r
A = ∑ i = 1 r λ i G i G i H = G i = G i 2 G i G k = 0 i = k ∑ i = 1 r G i = E 满足上述性质的矩阵G i
rank ( G i ) = n i
其中G i C n V λ i
例1：求正规矩阵A = 0 1 1 − 1 1 0 − 1 1 1 − 1 0 1 − 1 1 1 0
解：先求出A ∣ λ E − A ∣ = ( λ − 1 ) 3 ( λ + 3 ) λ 1 = λ 2 = λ 3 = 1 λ 4 = − 3 λ = 1 α 1 = 1 1 0 0 α 2 = 1 0 1 0 α 3 = − 1 0 0 1 λ = − 3 α 4 = 1 − 1 − 1 1 α 1 , α 2 , α 3 η 1 = 2 1 2 1 0 0 η 2 = 6 1 − 6 1 6 2 0 η 3 = − 12 1 12 1 12 1 12 3 α 4 η 4 = 2 1 − 2 1 − 2 1 2 1 λ = 1 G 1 = η 1 η 1 H + η 2 η 2 H + η 3 η 3 H = 4 3 4 1 4 1 − 4 1 4 1 4 3 − 4 1 4 1 4 1 − 4 1 4 3 4 1 − 4 1 4 1 4 1 4 3 λ = − 3 G 2 = η 4 η 4 H = 4 1 − 4 1 − 4 1 4 1 − 4 1 4 1 4 1 − 4 1 − 4 1 4 1 4 1 − 4 1 4 1 − 4 1 − 4 1 4 1 A A = G 1 − 3 G 2
例2：求正规矩阵A = 0 1 i − 1 0 0 i 0 0
解：∣ λ E − A ∣ = λ ( λ 2 + 2 ) λ 1 = − 2 i λ 2 = 2 i λ 3 = 0 α 1 = − 2 − i 1 α 2 = 2 − i 1 α 3 = 0 i 1 η 1 = − 2 1 − 2 i 2 1 η 2 = 2 1 − 2 i 2 1 η 3 = 0 2 i 2 1 A = − 2 i η 1 η 1 H + 2 i η 2 η 2 H + 0 η 3 η 3 H = − 2 i 2 1 4 2 i − 4 2 − 4 2 i 4 1 4 i − 4 2 − 4 i 4 1 + 2 i 2 1 − 4 2 i 4 2 4 2 i 4 1 4 i 4 2 − 4 i 4 1 + 0 0 0 0 0 2 1 − 2 i 0 2 i 2 1
（注意，即使特征值是0，那一项也必须写上）
6. 谱分解(可对角化矩阵)
可对角化矩阵的谱分解 ：设A n λ 1 , ⋯ , λ n α 1 , ⋯ , α n P = [ α 1 ⋯ α n ] P − 1 = [ β 1 T ⋯ β n T ] T A = P λ 1 ⋱ λ n P − 1 = [ α 1 ⋯ α n ] λ 1 ⋱ λ n β 1 T ⋮ β n T = λ 1 α 1 β 1 T + λ 2 α 2 β 2 T + ⋯ + λ n α n β n T λ i n i α i 1 , α i 2 , ⋯ , α i n i G i = α i 1 β i 1 T + α i 2 β i 2 T + ⋯ + α i n i β i n i T A = λ 1 G 1 + λ 2 G 2 + ⋯ + λ r G r
例：已知A = 4 − 3 − 3 6 − 5 − 6 0 0 0
解：∣ λ E − A ∣ = ( λ − 1 ) 2 ( λ + 2 ) λ 1 = λ 2 = 1 λ 3 = − 2 α 1 = 2 − 1 0 α 2 = 0 0 1 α 3 = − 1 1 1 P = 2 − 1 0 0 0 1 − 1 1 1 P − 1 = 1 − 1 1 1 − 2 2 0 1 0 ( P − 1 ) T = 1 1 0 − 1 − 2 1 1 2 0 β 1 T = 1 1 0 β 2 T = − 1 − 2 1 β 3 T = 1 2 0 G 1 = α 1 β 1 T + α 2 β 2 T = 2 − 1 − 1 2 − 1 − 2 0 0 1 G 2 = α 3 β 3 T = − 1 1 1 − 2 2 2 0 0 0 A = G 1 − 2 G 2
五、范数、序列、级数
1. 向量范数
向量范数 ：设V R C n V α α ∣∣ α ∣∣
非负性：当α = 0 ∣∣ α ∣∣ > 0 α = 0 ∣∣ α ∣∣ = 0
齐次性：∣∣ k α ∣∣ = ∣ k ∣ ∣∣ α ∣∣ k
三角不等式：任取α , β ∈ V ∣∣ α + β ∣∣ ≤ ∣∣ α ∣∣ + ∣∣ β ∣∣
只要满足上面性质的都叫范数，常用的有1-范数 （∣∣ α ∣ ∣ 1 = ∑ i = 1 n ∣ a i ∣ 2-范数 （∣∣ α ∣ ∣ 2 = ( ∑ i = 1 n ∣ a i ∣ 2 ) 2 1 = α H α ∞ ∣∣ α ∣ ∣ ∞ = 1 ≤ i ≤ n max ∣ a i ∣ α = ( a 1 , ⋯ , a n ) T ∈ C n
∣∣ α ∣ ∣ ∞ ≤ ∣∣ α ∣ ∣ 1 ≤ n ∣∣ α ∣ ∣ ∞
∣∣ α ∣ ∣ 2 ≤ ∣∣ α ∣ ∣ 1 ≤ n ∣∣ α ∣ ∣ 2
∣∣ α ∣ ∣ ∞ ≤ ∣∣ α ∣ ∣ 2 ≤ n ∣∣ α ∣ ∣ ∞
引理（Holder不等式 ）：设α = [ a 1 , ⋯ , a n ] T , β = [ b 1 , ⋯ , b n ] T ∈ C n ∑ i = 1 n ∣ a i b i ∣ ≤ ( ∑ i = 1 n ∣ a i ∣ p ) p 1 ( ∑ i = 1 n ∣ b i ∣ q ) q 1 p > 1 q > 1 p 1 + q 1 = 1
引理（Minkowski不等式 ）：设α = [ a 1 , ⋯ , a n ] T , β = [ b 1 , ⋯ , b n ] T ∈ C n ( ∑ i = 1 n ∣ a i + b i ∣ p ) p 1 ≤ ( ∑ i = 1 n ∣ a i ∣ p ) p 1 + ( ∑ i = 1 n ∣ b i ∣ p ) p 1 p ≥ 1
以上引理最常用的都是p = 2
p-范数 ：一般地，设向量α = [ a 1 , ⋯ , a n ] T p ≥ 1 ∣∣ α ∣ ∣ p = ( ∑ i = 1 n ∣ a i ∣ p ) p 1 α
p-范数三角不等式性质的证明用到了Minkowski不等式。
关于∞ α = [ a 1 , ⋯ , a n ] T x = 1 ≤ i ≤ n max ∣ a i ∣ y i = x ∣ a i ∣ ≤ 1 x = ∣ a i ∣ ∑ y i ≥ 1 ∣∣ α ∣ ∣ p = x ( ∑ i = 1 n y i p ) p 1 1 ≤ ∑ i = 1 n y i p ≤ n 1 ≤ ( ∑ i = 1 n y i p ) p 1 ≤ n p 1 p → ∞ lim ( ∑ i = 1 n y i p ) p 1 = 1 ∣∣ α ∣ ∣ ∞ = p → ∞ lim ∣∣ α ∣ ∣ p = x = 1 ≤ i ≤ n max ∣ a i ∣
向量范数的等价性定理 ：设∣∣ α ∣ ∣ a ∣∣ α ∣ ∣ b n V α d 1 d 2 d 1 ∣∣ α ∣ ∣ b ≤ ∣∣ α ∣ ∣ a ≤ d 2 ∣∣ α ∣ ∣ b ∀ α ∈ V V
比如：设∣∣ ⋅ ∣ ∣ b C m A ∈ C m × n rank ( A ) = n ∣∣ α ∣ ∣ a = ∣∣ A α ∣ ∣ b α ∈ C m ∣∣ ⋅ ∣ ∣ a C n
2. 矩阵范数
Ⅰ. 定义及常用矩阵范数
矩阵范数 ：对于任意一个矩阵A ∈ C m × n ∣∣ A ∣∣ A
非负性：当A = 0 ∣∣ A ∣∣ > 0 A = 0 ∣∣ A ∣∣ = 0
齐次性：∣∣ k A ∣∣ = ∣ k ∣ ∣∣ A ∣∣ k
三角不等式：任取A , B ∈ C m × n ∣∣ A + B ∣∣ ≤ ∣∣ A ∣∣ + ∣∣ B ∣∣
矩阵乘法的相容性：对于任意两个可以相乘的矩阵A B ∣∣ A B ∣∣ ≤ ∣∣ A ∣∣ ∣∣ B ∣∣
那么我们称∣∣ A ∣∣ A
比如，对于A ∈ C m × n ∣∣ A ∣∣ = ∑ i = 1 m ∑ j = 1 n ∣ a ij ∣ ∣∣ A ∣∣ A A ∈ C m × p B ∈ C p × n
∣∣ A B ∣∣ = ∑ i = 1 m ∑ j = 1 n ∣ ∑ k = 1 p a ik b kj ∣ ≤ ∑ i = 1 m ∑ j = 1 n ∑ k = 1 p ∣ a ik ∣∣ b kj ∣ ≤ ∑ i = 1 m ∑ j = 1 n [( ∑ k = 1 p ∣ a ik ∣ ) ( ∑ k = 1 p b kj )] = ( ∑ i = 1 m ∑ k = 1 p ∣ a ik ∣ ) ( ∑ j = 1 n ∑ k = 1 p ∣ b kj ∣ ) = ∣∣ A ∣∣ ∣∣ B ∣∣
再比如，对于A ∈ C n × n ∣∣ A ∣∣ = n i , j max ∣ a ij ∣ A , B ∈ C n × n
∣∣ A B ∣∣ = n i , j max ∣ ∑ k = 1 n a ik b kj ∣ ≤ n i , j max ∑ k = 1 n ∣ a ik ∣∣ b kj ∣ ≤ n ⋅ n i , k max ∣ a ik ∣ ⋅ k , j max ∣ b kj ∣ = n i , k max ∣ a ik ∣ ⋅ n k , j max ∣ b kj ∣ = ∣∣ A ∣∣ ∣∣ B ∣∣
又比如，对于A ∈ C m × n ∣∣ A ∣ ∣ F = ( ∑ i = 1 m ∑ j = 1 n ∣ a ij ∣ 2 ) 2 1 ∣∣ A ∣∣ A Frobenious范数 ，同样这里只给出性质4的证明。设A ∈ C m × l B ∈ C l × n
∣∣ A B ∣ ∣ F 2 = ∑ i = 1 m ∑ j = 1 n ∣ ∑ k = 1 l a ik b kj ∣ 2 ≤ ∑ i = 1 m ∑ j = 1 n ( ∑ k = 1 l ∣ a ik ∣∣ b kj ∣ ) 2 ≤ ∑ i = 1 m ∑ j = 1 n [( ∑ k = 1 l ∣ a ik ∣ 2 ) ( ∑ k = 1 l ∣ b kj ∣ 2 )] = ( ∑ i = 1 m ∑ k = 1 l ∣ a ik ∣ 2 ) ( ∑ j = 1 n ∑ k = 1 l ∣ b kj ∣ 2 ) = ∣∣ A ∣ ∣ F 2 ∣∣ B ∣ ∣ F 2
Frobenious范数的性质 ：
如果A = [ α 1 ⋯ α n ] ∣∣ A ∣ ∣ F 2 = ∑ i = 1 n ∣∣ α i ∣ ∣ 2 2
∣∣ A ∣ ∣ F 2 = tr ( A H A ) = ∑ i = 1 n λ i λ i A H A 酉不变性 ：对于任何m U n V ∣∣ A ∣ ∣ F = ∣∣ A H ∣ ∣ F = ∣∣ U A ∣ ∣ F = ∣∣ A V ∣ ∣ F = ∣∣ U A V ∣ ∣ F
性质2的证明：比如A = [ a 11 a 21 a 12 a 22 ] A T A = [ a 11 a 12 a 21 a 22 ] [ a 11 a 21 a 12 a 22 ] = [ a 11 2 + a 21 2 ⋯ ⋯ a 12 2 + a 22 2 ] A T A ∣∣ A ∣ ∣ F 2 tr ( A H A ) = tr ( A A H )
酉不变性的证明：设A = [ α 1 ⋯ α n ] A A H U A = [ U α 1 ⋯ U α n ] ∣∣ U A ∣ ∣ F 2 = ∣∣ U α 1 ∣ ∣ 2 2 + ⋯ + ∣∣ U α n ∣ ∣ 2 2 = α 1 H U H U α 1 + ⋯ + α n H U H U α n = α 1 H α 1 + ⋯ + α n H α = ∣∣ A ∣ ∣ F 2 ∣∣ A V ∣ ∣ F = ∣∣ V H A H ∣ ∣ F ∣∣ A V ∣ ∣ F ∣∣ A ∣ ∣ F
矩阵的谱半径 ：设A ∈ C n × n A n λ 1 , λ 2 , ⋯ , λ n ρ ( A ) = max { ∣ λ 1 ∣ , ∣ λ 2 ∣ , ⋯ , ∣ λ n ∣ } A
定理：n A ∈ C n × n ρ ( A ) ≤ ∣∣ A ∣∣
证明：∵ A X = λ X X = 0 ∴ ∣ λ ∣ ∣∣ X ∣∣ = ∣∣ λ X ∣∣ = ∣∣ A X ∣∣ ≤ ∣∣ A ∣∣ ∣∣ X ∣∣ ∣ λ ∣ ≤ ∣∣ A ∣∣
特别地，若A n ρ ( A ) = ∣∣ A ∣ ∣ 2
证明：设A λ 1 , λ 2 , ⋯ , λ n U A = U diag ( λ 1 , λ 2 , ⋯ , λ n ) U H A H A = U diag ( ∣ λ 1 ∣ 2 , ∣ λ 2 ∣ 2 , ⋯ , ∣ λ n ∣ 2 ) U H ∣∣ A ∣ ∣ 2 = j max ( λ j ( A H A ) ) 2 1 = j max ∣ λ j ∣ = ρ ( A )
其中λ j ( ⋅ ) λ j ⋅ ∣∣ A ∣ ∣ 2
矩阵范数的一些结论（设∣∣ ⋅ ∣∣ C n × n
∣∣ E ∣∣ ≥ 1 若A λ A ∣∣ A − 1 ∣ ∣ − 1 ≤ ∣ λ ∣ ≤ ∣∣ A ∣∣
证明：根据特征值的性质，λ − 1 A − 1 ∣ λ ∣ ≤ ∣∣ A ∣∣ ∣ λ − 1 ∣ ≤ ∣∣ A − 1 ∣∣ ∣ λ ∣ ≥ ∣∣ A − 1 ∣ ∣ − 1
矩阵范数的等价性定理 ：设∣∣ A ∣ ∣ α ∣∣ A ∣ ∣ β A d 1 d 2 d 1 ∣∣ A ∣ ∣ β ≤ ∣∣ A ∣ ∣ α ≤ d 2 ∣∣ A ∣ ∣ β ∀ A ∈ C m × n
矩阵范数与向量范数的相容 ：设∣∣ X ∣ ∣ α ∣∣ A ∣ ∣ β A X ∣∣ A X ∣ ∣ α ≤ ∣∣ A ∣ ∣ β ∣∣ X ∣ ∣ α ∣∣ A ∣ ∣ β ∣∣ X ∣ ∣ α
例：矩阵的Frobenious范数与向量的2-范数是相容的
证：A X n = 1 ∣∣ A X ∣ ∣ 2 = ∣∣ A X ∣ ∣ F ≤ ∣∣ A ∣ ∣ F ∣∣ X ∣ ∣ F = ∣∣ A ∣ ∣ F ∣∣ X ∣ ∣ 2
Ⅱ.
通过向量范数构造与之相容的矩阵范数
通过已知的范数构造的范数叫诱导范数，下面通过例题说明
例：设∣∣ X ∣ ∣ α ∣∣ A ∣ ∣ i = X = 0 max ∣∣ X ∣ ∣ α ∣∣ A X ∣ ∣ α ∣∣ A ∣ ∣ i ∣∣ X ∣ ∣ α
证：容易验证∣∣ A ∣ ∣ i
∣∣ A B ∣ ∣ i = X = 0 max ∣∣ X ∣ ∣ α ∣∣ A BX ∣ ∣ α = X = 0 max ( ∣∣ BX ∣ ∣ α ∣∣ A ( BX ) ∣ ∣ α ∣∣ X ∣ ∣ α ∣∣ BX ∣ ∣ α ) ≤ BX = 0 max ∣∣ BX ∣ ∣ α ∣∣ A ( BX ) ∣ ∣ α X = 0 max ∣∣ X ∣ ∣ α ∣∣ BX ∣ ∣ α ≤ BX = 0 max ∣∣ X ∣ ∣ α ∣∣ A X ∣ ∣ α X = 0 max ∣∣ X ∣ ∣ α ∣∣ BX ∣ ∣ α = ∣∣ A ∣ ∣ i ∣∣ B ∣ ∣ i
然后验证∣∣ A ∣ ∣ i ∣∣ X ∣ ∣ α
由∣∣ A ∣ ∣ i X = 0 ∣∣ A ∣ ∣ i ≥ ∣∣ X ∣ ∣ α ∣∣ A X ∣ ∣ α ∣∣ A X ∣ ∣ α ≤ ∣∣ A ∣ ∣ i ∣∣ X ∣ ∣ α X = 0 ∣∣ A X ∣ ∣ α = ∣∣ A ∣ ∣ i ∣∣ X ∣ ∣ α = 0
由上面例题所定义的矩阵范数称为由向量范数∣∣ X ∣ ∣ α 诱导范数 （或算子范数 ），由向量p-范数∣∣ X ∣ ∣ p 矩阵p-范数 ，即∣∣ A ∣ ∣ p = X = 0 max ∣∣ X ∣ ∣ p ∣∣ A X ∣ ∣ p ∣∣ A ∣ ∣ 1 ∣∣ A ∣ ∣ 2 ∣∣ A ∣ ∣ ∞
常用的矩阵p-范数的形式（设A ∈ C m × n
列和范数 ：∣∣ A ∣ ∣ 1 = j max ( ∑ i = 1 m ∣ a ij ∣ ) j = 1 , 2 , ⋯ , n 谱范数 ：∣∣ A ∣ ∣ 2 = j max ( λ j 2 1 ) λ j A H A j λ j ( A H A ) A H A ≥ 0 行和范数 ：∣∣ A ∣ ∣ ∞ = i max ( ∑ j = 1 n ∣ a ij ∣ ) i = 1 , 2 , ⋯ , n
比如A = 2 0 1 − 1 2 2 0 3 0 A H A = 5 0 0 0 9 6 0 6 9 ∣∣ A ∣ ∣ F = 23 ∣∣ A ∣ ∣ 1 = 5 ∣∣ A ∣ ∣ 2 = 15 ∣∣ A ∣ ∣ ∞ = 5
它们之间的关系：
∣∣ A H ∣ ∣ 1 = ∣∣ A T ∣ ∣ 1 = ∣∣ A ∣ ∣ ∞ ∣∣ A H ∣ ∣ 2 = ∣∣ A T ∣ ∣ 2 = ∣∣ A ∣ ∣ 2 ∣∣ A H A ∣ ∣ 2 = ∣∣ A ∣ ∣ 2 2 ∣∣ A ∣ ∣ 2 2 ≤ ∣∣ A ∣ ∣ 1 ∣∣ A ∣ ∣ ∞
第3点的证明：∣∣ A H A ∣ ∣ 2 2 = j max λ j [( A H A ) H ( A H A )] = j max λ j [( A H A ) 2 ] = [ j max λ j ( A H A ) ] 2 = ∣∣ A ∣ ∣ 2 4
第4点的证明：∣∣ A ∣ ∣ 2 2 = j max λ j ( A H A ) ≤ ∣∣ A H A ∣ ∣ 1 ≤ ∣∣ A H ∣ ∣ 1 ∣∣ A ∣ ∣ 1 = ∣∣ A ∣ ∣ ∞ ∣∣ A ∣ ∣ 1
（其中λ j ( ⋅ ) λ j ⋅
Ⅲ.
通过矩阵范数构造与之相容的向量范数
定理：设∣∣ A ∣ ∣ ∗ ∣∣ X ∣∣ ∣∣ A X ∣∣ ≤ ∣∣ A ∣ ∣ ∗ ∣∣ X ∣∣
证明：对于任意的非零向量α ∣∣ X ∣∣ = ∣∣ X α H ∣ ∣ ∗ ∣∣ A X ∣∣ = ∣∣ A X α H ∣ ∣ ∗ ≤ ∣∣ A ∣ ∣ ∗ ∣∣ X α H ∣ ∣ ∗ ≤ ∣∣ A ∣ ∣ ∗ ∣∣ X ∣∣
例：已知矩阵范数∣∣ A ∣ ∣ ∗ = ∣∣ A ∣∣ = ∑ i = 1 m ∑ j = 1 n ∣ a ij ∣
解：取α = [ 0 1 ⋯ 0 ] T X = [ x 1 x 2 ⋯ x n ] T ∣∣ X ∣∣ = ∣∣ X α H ∣ ∣ ∗ = ∑ i = 1 n ∣ x i ∣ = ∣∣ X ∣ ∣ 1
3. 矩阵序列与极限
矩阵序列及其收敛、极限 ：设矩阵序列{ A ( k ) } ( k ) k A ( k ) = [ a ij ( k ) ] ∈ C m × n nm { a ij ( k ) } i = 1 , 2 , ⋯ , m j = 1 , 2 , ⋯ , n { A ( k ) } 收敛 。进一步，如果k → ∞ lim a ij ( k ) = a ij k → ∞ lim A ( k ) = A = [ a ij ] A 矩阵序列{ A ( k ) } 。
矩阵序列的收敛定理 ：矩阵序列{ A ( k ) } A k → ∞ lim ∣∣ A ( k ) − A ∣∣ = 0 ∣∣ ⋅ ∣∣
矩阵序列的极限的运算性质：
一个收敛的矩阵序列的极限是唯一的.
设k → ∞ lim A ( k ) = A k → ∞ lim B ( k ) = B k → ∞ lim ( a A ( k ) + b B ( k ) ) = a A + b B a , b ∈ C
设A ( k ) ∈ C m × l B ( k ) ∈ C l × n k → ∞ lim A ( k ) = A k → ∞ lim B ( k ) = B k → ∞ lim A ( k ) B ( k ) = A B
设k → ∞ lim A ( k ) = A A ( k ) ∈ C m × n P ∈ C m × m Q ∈ C n × n k → ∞ lim P A ( k ) Q = P A Q
设k → ∞ lim A ( k ) = A { A ( k ) } A {( A ( k ) ) − 1 } k → ∞ lim ( A ( k ) ) − 1 = A − 1
例：证明对于幂序列{ A ( k ) } = { A , A 2 , ⋯ , A k , ⋯ } ∣∣ A ∣∣ < 1 k → ∞ lim A ( k ) = 0
证：k → ∞ lim ∣∣ A ( k ) − 0∣∣ = k → ∞ lim ∣∣ A k ∣∣ ≤ k → ∞ lim ∣∣ A ∣ ∣ k = 0 ≥ 0 ∴ k → ∞ lim A ( k ) = 0
幂序列收敛于零的充要条件 ：对于幂序列A , A 2 , ⋯ , A k , ⋯ k → ∞ lim A ( k ) = 0 ρ ( A ) < 1
证明：设A J = diag ( J 1 ( λ 1 ) , J 2 ( λ 2 ) , ⋯ , J r ( λ r )) J i ( ⋅ ) J A k = P diag ( J 1 k ( λ 1 ) , J 2 k ( λ 2 ) , ⋯ , J r k ( λ r )) P − 1 k → ∞ lim A k = 0 ⇔ k → ∞ lim J i k ( λ i ) = 0 J i k ( λ i ) = λ i k C k 1 λ i k − 1 λ i k ⋯ ⋱ ⋱ C k d i − 1 λ i k − d i + 1 ⋮ C k 1 λ i k − 1 λ i k d i × d i C k l = { l ! k ( k − 1 ) ⋯ ( k − l + 1 ) 0 , l ≤ k , l > k k → ∞ lim J i k ( λ i ) = 0 ∣ λ i ∣ < 1 k → ∞ lim A k = 0 ρ ( A ) < 1
Gelfand定理 ：设∣∣ ⋅ ∣∣ C n × n A ∈ C n × n ρ ( A ) = k → ∞ lim ∣∣ A k ∣ ∣ k 1
证明：谱半径不大于其任何一种范数，于是( ρ ( A ) ) k = ρ ( A k ) ≤ ∣∣ A k ∣∣ ⟶ ρ ( A ) ≤ ∣∣ A k ∣ ∣ k 1 ϵ > 0 A = [ ρ ( A ) + ϵ ] − 1 A ρ ( A ) < 1 k → ∞ lim A k = 0 k → ∞ ∣∣ A k − 0∣∣ → 0 K k > K ∣∣ A k ∣∣ ≤ 1 ∣∣ A k ∣∣ ≤ [ ρ ( A ) + ϵ ] k ⟶ ∣∣ A k ∣ ∣ k 1 ≤ ρ ( A ) + ϵ
4. 矩阵级数
矩阵级数的收敛 ：设A ( k ) = [ a ij ( k ) ] ∈ C m × n mn ∑ k = 1 ∞ a ij ( k ) i = 1 , 2 , ⋯ , m j = 1 , 2 , ⋯ , n ∑ k = 1 ∞ A ( k ) = A ( 1 ) + A ( 2 ) + ⋯ + A ( k ) + ⋯ 收敛 。特别地，如果∑ k = 1 ∞ a ij ( k ) 绝对收敛 。
矩阵级数的绝对收敛 ：设A ( k ) = [ a ij ( k ) ] ∈ C m × n ∑ k = 1 ∞ A ( k ) = A ( 1 ) + A ( 2 ) + ⋯ + A ( k ) + ⋯ ∑ k = 1 ∞ ∣∣ A ( k ) ∣∣ = ∣∣ A ( 1 ) ∣∣ + ∣∣ A ( 2 ) ∣∣ + ⋯ + ∣∣ A ( k ) ∣∣ + ⋯ ∣∣ ⋅ ∣∣
证明：先证明对其中一种矩阵范数成立，再利用矩阵范数的等价性推广到其他范数
充分性：取∣∣ A ( k ) ∣∣ = ∑ i = 1 m ∑ j = 1 n ∣ a ij ( k ) ∣ i , j ∣∣ A ( k ) ∣∣ ≥ ∣ a ij ( k ) ∣ ∑ k = 1 ∞ ∣∣ A ( k ) ∣∣ = ∣∣ A ( 1 ) ∣∣ + ∣∣ A ( 2 ) ∣∣ + ⋯ + ∣∣ A ( k ) ∣∣ + ⋯ i , j ∑ k = 1 ∞ ∣ a ij ( k ) ∣ = ∣ a ij ( 1 ) ∣ + ∣ a ij ( 2 ) ∣ + ⋯ + ∣ a ij ( k ) ∣ + ⋯
必要性：若∑ k = 1 ∞ A ( k ) = A ( 1 ) + A ( 2 ) + ⋯ + A ( k ) + ⋯ i , j ∑ k = 1 ∞ ∣ a ij ( k ) ∣ = ∣ a ij ( 1 ) ∣ + ∣ a ij ( 2 ) ∣ + ⋯ + ∣ a ij ( k ) ∣ + ⋯ < ∞ ∑ k = 1 ∞ ∣∣ A ( k ) ∣∣ = ∑ k = 1 ∞ ∑ i = 1 m ∑ j = 1 n ∣ a ij ( k ) ∣ = ∑ i = 1 m ∑ j = 1 n ∑ k = 1 ∞ ∣ a ij ( k ) ∣ < ∞ S n = ∑ k = 1 n ∣∣ A ( k ) ∣∣ < ∞
定理 ：对于两个绝对收敛的矩阵级数，它们的Cauchy积所组成的矩阵级数仍然绝对收敛。
即，设S 1 = A ( 1 ) + A ( 2 ) + ⋯ S 2 = B ( 1 ) + B ( 2 ) + ⋯ S 1 → A S 2 → B S 3 = A ( 1 ) B ( 1 ) + ( A ( 1 ) B ( 2 ) + A ( 2 ) B ( 1 ) ) + ( A ( 1 ) B ( 3 ) + A ( 2 ) B ( 2 ) + A ( 3 ) B ( 1 ) ) + ⋯ S 3 → A B
矩阵幂级数 ：设A ∈ C n × n ∑ k = 0 ∞ a k A k = a 0 E + a 1 A + a 2 A 2 + ⋯ + a k A k + ⋯
矩阵幂级数的绝对收敛 ：如果矩阵A ∣∣ A ∣∣ a 0 + a 1 x + a 2 x 2 + ⋯ + a k x k + ⋯ ∑ k = 0 ∞ a k A k
证明：∑ k = 0 ∞ ∣∣ a k A k ∣∣ ≤ ∑ k = 0 ∞ ∣ a k ∣ ∣∣ A k ∣∣ ∣∣ A k ∣∣ x ∣∣ A ∣∣ ∑ k = 0 ∞ a k x k
Cauchy-Hadamard定理 ：幂级数∑ k = 0 ∞ a k x k ∣ x ∣ < R ∣ x ∣ > R ∣ x ∣ = R R R 1 = k → ∞ lim ∣ a k a k + 1 ∣
将Cauchy-Hadamard定理推广到矩阵：设幂级数∑ k = 0 ∞ a k x k R A n ρ ( A ) < R ∑ k = 0 ∞ a k A k ρ ( A ) > R ∑ k = 0 ∞ a k A k ρ ( A ) = R
证明：设A J = diag ( J 1 ( λ 1 ) , J 2 ( λ 2 ) , ⋯ , J r ( λ r )) J i ( λ i ) λ i d i × d i A k = P diag ( J 1 k ( λ 1 ) , J 2 k ( λ 2 ) , ⋯ , J r k ( λ r )) P − 1 ∑ k = 0 ∞ a k A k = ∑ k = 0 ∞ a k ( P J k P − 1 ) = P diag ( ∑ k = 0 ∞ a k J 1 k ( λ 1 ) , ∑ k = 0 ∞ a k J 2 k ( λ 2 ) , ⋯ , ∑ k = 0 ∞ a k J r k ( λ r )) P − 1 ∑ k = 0 ∞ a k J i k ( λ i ) = ∑ k = 0 ∞ a k λ i k ∑ k = 0 ∞ a k C k 1 λ i k − 1 ∑ k = 0 ∞ a k λ i k ⋯ ⋱ ⋱ ∑ k = 0 ∞ a k C k d i − 1 λ i k − d i + 1 ⋮ ∑ k = 0 ∞ a k C k 1 λ i k − 1 ∑ k = 0 ∞ a k λ i k d i × d i C C k l = { l ! k ( k − 1 ) ⋯ ( k − l + 1 ) 0 , l ≤ k , l > k ∑ k = 0 ∞ a k A k ρ ( A ) < R ∑ k = 0 ∞ a k λ i k ∑ k = 0 ∞ a k C k l λ i k − 1 , ⋯ ρ ( A ) < R ∑ k = 0 ∞ a k A k
和初等函数的Taylor展开式类似，矩阵幂级数：
e A = ∑ n = 0 ∞ n ! A n = E + A + 2 ! A 2 + ⋯ + n ! A n + ⋯
sin A = ∑ n = 0 ∞ ( 2 n + 1 )! ( − 1 ) n A 2 n + 1 = A − 3 ! A 3 + 5 ! A 5 + ⋯ + ( − 1 ) n ( 2 n + 1 )! A 2 n + 1 + ⋯
cos A = ∑ n = 0 ∞ ( 2 n )! ( − 1 ) n A 2 n = E − 2 ! A 2 + 4 ! A 4 − ⋯ + ( − 1 ) n ( 2 n )! A 2 n + ⋯
arctan A = ∑ n = 1 ∞ 2 n − 1 ( − 1 ) n − 1 A 2 n − 1 = A − 3 A 3 + 5 A 5 − ⋯ + ( − 1 ) n 2 n + 1 A 2 n + 1 + ⋯ ρ ( A ) < 1
ln ( E + A ) = ∑ n = 0 ∞ n ( − 1 ) n + 1 A n = A − 2 A 2 + 3 A 3 − ⋯ + ( − 1 ) n − 1 n A n + ⋯ ρ ( A ) < 1
( E + A ) − 1 = E − A + A 2 − A 3 + ⋯ + ( − 1 ) n A n + ⋯ ρ ( A ) < 1
( E − A ) − 1 = E + A + A 2 + A 3 + ⋯ + A n + ⋯ ρ ( A ) < 1
定理：矩阵幂级数E + A + A 2 + A 3 + ⋯ + A n + ⋯ ρ ( A ) < 1 ( E − A ) − 1
例：已知A = [ 5 1 5 3 5 3 5 1 ] ∑ k = 0 ∞ k 2 A k
解：先求出其谱半径，∣ λ E − A ∣ = ( λ + 5 2 ) ( λ − 5 4 ) λ 1 = − 5 2 λ 2 = 5 4 ρ ( A ) = 5 4 ∑ k = 1 ∞ k 2 A 2 R = 1 ρ ( A ) < R
由于∑ k = 0 ∞ x k = ( 1 − x ) − 1 ∑ k = 0 ∞ k x k − 1 = ( 1 − x ) − 2 ⟶ ∑ k = 0 ∞ k x k = x ( 1 − x ) − 2 ∑ k = 0 ∞ k 2 x k = x ( 1 + x ) ( 1 − x ) − 3 ∑ k = 0 ∞ k 2 A k = A ( E + A ) ( E − A ) − 3 = 5 [ − 24 7 24 11 24 11 − 24 7 ]
六、矩阵函数
1. 矩阵多项式
矩阵多项式 ：设A ∈ C n × n x f ( x ) = a m x m + a m − 1 x m − 1 + ⋯ + a 1 x + a 0 f ( A ) = a m A m + a m − 1 A m − 1 + ⋯ + a 1 A + a 0 E A
当次数m A f ( A ) = P diag ( f ( J 1 ) , f ( J 2 ) , ⋯ , f ( J r )) P − 1 f ( A ) J i ( λ i ) = λ i 1 λ i ⋱ ⋱ 1 λ i J i k ( λ i ) = λ i E + 0 1 0 ⋱ ⋱ 1 0 k = λ i k C k 1 λ i k − 1 λ i k ⋯ ⋱ ⋱ C k d i − 1 λ i k − d i + 1 ⋮ C k 1 λ i k − 1 λ i k d i × d i C C k l = { l ! k ( k − 1 ) ⋯ ( k − l + 1 ) 0 , l ≤ k , l > k 1 ! ( λ i k ) ′ = C k 1 λ k − 1 2 ! ( λ i k ) ′′ = C k 2 λ i k − 2 ⋯ ( d i − 1 )! ( λ i k ) ( d i − 1 ) = C k d i − 1 λ i k − d i + 1 f ( J i ) = f ( λ i ) f ′ ( λ i ) f ( λ i ) ⋯ ⋱ ⋱ ( d i − 1 )! 1 f ( d i − 1 ) ( λ i ) ⋮ f ′ ( λ i ) f ( λ i ) d i × d i
定理 ：设A ∈ C n × n A n λ 1 , λ 2 , ⋯ , λ n f ( A ) f ( λ 1 ) , f ( λ 2 ) , ⋯ , f ( λ n )
例：已知多项式f ( x ) = x 4 − 2 x 3 + x − 1 A = 2 1 1 0 1 − 1 0 1 3 f ( A )
解：A = P J A P − 1 = 0 1 1 1 0 0 1 0 − 1 2 0 0 1 2 0 0 0 2 0 1 0 1 − 1 1 0 1 − 1 f ( A ) = 0 1 1 1 0 0 1 0 − 1 f ( 2 ) 0 0 f ′ ( 2 ) f ( 2 ) 0 0 0 f ( 2 ) 0 1 0 1 − 1 1 0 1 − 1
2. 最小多项式
化零多项式 ：设A ∈ C n × n x f ( x ) = a m x m + a m − 1 x m − 1 + ⋯ + a 1 x + a 0 f ( x ) f ( A ) = 0 n × n f ( x ) A
推论：若f ( λ ) A h ( λ ) f ( λ ) h ( λ ) A
Hamilton-Cayley定理 ：已知A ∈ C n × n f ( λ ) f ( A ) = 0 n × n
比如：A = 2 0 0 0 2 0 0 1 2 f A ( λ ) = ( λ − 2 ) 3 f A ( A ) = 0
并且容易看出g ( λ ) = ( λ − 2 ) 2 g ( A ) = 0 f ( A ) = 0
最小多项式 ：已知A ∈ C n × n A A m ( λ )
最小多项式的性质：
矩阵的任何一个化零多项式均能被m ( λ )
矩阵A
相似的矩阵有相同的最小多项式。
例：求Jordan块J i ( λ i ) = λ i 1 λ i ⋱ ⋱ 1 λ i d i × d i
解；J i f ( λ ) = ( λ − λ i ) d i m ( λ ) = ( λ − λ i ) k 1 ≤ k ≤ d i k < d i m ( J i ) = ( J i − λ i E ) k = J i ( λ i ) = 0 0 0 ⋯ 0 0 1 ⋱ ⋱ ⋱ ⋯ ⋱ ⋱ 0 0 0 ⋯ 1 ⋯ 0 0 = 0 m ( λ ) = ( λ − λ i ) d i
定理：设分块对角阵A = diag ( A 1 , A 2 , ⋯ , A r ) m 1 ( λ ) , m 2 ( λ ) , ⋯ , m r ( λ ) A 1 , A 2 , ⋯ , A r A [ m 1 ( λ ) , m 2 ( λ ) , ⋯ , m r ( λ )]
定理 ：设矩阵A J = diag ( J 1 ( λ 1 ) , J 2 ( λ 2 ) , ⋯ , J s ( λ s )) J i ( λ i ) d i A ( λ − λ 1 ) d 1 , ⋯ , ( λ − λ s ) d s
比如：A = 3 0 0 0 1 3 0 0 0 0 3 0 0 0 0 5 ( λ − 3 ) 2 ( λ − 5 )
3. 矩阵函数
定义：设A ∈ C n × n λ 1 , λ 2 , ⋯ , λ s A s A m ( λ ) = ( λ − λ 1 ) d 1 ( λ − λ 2 ) d 2 ⋯ ( λ − λ s ) d s d i ≥ 1 ∑ i = 1 s d i = m f ( x ) { f ( λ i ) , f ′ ( λ i ) , ⋯ , f ( d i − 1 ) ( λ i ) , i = 1 , 2 , ⋯ , s } m 函数f ( x ) A 。
比如：f ( x ) = ( x − 3 ) ( x − 4 ) 1 A = 8 3 4 − 3 − 2 2 6 0 − 2 A m ( λ ) = ( λ − 2 ) ( λ − 1 ) 2 λ = 2 f ( x ) f ( 2 ) = 2 1 λ = 1 f ( 1 ) f ′ ( 1 ) f ( 1 ) = 6 1 f ′ ( 1 ) = 36 5 f ( x ) A
如果某个矩阵B m ( λ ) = ( λ − 1 ) ( λ − 3 ) 2 f ( 3 ) f ( x ) B
矩阵函数 ：设矩阵A ∈ C n × n m ( λ ) = ( λ − λ 1 ) d 1 ( λ − λ 2 ) d 2 ⋯ ( λ − λ s ) d s f ( x ) A p ( λ ) f ( k ) ( λ i ) = p ( k ) ( λ i ) i = 1 , 2 , ⋯ , s k = 0 , 1 , ⋯ , d i − 1 f ( A ) = p ( A )
注1：满足上述定义的多项式p ( λ )
注2：矩阵函数f ( A ) A
定理 ：设g ( λ ) q ( λ ) A n g ( A ) q ( A ) g ( λ ) q ( λ ) A g ( k ) ( λ i ) = q ( k ) ( λ i ) i = 1 , 2 , ⋯ , s k = 0 , 1 , ⋯ , d i − 1
设A ∈ C n × n f ( A ) f ( A T ) f ( A T ) = [ f ( A ) ] T A A T f ( x ) A A T
矩阵函数的Jordan表示 ：和矩阵多项式一样，矩阵函数也可以写成f ( A ) = P f ( J ) P − 1 = P diag ( f ( J 1 ) , f ( J 2 ) , ⋯ , f ( J s )) P − 1 f ( J i ) = f ( λ i ) f ′ ( λ i ) f ( λ i ) ⋯ ⋱ ⋱ ( d i − 1 )! 1 f ( d i − 1 ) ( λ i ) ⋮ f ′ ( λ i ) f ( λ i ) d i × d i
定理 ：设A ∈ C n × n A n λ 1 , λ 2 , ⋯ , λ n f ( A ) f ( λ 1 ) , f ( λ 2 ) , ⋯ , f ( λ n )
例：设A = − 1 − 1 − 1 − 2 0 − 1 6 3 4 f ( A ) e t A sin A
解：首先算出A J = 1 0 0 0 1 0 0 1 1 P = − 1 1 0 2 1 1 2 0 1 f ( A ) f ( A ) = P f ( J ) P − 1 = − 1 1 0 2 1 1 2 0 1 f ( 1 ) 0 0 0 f ( 1 ) 0 0 f ′ ( 1 ) f ( 1 ) − 1 1 − 1 0 1 − 1 2 − 2 3 = f ( 1 ) − 2 f ′ ( 1 ) − f ′ ( 1 ) − f ′ ( 1 ) − 2 f ′ ( 1 ) f ( 1 ) − f ′ ( 1 ) − f ′ ( 1 ) 6 f ′ ( 1 ) 3 f ′ ( 1 ) f ( 1 ) + 3 f ′ ( 1 ) f ( x ) = e t x f ( 1 ) = e t f ′ ( 1 ) = t e t sin
矩阵函数多项式表示 ：设矩阵A ∈ C n × n m ( λ ) = ( λ − λ 1 ) d 1 ( λ − λ 2 ) d 2 ⋯ ( λ − λ s ) d s λ 1 , λ 2 , ⋯ , λ s A s d i ≥ 1 ∑ i = 1 s d i = m m − 1 p ( x ) = a m − 1 x m − 1 + a m − 2 x m − 2 + ⋯ + a 1 x + a 0 p ( k ) ( λ i ) = f ( k ) ( λ i ) i = 1 , 2 , ⋯ , s k = 1 , 2 , ⋯ , d i − 1 p ( x ) f ( A ) = a m − 1 A m − 1 + a m − 2 A m − 2 + ⋯ + a 1 A + a o E f ( A )
例：设A = 2 1 1 0 1 − 1 0 1 3 f ( A ) e t A
解：A J = 2 0 0 0 2 0 0 1 2 m ( λ ) = ( λ − 2 ) 2 p ( x ) = a 1 x + a 0 { p ( 2 ) = f ( 2 ) p ′ ( 2 ) = f ′ ( 2 ) { f ( 2 ) = 2 a 1 + a 0 f ′ ( 2 ) = a 1 { a 1 = f ′ ( 2 ) a 0 = f ( 2 ) − 2 f ′ ( 2 ) f ( A ) f ( A ) = p ( A ) = f ′ ( 2 ) A + [ f ( 2 ) − 2 f ′ ( 2 )] E f ( x ) = e t x f ( 2 ) = e 2 t f ′ ( 2 ) = t e 2 t e A t = e 2 t 1 t t 0 1 − t − t 0 t 1 + t
矩阵的幂级数表示 ：设A ∈ C n × n f ( x ) x f ( x ) = ∑ k = 0 ∞ c k x k R A ρ ( A ) < R ∑ k = 0 ∞ c k A k ∑ k = 0 ∞ c k A k = P diag ( ∑ k = 0 ∞ c k J 1 k ( λ 1 ) , ⋯ , ∑ k = 0 ∞ c k J r k ( λ r )) P − 1 = P diag ( f ( J 1 ) , ⋯ , f ( J r )) P − 1 = f ( A )
例：设A = 2 1 1 0 1 − 1 0 1 3 ∑ k = 0 ∞ 1 0 k + 1 k + 1 A k
解：算出A J = 2 0 0 0 2 0 0 1 2 ρ ( A ) = 2 f ( x ) = ∑ k = 0 ∞ 1 0 k + 1 k + 1 x k = ∑ k = 0 ∞ [( 10 x ) k + 1 ] ′ = [ ∑ k = 0 ∞ ( 10 x ) k + 1 ] ′ = [ − 1 + ( 1 − 10 x ) − 1 ] ′ = 10 1 ( 1 − 10 x ) − 2 ∣ x ∣ < R = 10 ∴ ∑ k = 0 ∞ 1 0 k + 1 k + 1 A k = f ( A ) = 10 1 ( E − 10 A ) − 2 = 128 5 4 1 1 0 3 − 1 0 1 5 − 2
4. 矩阵指数函数与矩阵三角函数
e A t = ∑ k = 0 ∞ k ! 1 A k t k
sin A t = ∑ k = 0 ∞ ( 2 k + 1 )! ( − 1 ) k A 2 k + 1 t 2 k + 1
cos A t = ∑ k = 0 ∞ ( 2 k )! ( − 1 ) k A 2 k t 2 k
定理：设A , B ∈ C n × n A B = B A
e A + B = e A e B = e B e A sin ( A + B ) = sin A cos B + cos A sin B sin ( 2 A ) − 2 sin A cos A cos ( A + B ) = cos A cos B − sin A sin B cos ( 2 A ) = cos 2 A − sin 2 A sin 2 A + cos 2 A = E
注：A B = B A A B e A + B e A e B e B e A
推论：
e λ A e μ A = e λ + μ A λ μ e 0 n × n = E n × n e A e − A = e − A e A = E e A ( e A ) − 1 = e − A
几个特殊的性质：
d t d ( e A t ) = A e A t = e A t A d t d [ sin ( A t )] = A [ cos ( A t )] = [ cos ( A t )] A d t d [ cos ( A t )] = − A [ sin ( A t )] = − [ sin ( A t )] A ∣ e A ∣ = e tr ( A ) e A = P e J P − 1 = P diag ( e J 1 , ⋯ , e J r ) P − 1 e J i = e λ i e λ i e λ i ⋯ ⋱ ⋱ d i − 1 1 e λ i ⋮ e λ i e λ i d i × d i ∣ e A ∣ = ∏ i = 1 r ∣ e J i ∣ = ∏ i = 1 r e d i λ i = e tr ( A ) 设A e i A e i A ( e i A ) H = ( cos A + i sin A ) [( cos A ) H − i ( sin A ) H ] = ( cos A + i sin A ) ( cos A − i sin A ) = E ( cos A ) H ( sin A ) H A A k
设A e A
七、函数矩阵、矩阵微分方程
1. 函数矩阵
函数矩阵 ：以未定元x A ( x ) = a 11 ( x ) a 21 ( x ) ⋯ a m 1 ( x ) a 12 ( x ) a 22 ( x ) ⋯ a m 2 ( x ) ⋯ ⋯ ⋯ ⋯ a 1 n ( x ) a 2 n ( x ) ⋯ a mn ( x ) a ij ( x ) [ a , b ]
函数矩阵的加、乘、数乘、转置和数字矩阵一致。如果A ( x ) B ( x ) = B ( x ) A ( x ) = E A ( x ) [ a , b ] B ( x ) A ( x ) A − 1 ( x )
比如A ( x ) = [ x 1 0 1 e x ] [ 3 , 5 ] A − 1 = [ x 0 − e x x e x 1 ]
可逆 ：n A ( x ) [ a , b ] ∣ A ( x ) ∣ [ a , b ] A − 1 ( x ) = ∣ A ( x ) ∣ 1 A ∗ ( x ) A ∗ ( x ) A ( x )
秩 ：区间[ a , b ] m × n A ( x )
注意：对于n
比如A ( x ) = [ 0 x − 1 x 2 ] ∣ A ( x ) ∣ = x [ a , b ]
极限 ：如果A ( x ) = ( a ij ( x ) ) m × n a ij ( x ) x = x 0 x → x 0 lim a ij ( x ) = a ij a ij A ( x ) x = x 0 x → x 0 lim A ( x ) = A A = ( a ij ) m × n
极限的运算性质（设x → x 0 lim A ( x ) = A x → x 0 lim B ( x ) = B
x → x 0 lim ( A ( x ) ± B ( x )) = A ± B x → x 0 lim ( k A ( x )) = k A x → x 0 lim ( A ( x ) B ( x )) = A B
连续 ：如果A ( x ) a ij ( x ) x = x 0 x → x 0 lim a ij ( x ) = a ij ( x 0 ) A ( x ) x = x 0 x → x 0 lim A ( x ) = A ( x 0 )
可导 ：如果A ( x ) = ( a ij ( x ) ) m × n a ij ( x ) x = x 0 [ a , b ] A ( x ) x = x 0 [ a , b ] A ′ ( x 0 ) = d x d A ( x ) ∣ x = x 0 = Δ x → 0 lim Δ x A ( x 0 + Δ x ) − A ( x 0 ) = a 11 ′ ( x 0 ) a 21 ′ ( x 0 ) ⋯ a m 1 ′ ( x 0 ) a 12 ′ ( x 0 ) a 22 ′ ( x 0 ) ⋯ a m 2 ′ ( x 0 ) ⋯ ⋯ ⋯ ⋯ a 1 n ′ ( x 0 ) a 2 n ′ ( x 0 ) ⋯ a mn ′ ( x 0 )
导数的运算性质：
A ( x ) x d A ( x ) = 0 设A ( x ) = ( a ij ( x ) ) m × n B ( x ) = ( b ij ( x ) ) m × n d x d [ A ( x ) + B ( x )] = x d A ( x ) + d x d B ( x )
设k ( x ) x A ( x ) k ( x ) A ( x ) d x d [ k ( x ) A ( x )] = d x d k ( x ) A ( x ) + k ( x ) d x d A ( x ) k ( x ) k d x d [ k A ( x )] = k d x d A ( x )
设A ( x ) B ( x ) A ( x ) B ( x ) d x d [ A ( x ) B ( x )] = d x d A ( x ) B ( x ) + A ( x ) d x d B ( x ) d x d A 2 ( x ) = d x d A ( x ) A ( x ) + A ( x ) d x d A ( x ) = 2 A ( x ) d x d A ( x )
如果A ( x ) A − 1 ( x ) d x d A − 1 ( x ) = − A − 1 ( x ) d x d A ( x ) A − 1 ( x ) d x d A − 1 ( x ) A − 1 ( x ) A − 1 ( x ) A ( x ) A − 1 ( x ) = E d x d [ A ( x ) A − 1 ( x )] = 0 d x d [ A ( x ) A − 1 ( x )] = d x d A ( x ) A − 1 ( x ) + A ( x ) d x d A − 1 ( x )
设A ( x ) x = f ( t ) t f ( t ) A ( t ) d t d A ( x ) = d x d A ( x ) f ′ ( t ) = f ′ ( t ) d x d A ( x )
可积 ：如果函数矩阵A ( x ) = ( a ij ) m × n a ij ( x ) [ a , b ] A ( x ) [ a , b ] ∫ a b A ( x ) d x = ∫ a b a 11 ( x ) d x ∫ a b a 21 ( x ) d x ⋯ ∫ a b a m 1 ( x ) d x ∫ a b a 12 ( x ) d x ∫ a b a 22 ( x ) d x ⋯ ∫ a b a m 2 ( x ) d x ⋯ ⋯ ⋯ ⋯ ∫ a b a 1 n ( x ) d x ∫ a b a 2 n ( x ) d x ⋯ ∫ a b a mn ( x ) d x
积分的运算性质：
∫ a b k A ( x ) d x = k ∫ a b A ( x ) d x k ∈ R ∫ a b [ A ( x ) + B ( x )] d x = ∫ a b A ( x ) d x + ∫ a b B ( x ) d x
函数向量的线性相关性 ：设有定义在区间[ a , b ] m α i ( x ) = ( a i 1 ( x ) , a i 2 ( x ) , ⋯ , a in ( x )) i = 1 , 2 , ⋯ , m k 1 , k 2 , ⋯ , k m x ∈ [ a , b ] k 1 α 1 ( x ) + k 2 α 2 ( x ) + ⋯ + k m α m ( x ) = 0 [ a , b ] α 1 ( x ) , α 2 ( x ) , ⋯ , α m ( x ) k 1 = k 2 = ⋯ = k m = 0 α 1 ( x ) , α 2 ( x ) , ⋯ , α m ( x )
Gram矩阵 ：设α 1 ( x ) , α 2 ( x ) , ⋯ , α m ( x ) m [ a , b ] α i ( x ) = ( a i 1 ( x ) , a i 2 ( x ) , ⋯ , a in ( x )) i = 1 , 2 , ⋯ , m g ij = ∫ a b α i ( x ) α j T ( x ) d x i , j = 1 , 2 , ⋯ , m g ij G = ( g ij ) m × m α 1 ( x ) , α 2 ( x ) , ⋯ , α m ( x ) det G
函数向量线性无关的充要条件 ：定义在区间[ a , b ] α 1 ( x ) , α 2 ( x ) , ⋯ , α m ( x )
例：α 1 ( x ) = ( 0 , x ) α 2 ( x ) = ( x , 0 ) α 1 ( x ) α 2 ( x )
解：g 11 = ∫ a b x 2 d x = 3 1 ( b 3 − a 3 ) g 12 = g 21 = 0 g 22 = ∫ a b x 2 d x = 3 1 ( b 3 − a 3 ) G = [ 3 1 ( b 3 − a 3 ) 0 0 3 1 ( b 3 − a 3 ) ] det G = 9 1 ( b 3 − a 3 ) 2 a = b α 1 ( x ) α 2 ( x ) [ a , b ]
Wronski矩阵 ：设α i ( x ) = ( a i 1 ( x ) , a i 2 ( x ) , ⋯ , a in ( x )) i = 1 , 2 , ⋯ , m m [ a , b ] m − 1 A ( x ) = α 1 ( x ) α 2 ( x ) ⋮ α m ( x ) = a 11 ( x ) a 21 ( x ) ⋯ a m 1 ( x ) a 12 ( x ) a 22 ( x ) ⋯ a m 2 ( x ) ⋯ ⋯ ⋯ ⋯ a 1 n ( x ) a 2 n ( x ) ⋯ a mn ( x ) W ( x ) = ( A ( x ) , A ′ ( x ) , ⋯ , A ( m − 1 ) ( x ) ) m × n = a 11 ( x ) a 21 ( x ) ⋯ a m 1 ( x ) a 12 ( x ) a 22 ( x ) ⋯ a m 2 ( x ) ⋯ ⋯ ⋯ ⋯ a 1 n ( x ) a 2 n ( x ) ⋯ a mn ( x ) ⋯ ⋯ ⋯ ⋯ a 11 ( m − 1 ) ( x ) a 21 ( m − 1 ) ( x ) ⋯ a m 1 ( m − 1 ) ( x ) a 12 ( m − 1 ) ( x ) a 22 ( m − 1 ) ( x ) ⋯ a m 2 ( m − 1 ) ( x ) ⋯ ⋯ ⋯ ⋯ a 1 n ( m − 1 ) ( x ) a 2 n ( m − 1 ) ( x ) a mn ( m − 1 ) ( x )
是一个m × mn W ( x ) α 1 ( x ) , α 2 ( x ) , ⋯ , α m ( x )
函数向量线性无关的充分不必要条件 ：设W ( x ) α 1 ( x ) , α 2 ( x ) , ⋯ , α m ( x ) [ a , b ] x 0 ∈ [ a , b ] W ( x 0 ) m α 1 ( x ) , α 2 ( x ) , ⋯ , α m ( x ) [ a , b ]
例：设α 1 ( x ) = ( 1 , x , x 2 ) α 2 ( x ) = ( e x , 1 , x ) α 1 ( x ) α 2 ( x )
解：A ( x ) = [ 1 e x x 1 x 2 x ] A ′ ( x ) = [ 0 e x 1 0 2 x 1 ] W ( x ) = [ 1 e x x 1 x 2 x 0 e x 1 0 2 x 1 ] W ( x ) α 1 ( x ) α 2 ( x )
2. 矩阵微分方程
形如⎩ ⎨ ⎧ d t d x 1 = a 11 ( t ) x 1 ( t ) + a 12 ( t ) x 2 ( t ) + ⋯ + a 1 n ( t ) x n ( t ) + f 1 ( t ) d t d x 2 = a 21 ( t ) x 1 ( t ) + a 22 ( t ) x 2 ( t ) + ⋯ + a 2 n ( t ) x n ( t ) + f 2 ( t ) ⋮ d t d x n = a n 1 ( t ) x 1 ( t ) + a n 2 ( t ) x 2 ( t ) + ⋯ + a nn ( t ) x n ( t ) + f n ( t ) d t d x ( t ) = A ( t ) x ( t ) + f ( t ) A ( t ) = a 11 ( t ) a 21 ( t ) ⋯ a m 1 ( t ) a 12 ( t ) a 22 ( t ) ⋯ a m 2 ( t ) ⋯ ⋯ ⋯ ⋯ a 1 n ( t ) a 2 n ( t ) ⋯ a mn ( t ) x ( t ) = x 1 ( t ) x 2 ( t ) ⋮ x n ( t ) f ( t ) = f 1 ( t ) f 2 ( t ) ⋮ f n ( t ) ⎩ ⎨ ⎧ x 1 ( t 0 ) = x 10 x 2 ( t 0 ) = x 20 ⋮ x n ( t 0 ) = x n 0 x ( t 0 ) = x 10 x 20 ⋮ x n 0
齐次微分方程组的解 ：设A n d t d x ( t ) = A x ( t ) x ( t 0 ) = x 0 x ( t ) = e A ( t − t 0 ) x 0
非齐次微分方程组的解 ：设A n d t d x ( t ) = A x ( t ) + f ( t ) x ( t 0 ) = x 0 x ( t ) = e A ( t − t 0 ) x 0 + ∫ t 0 t e A ( t − τ ) f ( τ ) d τ
证明：d t d ( e − A t x ( t )) = − A e − A t x ( t ) + e − A t d t d x ( t ) = e − A t ( d t d x ( t ) − A x ( t )) = e − A t f ( t ) e − A t x ( t ) − e − A t 0 x ( t 0 ) = ∫ t 0 t e − A τ f ( τ ) d τ x ( t ) = e A ( t − t 0 ) x ( t 0 ) + ∫ t 0 t e A ( t − τ ) f ( τ ) d τ
非齐次微分方程组的解 2：设A B n f ( t ) d t d x ( t ) = A x ( t ) + B f ( t ) x ( t 0 ) = x 0 x = e A ( t − t 0 ) x 0 + ∫ t 0 t e A ( t − τ ) B f ( τ ) d τ
渐进稳定 ：设A n t 0 x 0 { d t d x ( t ) = A x ( t ) x ( t 0 ) = x 0 x ( t ) t → ∞ lim x ( t ) = 0 d t d x ( t ) = A x ( t )
渐进稳定的充要条件 ：对任意的t 0 x 0 { d t d x ( t ) = A x ( t ) x ( t 0 ) = x 0 x ( t ) A
例1：设A = 1 − 1 − 1 − 2 0 − 1 6 3 4 d t d x ( t ) = A x ( t ) x ( 0 ) = 1 1 1
解：首先算出矩阵函数e A t = ( 1 − 2 t ) e t − t e t − t e t − 2 t e t ( 1 − t ) e t − t e t 6 t e t 3 t e t ( 1 + 3 t ) e t x ( t ) = e A ( t − t 0 ) x 0 = e A t x ( 0 ) = ( 1 + 2 t ) e t ( 1 + t ) e t ( 1 + t ) e t
例2：设A = 1 − 1 − 1 − 2 0 − 1 6 3 4 f ( t ) = − e t 0 e t d t d x ( t ) = A x ( t ) + f ( t ) x ( 0 ) = 1 1 1
解：由前面定理知，解为x ( t ) = e A ( t − t 0 ) x 0 + ∫ t 0 t e A ( t − τ ) f ( τ ) d τ t 0 = 0 e A t x ( 0 ) = ( 1 + 2 t ) e t ( 1 + t ) e t ( 1 + t ) e t e A ( t − τ ) = ( 1 − 2 t + 2 τ ) e t − τ − ( t − τ ) e t − τ − ( t − τ ) e t − τ − 2 ( t − τ ) e t − τ ( 1 − t + τ ) e t − τ − ( t − τ ) e t − τ 6 ( t − τ ) e t − τ 3 ( t − τ ) e t − τ ( 1 + 3 t − 3 τ ) e t − τ e A ( t − τ ) f ( τ ) = e t − 1 + 8 ( t − τ ) 4 ( t − τ ) 1 + 4 ( t − τ ) ∫ 0 t e A ( t − τ ) f ( τ ) d τ = e t 4 t 2 − t 2 t 2 2 t 2 + t x ( t ) = ( 1 + t + 4 t 2 ) e t ( 1 + t + 2 t 2 ) e t ( 1 + 2 t + 2 t 2 ) e t
八、矩阵的广义逆
1. 广义逆
定理：设A F s × n A X A = A rank ( A ) = r P Q A = P [ E r 0 0 0 ] Q X = Q − 1 [ E r C B D ] P − 1 B C D r × ( s − r ) ( n − r ) × r ( n − r ) × ( s − r )
证明：把解代入方程得A X A = P [ E r 0 0 0 ] Q Q − 1 [ E r C B D ] P − 1 P [ E r 0 0 0 ] Q = A B C D
另一方面，任取方程的一个解G A G A = A P [ E r 0 0 0 ] QGP [ E r 0 0 0 ] Q = P [ E r 0 0 0 ] Q P Q [ E r 0 0 0 ] QGP [ E r 0 0 0 ] = [ E r 0 0 0 ] QGP = [ H C B D ] [ E r 0 0 0 ] [ H C B D ] [ E r 0 0 0 ] = [ E r 0 0 0 ] [ H 0 0 0 ] = [ E r 0 0 0 ] H = E r G = Q − 1 [ E r C B D ] P − 1
广义逆 ：设A s × n A X A = A A A A −
特别地，如果A A − 1 A X A = A A −
例：已知A = 0 2 − 4 − 1 − 4 5 3 1 7 0 5 − 10 A −
解：由于广义逆要用到化为Simth标准型的行、列变换矩阵P Q [ A E 4 E 3 0 ] [ E r Q P 0 ] = 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 2 11 3 1 0 0 0 0 − 2 5 0 0 1 − 2 − 1 − 3 0 0 0 0 2 1 0 2 0 0 0 0 0 0 1 0 0 0 0 P Q P A Q = [ E 2 0 0 0 ] M = Q [ E 2 Y X Z ] P X ∈ C 2 × 1 Y ∈ C 2 × 2 Z ∈ C 2 × 1
广义逆的另一种表达式 ：任给A A = U [ Δ 0 0 0 ] V H A X A = A U [ Δ 0 0 0 ] V H X U [ Δ 0 0 0 ] V H = U [ Δ 0 0 0 ] V H X = V [ Δ − 1 C B D ] U H B C D
广义逆的性质 ：
rank ( A ) ≤ rank ( A − ) A − A A A − rank ( A ) = rank ( A − A ) = rank ( A A − ) ( A T ) − = ( A − ) T ( A H ) − = ( A − ) H 若A ∈ C n n × n A − = A − 1 A −
( k A ) − = k 1 A − k = 0 设S ∈ C m m × m T ∈ C n n × n B = S A T B − = ( S A T ) − = T − 1 A − S − 1
性质1的证明：rank ( A ) = rank ( A A − A ) ≤ rank ( A A − ) ≤ rank ( A − )
性质2的证明：( A − A ) 2 = A − ( A A − A ) = A − A ( A A − ) 2 = ( A A − A ) A − = A A − rank ( A ) = rank ( A A − A ) ≤ rank ( A A − ) ≤ rank ( A )
性质6的证明：S A T ( S A T ) X ( S A T ) = S A T X S T A TXS A = A TXS TXS A A − = TXS X = T − 1 A − S − 1
左逆 ：设A ∈ C m × n A L − 1 ∈ C n × m A L − 1 A = E n A L − 1 A
右逆 ：设A ∈ C m × n A R − 1 ∈ C n × m A A R − 1 = E m A R − 1 A
注：若A A A A
如果m = n A A − 1 = A L − 1 = A R − 1
左、右逆（如果存在）其实就是A 。
推论（设A ∈ C m × n
A ∈ C n m × n A A − A = E n A ∈ C m m × n A A A − = E m
充分性显然，下面证必要性：由于rank ( A ) = rank ( A − A ) = n A − A n × n A − A E n = A − A ( A − A ) − 1 = ( A − A ) ( A − A ) ( A − A ) − 1 ( A − A ) − 1 = A − A ( A − A ) − 1 ( A − A ) − 1 = ( A − A ) − 1
命题，有两个特殊的左、右逆：
如果A ∈ C n m × n ( A H A ) − 1 A H A L − 1
如果A ∈ C m m × n A H ( A A H ) − 1 A R − 1
自反广义逆 ：设A ∈ C m × n A A − A = A A − A A − = A A − ∈ C n × m A A r −
自反广义逆的充要条件 ：设A − ∈ C n × m A ∈ C m × n A − A rank ( A ) = rank ( A − )
自反广义逆的求法(1) ：设X , Y ∈ C n × m A ∈ C m × n Z = X A Y A
证明：A Z A = A ( X A Y ) A = A Y A = A Z A Z = ( X A Y ) A ( X A Y ) = X A X A Y = X A Y = Z
自反广义逆的求法(2) ：如果A ∈ C r m × n A = U [ Δ 0 0 0 ] V H U V A X = V [ Δ − 1 D B D Δ B ] U H B ∈ C r × ( m − r ) D ∈ C ( n − r ) × r
2. 伪逆矩阵
伪逆矩阵 ：设A ∈ C m × n A + ∈ C n × m A +
A A + A = A A + A A + = A + ( A A + ) H = A A + ( A + A ) H = A + A
则称A + A Penrose-Moore方程 ）
比如：设A = [ B 0 0 0 ] B A + = [ B − 1 0 0 0 ]
定理：A A + 唯一 的。
证明：设X Y A X = X A X = X ( A Y A ) X = X ( A Y ) H ( A X ) H = X ( A X A Y ) H = X ( A Y ) H = X A Y = X A Y A Y = ( X A ) H ( Y A ) H Y = ( Y A X A ) H Y = ( Y A ) H Y = Y A Y = Y
推论：若A ∈ C n n × n A + = A − 1
一般来说，( A B ) + = B + A + A = [ 0 1 ] B = [ 1 1 ] ( A B ) + = 1 + = 1 B + A + = [ 2 1 2 1 ] [ 0 1 ] = 2 1
伪逆矩阵的求法(1) ：设A ∈ C m × n A = BC A X = C H ( C C H ) − 1 ( B H B ) − 1 B H A
伪逆矩阵的求法(2) ：如果A ∈ C r m × n A = U [ Δ 0 0 0 ] V H U V A A + = V [ Δ − 1 0 0 0 ] U H
例：设A = [ − 1 2 0 0 1 − 2 ] A +
解：利用满秩分解得A = BC = [ − 1 2 ] [ 1 0 − 1 ] A + = C H ( C C H ) − 1 ( B H B ) − 1 B H = 1 0 − 1 [ 1 0 − 1 ] 1 0 − 1 − 1 ( [ − 1 2 ] [ − 1 2 ] ) − 1 [ − 1 2 ] = 10 1 − 1 0 1 2 0 − 2
推论：
若A ∈ C r m × r A + = ( A H A ) − 1 A H A + A = E r
若A ∈ C r r × m A + = A H ( A A H ) − 1 A A + = E r
比如假设A A = E A X = C H ( C C H ) − 1 ( B H B ) − 1 B H
伪逆矩阵的性质 ：
( A + ) + = A ( A H ) + = ( A + ) H ( A A H ) + = ( A H ) + A + = ( A + ) H A + ( A H A ) + = A + ( A H ) + = A + ( A + ) H A + = A H ( A A H ) + = ( A H A ) + A H
前两条性质只需验证满足Penrose-Moore方程即可，下面证明第3条性质：A + = A + A A + = ( A + A ) H A + = A H ( A + ) H A + = A H ( A A H ) + A + = A + A A + = A + ( A A + ) H = A + ( A + ) H A H = ( A H A ) + A H
例：设A ∈ C m × n P Q m n ( P A Q ) + = Q H A + P H = Q + A + P +
证：由于P Q P + = P − 1 Q + = Q − 1
3. 广义逆与线性方程组
非齐次线性方程A X = β ：设A ∈ C m × n β ∈ C m A X = β A A − β = A A − β
必要性：设有解且解为X = α A α = β β = A α = A A − A α = A A − β
充分性：设β = A A − β α = A − β A α = A ( A − β ) = β
注：如果存在A A − β = A A − β A Y β = A Y β
证明：A Y β = A Y ( A A − β ) = ( A A − β ) = β
非齐次线性方程A X = β ：设非齐次线性方程组A X = β X = A − β A − A
矩阵方程A XB = D ：设A ∈ C m × n B ∈ C s × t D ∈ C m × t A XB = D A B A − B − A A − D B − B = D
必要性：设方程有解，X A XB = D A B A − B − D = A XB = ( A A − A ) X ( B B − B ) = A A − D B − B
充分性：设存在A B A − B − A A − D B − B = D A − D B − A XB = D
矩阵方程A XB = D ：设A ∈ C m × n B ∈ C s × t D ∈ C m × t A XB = D X = A − D B − + Y − A − A Y B B − A − B − A B Y n × s
证明：直接代入即可，A XB = A ( A − D B − + Y − A − A Y B B − ) B = A A − D B − B + A Y B − A A − A Y B B − B = D + A Y B − A Y B = D
证明：直接代入即可，A XB = A ( A − D B − + Y − A − A Y B B − ) B = A A − D B − B + A Y B − A A − A Y B B − B = D + A Y B − A Y B = D
另一方面，任取方程一解G A GB = D G = A − D B + G − A − D B = A − D B + G − A − A GB B −
推论1 ：设A ∈ C m × n D ∈ C m × s A X = D A A − A A − D = D X = A − D + Y − A − A Y A − A Y n × s B = E
推论2 ：设A ∈ C m × n b ∈ C m A X = b X = A − b + Y − A − A Y = A − b + ( E n − A − A ) Y Y ∈ C n A − A
推论2解的形式类似”特解+齐次方程A X = 0 ( E n − A − A ) A X = 0 A ( E n − A − A ) = A − A A − A = 0 Y
4. 最小二乘问题
线性最小二乘问题：设A ∈ C m × n b ∈ C m 线性最小二乘问题 包含求解下面集合：S = { y ∈ C n : ∣∣ A y − b ∣ ∣ 2 = x ∈ C n min ∣∣ A x − b ∣ ∣ 2 } A x = b 相容 的，否则称它是不相容 的。称相容方程组A x = b A x = b 最小模解 。当A x = b x 0 ∈ C n x ∈ C n ∣∣ A x 0 − b ∣ ∣ 2 ≤ ∣∣ A x = b ∣ ∣ 2 x 0 A x = b 最小二乘解 。
范数最小值是0，但是当它取不到0时（找不到最小模解时），仍然可以找到一个最小值（对应的x
定理 ：下面命题是等价的（设A ∈ C m × n b ∈ C m
A x = b rank ( A ) = rank ( [ A b ] ) b ∈ R ( A ) b A b = A A − b A − A
定理 ：设A ∈ C m × n b ∈ C m A x = b
当rank ( A ) = n A + b A x = b ⟶ A H A x = A H b ⟶ x = ( A H A ) − 1 A H b = A + b
当rank ( A ) < n x = A − b + Y − A − A Y = A − b + ( E n − A − A ) Y Y ∈ C n
最小模解的形式 ：对于相容线性方程组A x = b x = A + b
证明：当rank ( A ) = n rank ( A ) < n x = A + b
方程的通解为x = A − b + ( E n − A − A ) Y x = A + b + ( E n − A + A ) Y ∣∣ x ∣ ∣ 2 2 = ∣∣ A + b ∣ ∣ 2 2 + ∣∣ ( E n − A + A ) Y ∣ ∣ 2 2 + Y H ( E n − A + A ) A + b + ( A + b ) H ( E n − A + A ) Y = ∣∣ A + b ∣ ∣ 2 2 + ∣∣ ( E n − A + A ) Y ∣ ∣ 2 2
(因为对于Y H ( E n − A + A ) A + b = Y H ( A + − A + A A + ) b = 0 ( A + b ) H ( E n − A + A ) Y = [( E n − A + A ) ( A + b ) ] H Y = 0
由于Y Y = 0 ∣∣ x ∣ ∣ 2 2 ∣∣ A + b ∣ ∣ 2 2
最小二乘解的讨论 ：对于不相容线性方程组A x = b x ∈ C n min ∣∣ A x − b ∣ ∣ 2 = ∣∣ A A + b − b ∣ ∣ 2 A x = A A + b
证明：
∣∣ A x − b ∣ ∣ 2 2 = ∣∣ ( A x − A A + b ) + ( A A + b − b ) ∣ ∣ 2 2 = ∣∣ A A + b − b ∣ ∣ 2 2 + ∣∣ A x − A A + b ∣ ∣ 2 2 + ( A A + b − b ) H ( A x − A A + b ) + ( A x − A A + b ) H ( A A + b − b ) = ∣∣ A A + b − b ∣ ∣ 2 2 + ∣∣ A x − A A + b ∣ ∣ 2 2 ≥ ∣∣ A A + b − b ∣ ∣ 2 2
（其中( A A + b − b ) H ( A x − A A + b ) = b H ( A A + − E ) A ( x − A + b ) = 0 ( A x − A A + b ) H ( A A + b − b ) = ( A x − A A + b ) H [ b ( A A + − E ) ] H = [ b ( A A + − E ) A ( x − A + b ) ] H = 0 x = A + b A x = A A + b ∣∣ A x − A A + b ∣ ∣ 2 2
这样就把求解最小二乘解的问题转化为求解相容方程A x = A A + b
定理 ：设A ∈ C m × n b ∈ C m A x = b
当rank ( A ) = n A x = b A + b A x = A A + b A + b
当rank ( A ) < n A x = b x = A + ( A A + b ) + ( E n − A + A ) Y = A + b + ( E n − A + A ) Y Y ∈ C n x 0 ∣∣ u ∣ ∣ 2 ≤ ∣∣ x 0 ∣ ∣ 2 u 最佳最小二乘解 。
最小二乘解的形式 ：x ∗ = A + b A x = b
证明：当rank ( A ) = n rank ( A ) < n x ∗ = A + b
∣∣ x ∣ ∣ 2 2 = ∣∣ A + b + ( E n − A + A ) Y ∣ ∣ 2 2 = ∣∣ A + b ∣ ∣ 2 + ∣∣ ( E n − A + A ) Y ∣ ∣ 2 2 0
例1：已知矩阵A = 1 0 2 2 0 4 0 2 0 b = 1 1 3 A x = b x ∈ C 3 min ∣∣ b − A x ∣ ∣ 2
解：由于rank ( A ) = 2 rank ( [ A b ] ) = 3 A x = b A A + = 50 1 2 4 0 0 0 25 4 8 0 x = A + b + ( E n − A + A ) Y = 50 1 14 28 25 + 5 4 − 5 2 0 − 5 2 5 1 0 0 0 0 y 1 y 2 y 3 = 50 1 14 28 25 + k − 2 1 0 ∀ k ∈ C − 2 1 0 A x = 0 A x = A A + b A x = 0 x ∗ = A + b = 50 1 14 28 25 x ∈ C 3 min ∣∣ b − A x ∣ ∣ 2 = ∣∣ b − A A + b ∣ ∣ 2 = 5 1
例2：已知一组数据
t i -2
1
3
5
y i -1
4
7
12
假设它们近似满足线性关系y i = a t i + b + δ i i = 1 , 2 , 3 , 4 a b δ 1 2 + δ 2 2 + δ 3 2 + δ 4 2
解：设A = − 2 1 3 5 1 1 1 1 y = − 1 4 7 12 δ = δ 1 δ 2 δ 3 δ 4 x = [ a b ] min δ T δ = x min ∣∣ y − A x ∣ ∣ 2 2 A x = y A x = A + y = ( A T A ) − 1 A T y = 107 1 [ 194 249 ] ≈ [ 1.813 2.327 ]
九、Kronecker积
1. Kronecker积
定义：设A = ( a ij ) m × n B = ( b ij ) p × q a 11 B a 21 B ⋮ a m 1 B a 12 B a 22 B ⋮ a m 2 B ⋯ ⋯ ⋯ a 1 n B a 2 n B ⋮ a mn B m p × n q A B Kronecker积 ，或A B 直积 A ⊗ B
比如：X = ( x 1 , x 2 , x 3 ) T Y = ( y 1 , y 2 ) T X ⊗ Y = ( x 1 y 1 , x 1 y 2 , x 2 y 1 , x 2 y 2 , x 3 y 1 , x 3 y 2 ) T Y ⊗ X = ( y 1 x 1 , y 1 x 2 , y 1 x 3 , y 2 x 1 , y 2 x 2 , y 2 x 3 ) T X ⊗ Y = Y ⊗ X
如果X , Y ∈ C n X Y T = X ⊗ Y T
Kronecker积的运算性质 ：
k ( A ⊗ B ) = ( k A ) ⊗ B = A ⊗ ( k B ) ∀ k ∈ C A ⊗ ( B + C ) = A ⊗ B + A ⊗ C ( B + C ) ⊗ A = B ⊗ A + C ⊗ A ( A + B ) ⊗ ( C + D ) = A ⊗ C + A ⊗ D + B ⊗ C + B ⊗ D A ⊗ ( B ⊗ C ) = ( A ⊗ B ) ⊗ C E m ⊗ E n = E n ⊗ E m = E mn
定理 ：设A = ( a ij ) m × n B = ( b ij ) l × r C = ( c ij ) n × p D = ( d ij ) r × s ( A ⊗ B ) ( C ⊗ D ) = A C ⊗ B D
推论：若A = ( a ij ) m × m B = ( b ij ) n × n A ⊗ B = ( A ⊗ E n ) ( E m ⊗ B ) = ( E m ⊗ B ) ( A ⊗ E n )
定理 ：设A = ( a ij ) m × n B = ( b ij ) p × q
( A ⊗ B ) T = A T ⊗ B T ( A ⊗ B ) H = A H ⊗ B H ( A ⊗ B ) + = A + ⊗ B +
从第3条可知，如果A B m n A ⊗ B ( A ⊗ B ) − 1 = A − 1 ⊗ B − 1 ∵ ( A ⊗ B ) ( A − 1 ⊗ B − 1 ) = ( A A − 1 ) ⊗ ( B B − 1 ) = E m ⊗ E n = E mn ∴ ( A ⊗ B ) − 1 = A − 1 ⊗ B − 1
定理 ：若A = ( a ij ) m × m B = ( b ij ) n × n tr ( A ⊗ B ) = tr ( A ) ⋅ tr ( B )
证明：tr ( A ⊗ B ) = a 11 ⋅ tr ( B ) + a 22 ⋅ tr ( B ) + ⋯ + a mm ⋅ tr ( B ) = tr ( A ) ⋅ tr ( B )
定理 ：设A = ( a ij ) m × n B = ( b ij ) p × q rank ( A ⊗ B ) = rank ( A ) ⋅ rank ( B )
证明：设rank ( A ) = r rank ( B ) = s M N P Q M A N = [ E r 0 0 0 ] = Δ A 1 PBQ = [ E s 0 0 0 ] = Δ B 1 M P N Q A ⊗ B = ( M − 1 A 1 N − 1 ) ⊗ ( P − 1 B 1 Q − 1 ) = ( M − 1 ⊗ P − 1 ) ( A 1 ⊗ B 1 ) ( N − 1 ⊗ Q − 1 ) = ( M ⊗ P ) − 1 ( A 1 ⊗ B 1 ) ( N ⊗ Q ) − 1 ∴ rank ( A ⊗ B ) = rank ( A 1 ⊗ B 1 ) = rank ( A ) ⋅ rank ( B )
定理 ：若A = ( a ij ) m × m B = ( b ij ) n × n ∣ A ⊗ B ∣ = ∣ A ∣ n ∣ B ∣ m
证明：假设A B J 1 J 2 P − 1 A P = J 1 Q − 1 BQ = J 2 A ⊗ B = ( P J 1 P − 1 ) ⊗ ( Q J 2 Q − 1 ) = ( P ⊗ Q ) ( J 1 ⊗ J 2 ) ( P − 1 ⊗ Q − 1 ) = ( P ⊗ Q ) ( J 1 ⊗ J 2 ) ( P ⊗ Q ) − 1 ( A ⊗ B ) ( J 1 ⊗ J 2 ) A B { λ i , i = 1 , 2 , ⋯ , m } { μ j , j = 1 , 2 , ⋯ , n } ∣ A ⊗ B ∣ = ∣ J 1 ⊗ J 2 ∣ = ∏ j = 1 n ( λ 1 μ j ) ⋅ ∏ j = 1 n ( λ 2 μ j ) ⋯ ∏ j = 1 n ( λ m μ j ) = ( λ 1 n λ 2 n ⋯ λ m n ) ( ∏ j = 1 n μ j ) m = ∣ A ∣ n ∣ B ∣ m
2. Kronecker积的特征值
定理：设x 1 , x 2 , ⋯ , x n n m y 1 , y 2 , ⋯ , y q q p n q m p x i ⊗ y j i = 1 , 2 , ⋯ , n j = 1 , 2 , ⋯ , q n q m p x i ⊗ y j x 1 , x 2 , ⋯ , x n y 1 , y 2 , ⋯ , y q
定理(特征值) ：设λ 1 , λ 2 , ⋯ , λ m m A μ 1 , μ 2 , ⋯ , μ n n B mn f ( A , B ) = ∑ i , j = 0 l c ij A i ⊗ B j f ( λ r , μ s ) r = 1 , 2 , ⋯ , m s = , 1 , 2 , ⋯ , n c ij ∈ C
考虑由变量x , y f ( x , y ) = ∑ i , j = 0 l c ij x i y j A m B n f ( A , B ) = ∑ i , j = 0 l c ij A i ⊗ B j
比如：设f ( x , y ) = 2 x + x y 3 f ( A , B ) = 2 A ⊗ E + A ⊗ B 3
证明：设A B J A J B P − 1 A P = J A Q − 1 BQ = J B J A J b λ 1 , λ 2 , ⋯ , λ m μ 1 , μ 2 , ⋯ , μ n f ( A , B ) = ( P ⊗ Q ) f ( J A , J B ) ( P ⊗ Q ) − 1 f ( A , B ) ∼ f ( J A , J B ) f ( J A , J B ) = ∑ i , j = 0 l c ij ( J A i ⊗ J B j ) f ( λ r , μ s ) = ∑ i , j = 0 l c ij λ r i μ s j
推论：设λ 1 , λ 2 , ⋯ , λ m m A μ 1 , μ 2 , ⋯ , μ n n B
A ⊗ B λ r μ s r = 1 , 2 , ⋯ , m s = , 1 , 2 , ⋯ , n A ⊗ E n + E m ⊗ B λ r + μ s r = 1 , 2 , ⋯ , m s = , 1 , 2 , ⋯ , n
定理(特征向量) ：设x 1 , x 2 , ⋯ , x p m A λ 1 , λ 2 , ⋯ , λ m p ≤ m y 1 , y 2 , ⋯ , y q n B μ 1 , μ 2 , ⋯ , μ n q ≤ n x r ⊗ y s r = 1 , 2 , ⋯ , p s = , 1 , 2 , ⋯ , q f ( A , B ) f ( λ r , μ s ) f ( A , B ) = ∑ i , j = 0 l c ij A i ⊗ B j
证明：根据前面定理，x 1 , x 2 , ⋯ , x p y 1 , y 2 , ⋯ , y q x r ⊗ y s r = 1 , 2 , ⋯ , p s = , 1 , 2 , ⋯ , q A i x r = λ r i x r B j y s = μ s j y s
f ( A , B ) ( x r ⊗ y s ) = ( ∑ i , j = 0 l c ij A i ⊗ B j ) ( x r ⊗ y s ) = ∑ i , j = 0 l c ij ( A i ⊗ B j ) ( x r ⊗ y s ) = ∑ i , j = 0 l c ij ( A i x r ⊗ B j y s ) = ∑ i , j = 0 l c ij λ r i μ s j ( x r ⊗ y s ) = f ( λ r , μ s ) ( x r ⊗ y s )
3. 矩阵的列展开与行展开
矩阵的行、列展开 ：将A = ( a ij ) m × n mn A rs ( A ) rs ( A ) = ( a 11 , a 12 , ⋯ , a 1 n , a 21 , a 22 , ⋯ , a 2 n , ⋯ , a m 1 , a m 2 , ⋯ , a mn ) A mn A cs ( A ) cs ( A ) = ( a 11 , a 21 , ⋯ , a m 1 , a 12 , a 22 , ⋯ , a m 2 , ⋯ , a 1 n , a 2 n , ⋯ , a mn ) T
根据定义，容易看出：
rs ( A T ) = ( cs ( A ) ) T cs ( A T ) = ( rs ( A ) ) T 设A ∈ C m × n B ∈ C n × m tr ( A B ) = ∑ i = 1 m ∑ k = 1 n a ik b ki = rs ( A ) cs ( B )
定理 ：设A = ( a ij ) m × n B = ( b ij ) n × p C = ( c ij ) p × q
rs ( A BC ) = rs ( B ) ( A T ⊗ C )
cs ( A BC ) = ( C T ⊗ A ) cs ( B )
证明：记A = α 1 α 2 ⋮ α m B = β 1 β 2 ⋮ β n C = [ ν 1 ν 2 ⋯ ν q ] rs ( A BC ) = ⎩ ⎨ ⎧ α 1 α 2 ⋮ α m BC ⎭ ⎬ ⎫ = ( α 1 BC , α 2 BC , ⋯ , α m BC ) α i BC = α i β 1 β 2 ⋮ β n [ ν 1 ν 2 ⋯ ν q ] = α i β 1 ν 1 β 2 ν 1 ⋮ β n ν 1 β 2 ν 2 β 2 ν 2 ⋮ β n ν 2 ⋯ ⋯ ⋯ β 1 ν q β 2 ν q ⋮ β n ν q = [ ∑ j = 1 n a ij β j ν 1 ∑ j = 1 n a ij β j ν 2 ⋯ ∑ j = 1 n a ij β j ν q ] α i BC = [ β 1 β 2 ⋯ β n ] a i 1 ν 1 a i 2 ν 1 ⋮ a in ν 1 a i 1 ν 2 a i 2 ν 2 ⋮ a in ν 2 ⋯ ⋯ ⋯ a i 1 μ q a i 2 ν q ⋮ a in ν q = rs ( B ) a i 1 C a i 2 C ⋮ a in C = rs ( B ) ( α i T ⊗ C ) rs ( A BC ) = ( α 1 BC , α 2 BC , ⋯ , α m BC ) = rs ( B ) ( α 1 T ⊗ C , α 2 T ⊗ C , ⋯ , α m T ⊗ C ) = rs ( B ) ( A T ⊗ C )
由rs ( A T ) = ( cs ( A ) ) T ( cs ( A BC ) ) T = rs (( A BC ) T ) = rs ( C T B T A T ) = rs ( B T ) ( C ⊗ A T )
推论1：设A ∈ C m × r B ∈ C r × n rs ( A B ) = rs ( A ) ( E m ⊗ B ) = rs ( B ) ( A T ⊗ E n )
推论2：设A ∈ C m × m B ∈ C n × n X ∈ C m × n
cs ( A X ) = ( E n ⊗ A ) cs ( X ) cs ( XB ) = ( B T ⊗ E m ) cs ( X ) cs ( A X + XB ) = ( E n ⊗ A + B T ⊗ E m ) cs ( X )
证明：cs ( A X ) = cs ( A X E n ) cs ( XB ) cs ( A X + XB ) = cs ( A X ) + cs ( XB )
4. 线性矩阵代数方程
Ⅰ. 方程A 1 X B 1 + A 2 X B 2 + ⋯ + A p X B p = C
问题：设A j ∈ C m × m B j ∈ C n × n j = 1 , 2 , ⋯ , p A 1 X B 1 + A 2 X B 2 + ⋯ + A p X B p = C
解：对此方程两边分别作列展开得( B 1 T ⊗ A 1 ) cs ( X ) + ( B 2 T ⊗ A 2 ) cs ( X ) + ⋯ + ( B p T ⊗ A p ) cs ( X ) = cs ( C ) x = cs ( X ) c = cs ( C ) G = ∑ j = 1 p ( B j T ⊗ A j ) G x = c
定理 ：X ∈ C m × n A 1 X B 1 + A 2 X B 2 + ⋯ + A p X B p = C x = cs ( X ) G x = c
推论1：方程A 1 X B 1 + A 2 X B 2 + ⋯ + A p X B p = C 有解 的充要条件是rank ( G , c ) = rank ( G )
推论2：方程A 1 X B 1 + A 2 X B 2 + ⋯ + A p X B p = C 有唯一解 的充要条件是G
Ⅱ. 方程A X + XB = C
定理 ：设A ∈ C m × m B ∈ C n × n A X + XB = C 有唯一解 的充要条件是A B λ i ( A ) + λ j ( B ) = 0 i = , 1 , 2 , ⋯ , m j = 1 , 2 , ⋯ , n A − B
证明：A X + XB = C cs ( A X + XB ) = ( E n ⊗ A + B T ⊗ E m ) cs ( X ) = cs ( C ) G = ( E n ⊗ A + B T ⊗ E m ) G G λ j ( B ) + λ i ( A )
其中λ ( ⋅ ) λ ⋅
推论：设A ∈ C m × m B ∈ C n × n A X + XB = 0 有非零解 的充要条件是存在某个1 ≤ i ≤ m 1 ≤ j ≤ n λ i ( A ) + λ j ( B ) = 0
证明：因为A X + XB = 0 λ i ( A ) + λ j ( B ) = 0
Ⅲ. 方程X + A XB = C
定理 ：设A ∈ C m × m B ∈ C n × n X + A XB = C 有唯一解 的充要条件是A B λ i ( A ) λ j ( B ) = − 1 i = , 1 , 2 , ⋯ , m j = 1 , 2 , ⋯ , n
证明：方程两边列展开化为线性方程，变成cs ( X + A XB ) = cs ( E m X E n + A XB ) = ( E n ⊗ E m + B T ⊗ A ) cs ( X ) = cs ( C ) G = ( E n ⊗ E m + B T ⊗ A ) G 1 + λ i ( A ) λ j ( B ) = 0
例：已知A = [ 1 0 1 2 ] B = [ 3 1 0 3 ] C = [ − 4 7 4 0 ] X + A XB = C
解：方程两边列展开后得到对应的线性方程组为( E 2 ⊗ E 2 + B T ⊗ A ) cs ( X ) = cs ( C ) 4 0 0 0 3 7 0 0 1 0 4 0 1 2 3 7 x 1 x 2 x 3 x 4 = − 4 7 4 0 x 1 x 2 x 3 x 4 = − 2 1 1 0 X 2 × 2 X X = [ − 2 1 1 0 ]