4.5 Hermite 矩阵、对称矩阵的相合与同时对角化 任一二阶线性偏微分算子 L L L
L f = ∑ i , j = 1 n a i j ( x ) ∂ 2 f ( x ) ∂ x i ∂ x j + 诸 低 阶 项 , x = [ x i ] ∈ D ⊂ R n , (4.5.1) L f = \sum_ {i, j = 1} ^ {n} a _ {i j} (x) \frac {\partial^ {2} f (x)}{\partial x _ {i} \partial x _ {j}} + \text {诸 低 阶 项}, \quad x = [ x _ {i} ] \in D \subset \mathbf {R} ^ {n}, \tag {4.5.1} L f = i , j = 1 ∑ n a ij ( x ) ∂ x i ∂ x j ∂ 2 f ( x ) + 诸 低 阶 项 , x = [ x i ] ∈ D ⊂ R n , ( 4.5.1 ) 其中，假定系数 a i j ( x ) a_{ij}(x) a ij ( x ) D ⊂ R n D \subset \mathbb{R}^n D ⊂ R n f f f D D D x ∈ D x \in D x ∈ D Λ ( x ) = [ a i j ( x ) ] \Lambda(x) = [a_{ij}(x)] Λ ( x ) = [ a ij ( x )] f f f
如果作自变量到新变量 s = [ s t ] ∈ D ⊂ R n s = [s_t] \in D \subset \mathbb{R}^n s = [ s t ] ∈ D ⊂ R n s t = s t [ x ] = s t ( x 1 , … , x n ) s_t = s_t[x] = s_t(x_1, \dots, x_n) s t = s t [ x ] = s t ( x 1 , … , x n )
S ( x ) = [ ∂ S i ( x ) ∂ x j ] ∈ M n S (x) = \left[ \frac {\partial_ {S _ {i}} (x)}{\partial x _ {j}} \right] \in M _ {n} S ( x ) = [ ∂ x j ∂ S i ( x ) ] ∈ M n 在 D D D x = x ( s ) x = x(s) x = x ( s ) L L L
L f = ∑ i , j n [ ∑ p , q = 1 n ∂ s i ∂ x p a p i − ∂ s j ∂ x q ] ∂ 2 f ∂ s i + 诸 低 阶 项 = ∑ i , j = 1 n b i j ∂ 2 f ∂ s i ∂ s j + 诸 低 阶 项 . (4.5.2) \begin{array}{l} L f = \sum_ {i, j} ^ {n} \left[ \sum_ {p, q = 1} ^ {n} \frac {\partial s _ {i}}{\partial x _ {p}} a _ {p i} - \frac {\partial s _ {j}}{\partial x _ {q}} \right] \frac {\partial^ {2} f}{\partial s _ {i}} + \text {诸 低 阶 项} \\ = \sum_ {i, j = 1} ^ {n} b _ {i j} \frac {\partial^ {2} f}{\partial s _ {i} \partial s _ {j}} + \text {诸 低 阶 项}. \tag {4.5.2} \\ \end{array} L f = ∑ i , j n [ ∑ p , q = 1 n ∂ x p ∂ s i a p i − ∂ x q ∂ s j ] ∂ s i ∂ 2 f + 诸 低 阶 项 = ∑ i , j = 1 n b ij ∂ s i ∂ s j ∂ 2 f + 诸 低 阶 项 . ( 4.5.2 ) 因此，（在坐标 s = ⌈ s i ⌉ s = \lceil s_i\rceil s = ⌈ s i ⌉ B B B x − [ x , ] x - [x,] x − [ x , ] A A A
B = S A S T (4.5.3) B = S A S ^ {T} \tag {4.5.3} B = S A S T ( 4.5.3 ) 表示，其中 S S S
如果微分算子 L L L L = V ‾ 2 L = \overline{V}^2 L = V 2 L L L A A A B B B
另一个像(4.5.3)那样的变换例子来源于概率与统计。假定在有期望算子 E E E X 1 , X 2 , ⋯ , X n X_{1}, X_{2}, \cdots, X_{n} X 1 , X 2 , ⋯ , X n μ i = E ( X i ) \mu_{i} = E(X_{i}) μ i = E ( X i ) A = [ a i j ] = ( E [ ( X i − μ i ) ( X ˉ j − μ j ) ] ) ≡ Cov ( X ) A = [a_{ij}] = (E[(X_{i} - \mu_{i})(\bar{X}_{j} - \mu_{j})]) \equiv \operatorname{Cov}(X) A = [ a ij ] = ( E [( X i − μ i ) ( X ˉ j − μ j )]) ≡ Cov ( X ) X = [ X 1 , ⋯ , X n ] T X = [X_{1}, \cdots, X_{n}]^{T} X = [ X 1 , ⋯ , X n ] T S = [ s i j ] ∈ M n S = [s_{ij}] \in M_{n} S = [ s ij ] ∈ M n S X SX SX X X X S X SX SX
E ( ( S X ) 1 ) = E ( ∑ k n s k X k ) − ∑ k = 1 n s k E ( X k ) = ∑ k = 1 n s k f t k , E ((S X) _ {1}) = E \left(\sum_ {k} ^ {n} s _ {k} X _ {k}\right) - \sum_ {k = 1} ^ {n} s _ {k} E \left(X _ {k}\right) = \sum_ {k = 1} ^ {n} s _ {k f t _ {k}}, E (( SX ) 1 ) = E ( k ∑ n s k X k ) − k = 1 ∑ n s k E ( X k ) = k = 1 ∑ n s k f t k , 而 S X SX SX
Cov ( S X ) = ( E [ ( ( S X ) , − E ( ( S X ) , ) ) ( S X ‾ ) , − E ( ( S X ‾ ) , ) ) ] ) = ( E [ ( ∑ p = 1 n s r p ( X p − μ p ) ) ( ∑ q = 1 n s r q ( X q − μ q ) ) ] ) = ( ∑ p , q = 1 n s i p E [ ( X p − μ p ) ( X ˉ q − μ ˉ q ) ] s ˉ i q ) − ( ∑ p , q = 1 n s i p a p q s m q ) = S A S ′ . \begin{array}{l} \operatorname {C o v} (S X) = \left(E \left[ ((S X), - E ((S X),)) (\overline {{S X}}), - E ((\overline {{S X}}),)) \right]\right) \\ = \left(E \left[ \left(\sum_ {p = 1} ^ {n} s _ {r p} \left(X _ {p} - \mu_ {p}\right)\right) \left(\sum_ {q = 1} ^ {n} s _ {r q} \left(X _ {q} - \mu_ {q}\right)\right) \right]\right) \\ = \left(\sum_ {p, q = 1} ^ {n} s _ {i p} E \left[ \left(X _ {p} - \mu_ {p}\right) \left(\bar {X} _ {q} - \bar {\mu} _ {q}\right) \right] \bar {s} _ {i q}\right) - \left(\sum_ {p, q = 1} ^ {n} s _ {i p} a _ {p q} s _ {m q}\right) \\ = S A S ^ {\prime}. \\ \end{array} Cov ( SX ) = ( E [ (( SX ) , − E (( SX ) , )) ( SX ) , − E (( SX ) , )) ] ) = ( E [ ( ∑ p = 1 n s r p ( X p − μ p ) ) ( ∑ q = 1 n s r q ( X q − μ q ) ) ] ) = ( ∑ p , q = 1 n s i p E [ ( X p − μ p ) ( X ˉ q − μ ˉ q ) ] s ˉ i q ) − ( ∑ p , q = 1 n s i p a pq s m q ) = S A S ′ . 这说明
Cov ( S X ) − SCov ( X ) S ′ . (4.5.3’) \operatorname {C o v} (S X) - \operatorname {S C o v} (X) S ^ {\prime}. \tag {4.5.3'} Cov ( SX ) − SCov ( X ) S ′ . ( 4.5.3’ ) 因此，随机向量的协方差矩阵的变化规律与 ( 1.5.3 T ) (1.5.3^{T}) ( 1.5. 3 T ) S S S ( 4.5.3 T ) (4.5.3^{T}) ( 4.5. 3 T )
作为最后一个例子，考虑一般二次型
Q 1 ( t ) = ∑ i , j n a i j r i x j − t ′ A i j , t = [ r i ] ∈ C n , Q _ {1} (t) = \sum_ {i, j} ^ {n} a _ {i j} r _ {i} x _ {j} - t ^ {\prime} A _ {i j}, \quad t = [ r _ {i} ] \in \mathbb {C} ^ {n}, Q 1 ( t ) = i , j ∑ n a ij r i x j − t ′ A ij , t = [ r i ] ∈ C n , 以及Hermite型
H B ( r ) − ∑ i , j = 1 n b i j x i x j = r B i , r − [ r i j ∈ C ′ , H _ {B} (r) - \sum_ {i, j = 1} ^ {n} b _ {i j} x _ {i} x _ {j} = r B _ {i}, \quad r - [ r _ {i j} \in \mathbf {C} ^ {\prime}, H B ( r ) − i , j = 1 ∑ n b ij x i x j = r B i , r − [ r ij ∈ C ′ , 其中 A = [ a n ] ∥ B − [ b n ] A = [a_{n}] \parallel B - [b_{n}] A = [ a n ] ∥ B − [ b n ] S ∈ M n S \in M_{n} S ∈ M n
Q 1 ( S r ) − ( S r ) T A ( S t ) J ′ ( S ′ \ S ) J = Q s ′ i , ( r ) . Q _ {1} \left(S _ {r}\right) - \left(S r\right) ^ {T} A \left(S _ {t}\right) \quad J ^ {\prime} \left(S ^ {\prime} \backslash S\right) J = Q _ {s ^ {\prime} i, (r)}. Q 1 ( S r ) − ( S r ) T A ( S t ) J ′ ( S ′ \ S ) J = Q s ′ i , ( r ) . 219
H A ( S r ) = ( S x ) ∗ B ( S x ) = x ∗ ( S ∗ B S ) r = H s ∗ ( x ) . H _ {A} (S r) = (S x) ^ {*} B (S x) = x ^ {*} (S ^ {*} B S) r = H _ {s ^ {*}} (x). H A ( S r ) = ( S x ) ∗ B ( S x ) = x ∗ ( S ∗ BS ) r = H s ∗ ( x ) . 在这个例子中， A A A B B B S S S x \pmb{x} x
4.5.4 定义 设 A , B ∈ M n A, B \in M_n A , B ∈ M n S S S
(a) B = S A S ∗ B = S A S^{*} B = S A S ∗ B B B A A A B = S A S ′ B = SAS^{\prime} B = S A S ′ B B B r \pmb{r} r
显然，这两个相合概念肯定有密切关系；如果 S S S
练习 证明相合的矩阵有相同的秩.
值得提出的是，如果 A A A A A A S A S ⊺ \mathsf{SAS}^{\intercal} SAS ⊺ A A A S A S ⊺ \mathsf{SAS}^{\intercal} SAS ⊺
4.5.5 定理 *相合和'相合都是等价关系。即对任 A ∈ M n A \in M_{n} A ∈ M n
(a) A A A A A A A A A B B B B B B A A A A A A B B B B B B C C C A A A C C C
证明：对于(a)，我们把 A A A A = I A I ∗ A = IAI^{*} A = I A I ∗ A = S B S ∗ A = SBS^{*} A = SB S ∗ S S S B = S − 1 A ( S − 1 ) ∗ B = S^{-1}A(S^{-1})^{*} B = S − 1 A ( S − 1 ) ∗ A = S 1 B S 1 ∗ A = S_{1}BS_{1}^{*} A = S 1 B S 1 ∗ B = S 2 C S 2 ∗ B = S_{2}CS_{2}^{*} B = S 2 C S 2 ∗ A = ( S 1 S 2 ) C ( S 1 S 2 ) ∗ A = (S_{1}S_{2})C(S_{1}S_{2})^{*} A = ( S 1 S 2 ) C ( S 1 S 2 ) ∗ τ \pmb{\tau} τ
因此，所有 n × n n \times n n × n
通过识别相合关系的诸不变量可以辨认并划分各微分算子，这个实际问题促使我们去分析研究由（经实矩阵S)相合于某个矩阵的诸实对称矩阵组成的等价类的标准代表元问题。结果证明，这个问题有一个简单的解答：只要计算正、负特征值和零特征值的个数。因为这个理由，引进下述专有名词。
4.5.6 定义 设 A ∈ M n A \in M_n A ∈ M n A A A
i ( A ) = ( i 1 ( A ) , i − ( A ) , i 0 ( A ) ) , i (A) = \left(i _ {1} (A), i _ {-} (A), i _ {0} (A)\right), i ( A ) = ( i 1 ( A ) , i − ( A ) , i 0 ( A ) ) , 其中， i − ( A ) i_{-}(A) i − ( A ) A A A i ( A ) i(A) i ( A ) A A A i 3 ( A ) i_{3}(A) i 3 ( A ) A A A A A A i + ( A ) + i ( A ) i_{+}(A) + i(A) i + ( A ) + i ( A ) A A A i 1 ( A ) − i − ( A ) i_{1}(A) - i_{-}(A) i 1 ( A ) − i − ( A )
练习 证明，如果知道 A A A A ∈ M n A \in M_{n} A ∈ M n
如果 Λ ∈ M n \Lambda \in M_{n} Λ ∈ M n A = U Λ U ∗ A = U\Lambda U^{*} A = U Λ U ∗ Λ = d i a g ( λ 1 , … , λ n ) \Lambda = \mathrm{diag}(\lambda_1,\dots ,\lambda_n) Λ = diag ( λ 1 , … , λ n ) U U U Λ \pmb{\Lambda} Λ λ 1 , λ 2 , … , λ t + > 0 , λ t + + 1 , … , λ t − − t + < 0 \lambda_1,\lambda_2,\dots ,\lambda_{t_+} > 0,\lambda_{t_+ + 1},\dots ,\lambda_{t_{-} - t_+} < 0 λ 1 , λ 2 , … , λ t + > 0 , λ t + + 1 , … , λ t − − t + < 0 λ t − t − + t − = ⋯ = λ n = 0 \lambda_{t_{-}t_{-} + t_{-}} = \dots = \lambda_{n} = 0 λ t − t − + t − = ⋯ = λ n = 0
D = diag ( + λ 1 , … , + λ t + , + − λ t + 1 , … , + − λ t + t + , 1 , … , 1 ) D = \operatorname {d i a g} (+ \sqrt {\lambda_ {1}}, \dots , + \sqrt {\lambda_ {t _ {+}}}, + \sqrt {- \lambda_ {t _ {+ 1}}}, \dots , + \sqrt {- \lambda_ {t _ {+ t _ {+}}}}, 1, \dots , 1) D = diag ( + λ 1 , … , + λ t + , + − λ t + 1 , … , + − λ t + t + , 1 , … , 1 ) 则 D D D
Λ = D [ 1 ⋱ 1 0 − 1 ⋱ 1 0 0 ⋱ 0 ] D , \Lambda = D \left[ \begin{array}{c c c c c c c c c} 1 & & & & & & & & \\ & \ddots & & & & & & \\ & & 1 & & & & & 0 & \\ & & & - 1 & & & & \\ & & & & \ddots & & & \\ & & & & & 1 & & \\ & & 0 & & & & 0 & \\ & & & & & & \ddots & \\ & & & & & & & 0 \end{array} \right] D, Λ = D 1 ⋱ 1 0 − 1 ⋱ 1 0 ⋱ 0 0 D , 其中所展示的矩阵恰好有 i + ( A ) i_{+}(A) i + ( A ) i ( A ) i(A) i ( A ) i 0 ( A ) i_0(A) i 0 ( A ) A A A
A = U Δ U ∗ = S [ 1 ⋱ 0 1 1 ⋱ 1 0 0 ⋱ 0 ] S ∗ = S I ( A ) S ∗ , (4.5.7) A = U \Delta U ^ {*} = S \left[ \begin{array}{c c c c c c c c} 1 & & & & & & & \\ & \ddots & & & & & 0 & \\ & & 1 & & & & & \\ & & & 1 & & & & \\ & & & & \ddots & & & \\ & & & & & 1 & & \\ & & & & & & 0 & \\ & & 0 & & & & \ddots & \\ & & & & & & & 0 \end{array} \right] S ^ {*} = S I (A) S ^ {*}, \tag {4.5.7} A = U Δ U ∗ = S 1 ⋱ 1 0 1 ⋱ 1 0 0 ⋱ 0 S ∗ = S I ( A ) S ∗ , ( 4.5.7 ) 其中 S = U D S = UD S = U D I ( A ) I(A) I ( A ) A A A A A A
4.5.8 定理 设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n S ∈ M n S \in M_{n} S ∈ M n A = S B S ∗ A = SBS^{*} A = SB S ∗ A A A B B B
证明：如果 A A A B B B S S S A A A B B B
假定 A A A B B B S ∈ M n S \in M_{n} S ∈ M n A = S B S ∗ A = SBS^{*} A = SB S ∗ i 11 ( A ) − i 0 ( B ) i_{11}(A) - i_{0}(B) i 11 ( A ) − i 0 ( B ) i + ( A ) = i − ( B ) i_{+}(A) = i_{-}(B) i + ( A ) = i − ( B ) v 1 , v 2 , ⋯ , v r + ( A ) v_{1}, v_{2}, \cdots, v_{r_{+}(A)} v 1 , v 2 , ⋯ , v r + ( A ) A A A λ 1 ( A ) , ⋯ , λ r + ( A ) ( A ) \lambda_{1}(A), \cdots, \lambda_{r_{+}(A)}(A) λ 1 ( A ) , ⋯ , λ r + ( A ) ( A ) S 1 ( A ) − Span { v 1 , ⋯ , v r + ( A ) } S_{1}(A) - \operatorname{Span}\{v_{1}, \cdots, v_{r_{+}(A)}\} S 1 ( A ) − Span { v 1 , ⋯ , v r + ( A ) }
S 1 ( A ) S_{1}(A) S 1 ( A ) i + ( A ) i_{+}(A) i + ( A ) x = α 1 v 1 + ⋯ + α i , ( A ) v i , ( A ) ≠ 0 x = \alpha_{1}v_{1} + \dots +\alpha_{i,(A)}v_{i,(A)}\neq 0 x = α 1 v 1 + ⋯ + α i , ( A ) v i , ( A ) = 0 x ∗ A x = λ 1 ( A ) ∣ α 1 ∣ 2 + … x^{*}Ax = \lambda_{1}(A)\mid \alpha_{1}\mid^{2} + \dots x ∗ A x = λ 1 ( A ) ∣ α 1 ∣ 2 + … + λ i , ( A ) ( A ) ∣ α i , ( A ) ∣ 2 > 0. +\lambda_{i,(A)}(A)\mid \alpha_{i,(A)}\mid^{2} > 0. + λ i , ( A ) ( A ) ∣ α i , ( A ) ∣ 2 > 0.
x ∗ SBS ∗ x = ( S ∗ x ) ′ B ( S ∗ x ) > 0 , x ^ {*} \operatorname {S B S} ^ {*} x = (S ^ {*} x) ^ {\prime} B (S ^ {*} x) > 0, x ∗ SBS ∗ x = ( S ∗ x ) ′ B ( S ∗ x ) > 0 , 因而，对于具有维数 i + ( A ) i_{+}(A) i + ( A ) Span { S ∗ v 1 , … , S ∗ v i + ( A ) } \operatorname{Span}\{S^{*}v_{1},\dots ,S^{*}v_{i_{+}(A)}\} Span { S ∗ v 1 , … , S ∗ v i + ( A ) } y y y y ∗ B y > 0 y^{*}By > 0 y ∗ B y > 0 i + ( B ) ⩾ i + ( A ) i_{+}(B)\geqslant i_{+}(A) i + ( B ) ⩾ i + ( A ) A A A B B B i + ( B ) = i . ( A ) i_{+}(B) = i.(A) i + ( B ) = i . ( A )
练习 设 A ∈ M n A \in M_{n} A ∈ M n A ∗ A^{*} A ∗ A A A
练习 设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n A A A B B B
练习 设 A 1 B ∈ M n A_{1}B \in M_{n} A 1 B ∈ M n A A A B B B A A A B B B
练习 在由 n × n n \times n n × n n × n n \times n n × n
由于Sylvester定理保证在相合下Hermite矩阵的诸特征值的符号不变，从而完全解决了在相合下从Hermite矩阵的每个等价类选取一个代表元的问题。但是在'相合下诸特征值的大小如何变化呢？利用Weyl定理(4.3.1)最简单的形式，可以给出Sylvester定理的数量形式。
4.5.9 定理（Ostrowski）设 A , S ∈ M n A, S \in M_n A , S ∈ M n A A A S S S A A A S S ∗ SS^* S S ∗ k = 1 , 2 , … k = 1, 2, \dots k = 1 , 2 , … θ k \theta_k θ k λ i ( S S ∗ ) ⩽ θ k ⩽ λ n ( S S ∗ ) \lambda_i(SS^*) \leqslant \theta_k \leqslant \lambda_n(SS^*) λ i ( S S ∗ ) ⩽ θ k ⩽ λ n ( S S ∗ )
λ k ( S A S ∗ ) = θ k λ k ( A ) . (4.5.10) \lambda_ {k} (S A S ^ {*}) = \theta_ {k} \lambda_ {k} (A). \tag {4.5.10} λ k ( S A S ∗ ) = θ k λ k ( A ) . ( 4.5.10 ) 证明：首先，若 S S ∗ x = λ x SS^*x = \lambda x S S ∗ x = λ x x ≠ 0 x \neq 0 x = 0 λ = x ∗ S S ∗ x / x ∗ x = ( S ∗ x ) ∗ ( S ∗ x ) / x ∗ x > 0 \lambda = x^*SS^*x / x^*x = (S^*x)^*(S^*x) / x^*x > 0 λ = x ∗ S S ∗ x / x ∗ x = ( S ∗ x ) ∗ ( S ∗ x ) / x ∗ x > 0 S S ∗ SS^* S S ∗ k k k 1 ⩽ k ⩽ n 1 \leqslant k \leqslant n 1 ⩽ k ⩽ n A − λ k ( A ) I A - \lambda_k(A)I A − λ k ( A ) I k k k S ( A − λ k ( A ) I ) S ∗ = S A S ∗ − λ k ( A ) S S ∗ S(A - \lambda_k(A)I)S^* = SAS^* - \lambda_k(A)SS^* S ( A − λ k ( A ) I ) S ∗ = S A S ∗ − λ k ( A ) S S ∗ k k k S A S ∗ − λ k ( A ) S S ∗ SAS^* - \lambda_k(A)SS^* S A S ∗ − λ k ( A ) S S ∗ k k k
λ k ( S A S ∗ ) + λ 1 ( − λ k ( A ) S S ∗ ) ⩽ λ k ( S A S ∗ − λ k ( A ) S S ∗ ) = 0 ⩽ λ k ( S A S ′ ) + λ n ( − λ k ( A ) S S ′ ) , \begin{array}{l} \lambda_ {k} \left(S A S ^ {*}\right) + \lambda_ {1} \left(- \lambda_ {k} (A) S S ^ {*}\right) \leqslant \lambda_ {k} \left(S A S ^ {*} - \lambda_ {k} (A) S S ^ {*}\right) = 0 \\ \leqslant \lambda_ {k} (S A S ^ {\prime}) + \lambda_ {n} (- \lambda_ {k} (A) S S ^ {\prime}), \\ \end{array} λ k ( S A S ∗ ) + λ 1 ( − λ k ( A ) S S ∗ ) ⩽ λ k ( S A S ∗ − λ k ( A ) S S ∗ ) = 0 ⩽ λ k ( S A S ′ ) + λ n ( − λ k ( A ) S S ′ ) , 或
λ k ( S A S ∗ ) ⩽ − λ 1 ( − λ k ( A ) S S ∗ ) = λ n ( λ k ( A ) S S ∗ ) = { λ k ( A ) λ n ( S S ∗ ) , 如 果 λ k ( A ) ⩾ 0 , λ k ( A ) λ 1 ( S S ∗ ) , 如 果 λ l ( A ) ⩽ 0 , \begin{array}{l} \lambda_ {k} (S A S ^ {*}) \leqslant - \lambda_ {1} (- \lambda_ {k} (A) S S ^ {*}) = \lambda_ {n} (\lambda_ {k} (A) S S ^ {*}) \\ = \left\{ \begin{array}{l l} \lambda_ {k} (A) \lambda_ {n} \left(S S ^ {*}\right), & \text {如 果} \lambda_ {k} (A) \geqslant 0, \\ \lambda_ {k} (A) \lambda_ {1} \left(S S ^ {*}\right), & \text {如 果} \lambda_ {l} (A) \leqslant 0, \end{array} \right. \\ \end{array} λ k ( S A S ∗ ) ⩽ − λ 1 ( − λ k ( A ) S S ∗ ) = λ n ( λ k ( A ) S S ∗ ) = { λ k ( A ) λ n ( S S ∗ ) , λ k ( A ) λ 1 ( S S ∗ ) , 如 果 λ k ( A ) ⩾ 0 , 如 果 λ l ( A ) ⩽ 0 , H
λ k ( S A S ∗ ) ⩾ λ n ( − λ k ( A ) S S ∗ ) = λ 1 ( λ k ( A ) S S ∗ ) = { λ k ( A ) λ 1 ( S S ∗ ) , 如 果 λ k ( A ) ⩾ 0 , λ k ( A ) λ n ( S S ∗ ) , 如 果 λ k ( A ) ⩽ 0. \begin{array}{l} \lambda_ {k} \left(S A S ^ {*}\right) \geqslant \lambda_ {n} \left(- \lambda_ {k} (A) S S ^ {*}\right) = \lambda_ {1} \left(\lambda_ {k} (A) S S ^ {*}\right) \\ = \left\{ \begin{array}{l l} \lambda_ {k} (A) \lambda_ {1} (\mathrm {S S} ^ {*}), & \text {如 果} \lambda_ {k} (A) \geqslant 0, \\ \lambda_ {k} (A) \lambda_ {n} (\mathrm {S S} ^ {*}), & \text {如 果} \lambda_ {k} (A) \leqslant 0. \end{array} \right. \\ \end{array} λ k ( S A S ∗ ) ⩾ λ n ( − λ k ( A ) S S ∗ ) = λ 1 ( λ k ( A ) S S ∗ ) = { λ k ( A ) λ 1 ( SS ∗ ) , λ k ( A ) λ n ( SS ∗ ) , 如 果 λ k ( A ) ⩾ 0 , 如 果 λ k ( A ) ⩽ 0. 在任何一种情形 [ λ k ( A ) ⩾ 0 [\lambda_k(A) \geqslant 0 [ λ k ( A ) ⩾ 0 λ k ( A ) ⩽ 0 ] \lambda_k(A) \leqslant 0] λ k ( A ) ⩽ 0 ] λ k ( S A S ∗ ) = θ k λ k ( A ) \lambda_k(SAS^*) = \theta_k\lambda_k(A) λ k ( S A S ∗ ) = θ k λ k ( A ) λ 1 ( S S ∗ ) ⩽ θ k ⩽ λ n ( S S ∗ ) \lambda_1(SS^*) \leqslant \theta_k \leqslant \lambda_n(SS^*) λ 1 ( S S ∗ ) ⩽ θ k ⩽ λ n ( S S ∗ ) θ k \theta_k θ k
在 Ostrowki 定理中，如果 A = I ∈ M n A = I \in M_{n} A = I ∈ M n λ k ( A ) = 1 \lambda_{k}(A) = 1 λ k ( A ) = 1 θ k = λ k ( S S ∗ ) \theta_{k} = \lambda_{k}(SS^{*}) θ k = λ k ( S S ∗ ) S ∈ M n S \in M_{n} S ∈ M n λ 1 ( S S ∗ ) = λ n ( S S ∗ ) = 1 \lambda_{1}(SS^{*}) = \lambda_{n}(SS^{*}) = 1 λ 1 ( S S ∗ ) = λ n ( S S ∗ ) = 1 θ k = 1 \theta_{k} = 1 θ k = 1 θ k \theta_{k} θ k A A A S S S
通过简单的连续性论证，Ostrowski定理可以推广到包括 S S S ε > 0 \varepsilon >0 ε > 0 S + ε I S + \varepsilon I S + ε I S S S λ k ( ( S + ε I ) A ( S + ε I ) ∗ ) = θ k λ k ( A ) \lambda_{k}((S + \varepsilon I)A(S + \varepsilon I)^{*}) = \theta_{k}\lambda_{k}(A) λ k (( S + ε I ) A ( S + ε I ) ∗ ) = θ k λ k ( A ) λ 1 ( ( S + ε I ) ( S + ε I ) ∗ ) ⩽ θ k ⩽ λ n ( ( S + ε I ) ( S + ε I ) ∗ ) \lambda_1((S + \varepsilon I)(S + \varepsilon I)^*)\leqslant \theta_k\leqslant \lambda_n((S + \varepsilon I)(S + \varepsilon I)^*) λ 1 (( S + ε I ) ( S + ε I ) ∗ ) ⩽ θ k ⩽ λ n (( S + ε I ) ( S + ε I ) ∗ ) ε → 0 \varepsilon \to 0 ε → 0 0 ⩽ θ k ⩽ λ n ( S S ∗ ) 0\leqslant \theta_{k}\leqslant \lambda_{n}(SS^{*}) 0 ⩽ θ k ⩽ λ n ( S S ∗ )
4.5.11 推论 设 A 1 S ∈ M n A_{1}S \in M_{n} A 1 S ∈ M n A A A A A A S S ∗ SS^{*} S S ∗ k = 1 , 2 , … , n k = 1, 2, \dots, n k = 1 , 2 , … , n θ k \theta_{k} θ k λ 1 ( S S ∗ ) ⩽ θ k ⩽ λ n ( S S ∗ ) \lambda_{1}(SS^{*}) \leqslant \theta_{k} \leqslant \lambda_{n}(SS^{*}) λ 1 ( S S ∗ ) ⩽ θ k ⩽ λ n ( S S ∗ )
λ k ( S A S ∗ ) = θ k λ k ( A ) . \lambda_ {k} (S A S ^ {*}) = \theta_ {k} \lambda_ {k} (A). λ k ( S A S ∗ ) = θ k λ k ( A ) . 特别是 S A S ∗ SAS^{*} S A S ∗ A A A
求复对称矩阵在 7 ^7 7
4.5.12 定理 设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n S ∈ M n S \in M_{n} S ∈ M n A = S B S T A = SBS^{T} A = SB S T A A A B B B
证明：如果 A = S B S T A = SBS^{\mathrm{T}} A = SB S T S S S A A A B B B
A = U 1 Σ 1 U 1 T = U 1 I ( Σ 1 ) D 1 2 U 1 T = ( U 1 D 1 ) I ( Σ 1 ) ( U 1 D 1 ) T , A = U _ {1} \Sigma_ {1} U _ {1} ^ {T} = U _ {1} I (\Sigma_ {1}) D _ {1} ^ {2} U _ {1} ^ {T} = (U _ {1} D _ {1}) I (\Sigma_ {1}) (U _ {1} D _ {1}) ^ {T}, A = U 1 Σ 1 U 1 T = U 1 I ( Σ 1 ) D 1 2 U 1 T = ( U 1 D 1 ) I ( Σ 1 ) ( U 1 D 1 ) T , 其中， I ( Σ 1 ) I(\Sigma_1) I ( Σ 1 ) Σ 1 \Sigma_1 Σ 1 A A A U 1 U_{1} U 1 Σ 1 = d i a g ( σ 1 , \Sigma_{1} = \mathrm{diag}(\sigma_{1}, Σ 1 = diag ( σ 1 , σ 2 , … , σ n ) \sigma_{2},\dots ,\sigma_{n}) σ 2 , … , σ n ) σ i ⩾ 0 \sigma_{i}\geqslant 0 σ i ⩾ 0 D 1 = d i a g ( d 1 , d 2 , … , d n ) D_{1} = \mathrm{diag}(d_{1},d_{2},\dots ,d_{n}) D 1 = diag ( d 1 , d 2 , … , d n )
d t = { σ t , 如 果 σ t > 0 , 1 , 如 果 σ t = 0. d _ {t} = \left\{ \begin{array}{l l} \sqrt {\sigma_ {t}}, & \text {如 果} \sigma_ {t} > 0, \\ 1, & \text {如 果} \sigma_ {t} = 0. \end{array} \right. d t = { σ t , 1 , 如 果 σ t > 0 , 如 果 σ t = 0. 注意， D 1 D_{1} D 1 B = ( U 2 D 2 ) I ( Σ 2 ) ( U 2 D 2 ) T B = (U_2D_2)I(\Sigma_2)(U_2D_2)^T B = ( U 2 D 2 ) I ( Σ 2 ) ( U 2 D 2 ) T rank A = rank B \operatorname{rank} A = \operatorname{rank} B rank A = rank B I ( Σ 1 ) = I ( Σ 2 ) I(\Sigma_{1}) = I(\Sigma_{2}) I ( Σ 1 ) = I ( Σ 2 )
I ( Σ 1 ) = ( U 1 D 1 ) − 1 A [ ( U 1 D 1 ) T ] − 1 = I ( Σ 2 ) = ( U 2 D 2 ) − 1 B [ ( U 2 D 2 ) T ] − 1 , I \left(\Sigma_ {1}\right) = \left(U _ {1} D _ {1}\right) ^ {- 1} A \left[ \left(U _ {1} D _ {1}\right) ^ {T} \right] ^ {- 1} = I \left(\Sigma_ {2}\right) = \left(U _ {2} D _ {2}\right) ^ {- 1} B \left[ \left(U _ {2} D _ {2}\right) ^ {T} \right] ^ {- 1}, I ( Σ 1 ) = ( U 1 D 1 ) − 1 A [ ( U 1 D 1 ) T ] − 1 = I ( Σ 2 ) = ( U 2 D 2 ) − 1 B [ ( U 2 D 2 ) T ] − 1 , 因此
A = ( U 1 D 1 ) ( U 2 D 2 ) − 1 B [ ( U 1 D 1 ) ( U 2 D 2 ) − 1 ] t . A = \left(U _ {1} D _ {1}\right) \left(U _ {2} D _ {2}\right) ^ {- 1} B \left[ \left(U _ {1} D _ {1}\right) \left(U _ {2} D _ {2}\right) ^ {- 1} \right] ^ {t}. A = ( U 1 D 1 ) ( U 2 D 2 ) − 1 B [ ( U 1 D 1 ) ( U 2 D 2 ) − 1 ] t . 由此得出 A A A B ′ B^{\prime} B ′
练习 在相合下， n × n n \times n n × n n × n n \times n n × n
练习 设 A ∈ M n A \in M_{n} A ∈ M n S ∈ M n S \in M_{n} S ∈ M n A = S S T A = SS^{T} A = S S T A A A
练习 设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n X , Y ∈ M n X, Y \in M_{n} X , Y ∈ M n A = X B Y A = XBY A = XB Y A A A B B B S ∈ M n S \in M_{n} S ∈ M n A = S B S T A = SBS^{T} A = SB S T A A A B T B^{T} B T
如果 A = Z B Y A = ZBY A = ZB Y A A A B B B
上述结果相当于关于复矩阵的 7 {}^{7} 7
4.5.13 定理 设 A , S ∈ M n A, S \in M_{n} A , S ∈ M n A − 1 A T A^{-1} A^{T} A − 1 A T A = U Σ U T A = U \Sigma U^{T} A = U Σ U T S A S T = V M V T SAS^{T} = VMV^{T} S A S T = V M V T A A A S A S T SAS^{T} S A S T U U U V V V Σ = diag ( σ 1 , σ 2 , … , σ n ) \Sigma = \operatorname{diag}(\sigma_{1}, \sigma_{2}, \dots, \sigma_{n}) Σ = diag ( σ 1 , σ 2 , … , σ n ) M = diag ( μ 1 , μ 2 , … , μ n ) M = \operatorname{diag}(\mu_{1}, \mu_{2}, \dots, \mu_{n}) M = diag ( μ 1 , μ 2 , … , μ n ) σ i , μ i ⩾ 0 \sigma_{i}, \mu_{i} \geqslant 0 σ i , μ i ⩾ 0 λ i ( S S ∗ ) \lambda_{i}(SS^{*}) λ i ( S S ∗ ) S S ∗ SS^{*} S S ∗ σ i , μ i \sigma_{i}, \mu_{i} σ i , μ i λ i ( S S ∗ ) \lambda_{i}(SS^{*}) λ i ( S S ∗ ) k = 1 , 2 , … , n k = 1, 2, \dots, n k = 1 , 2 , … , n λ 1 ( S S ∗ ) ⩽ θ k ⩽ λ n ( S S ∗ ) \lambda_{1}(SS^{*}) \leqslant \theta_{k} \leqslant \lambda_{n}(SS^{*}) λ 1 ( S S ∗ ) ⩽ θ k ⩽ λ n ( S S ∗ ) θ k \theta_{k} θ k μ k − θ k σ k \mu_{k} - \theta_{k} \sigma_{k} μ k − θ k σ k S S S θ k > 0 \theta_{k} > 0 θ k > 0
证明：数 μ 2 \mu^2 μ 2 B B ∗ BB^{*} B B ∗ B = S A S ′ B = SAS^{\prime} B = S A S ′
μ k 2 = λ k ( B B ′ ) = λ k ( S A S T S ‾ A ˉ S ∗ ) = λ k ( S [ A S T S ˉ A ∣ S ∗ ] − θ k λ k ( A S T S A ˉ ) \mu_ {k} ^ {2} = \lambda_ {k} (B B ^ {\prime}) = \lambda_ {k} (\text {S A S} ^ {T} \overline {{\text {S}}} \bar {A} S ^ {*}) = \lambda_ {k} (\text {S} [ A S ^ {T} \bar {S} A | S ^ {*} ] - \theta_ {k} \lambda_ {k} (A S ^ {T} S \bar {A}) μ k 2 = λ k ( B B ′ ) = λ k ( S A S T S A ˉ S ∗ ) = λ k ( S [ A S T S ˉ A ∣ S ∗ ] − θ k λ k ( A S T S A ˉ ) 对适合 λ 1 ( S S ∗ ) ⩽ θ ^ k ⩽ λ n ( S S ∗ ) \lambda_1(SS^*)\leqslant \hat{\theta}_k\leqslant \lambda_n(SS^*) λ 1 ( S S ∗ ) ⩽ θ ^ k ⩽ λ n ( S S ∗ ) θ ^ k \hat{\theta}_k θ ^ k λ k \lambda_{k} λ k
μ k 2 = θ ^ k λ k ( A S t S ˙ A ˉ ) = θ ^ k λ k ( S A A S T ) − θ ^ k λ k ( S A A ˉ S ⋅ ) . \mu_ {k} ^ {2} = \hat {\theta} _ {k} \lambda_ {k} (A S ^ {t} \dot {S} \bar {A}) = \hat {\theta} _ {k} \lambda_ {k} (S A A S ^ {T}) - \hat {\theta} _ {k} \lambda_ {k} (S A \bar {A} S ^ {\cdot}). μ k 2 = θ ^ k λ k ( A S t S ˙ A ˉ ) = θ ^ k λ k ( S AA S T ) − θ ^ k λ k ( S A A ˉ S ⋅ ) . 再应用(4.5.11)，则对适合 λ 1 ( S S ∗ ) ⩽ θ ^ k ⩽ λ n ( S S ∗ ) \lambda_1(SS^*)\leqslant \hat{\theta}_k\leqslant \lambda_n(SS^*) λ 1 ( S S ∗ ) ⩽ θ ^ k ⩽ λ n ( S S ∗ ) θ ^ k \hat{\theta}_k θ ^ k
μ k ′ = θ ^ k θ ˉ k λ k ( Λ A ˉ ) − θ ^ k θ ˉ k σ k ′ . \mu_ {k} ^ {\prime} = \hat {\theta} _ {k} \bar {\theta} _ {k} \lambda_ {k} (\Lambda \bar {A}) - \hat {\theta} _ {k} \bar {\theta} _ {k} \sigma_ {k} ^ {\prime}. μ k ′ = θ ^ k θ ˉ k λ k ( Λ A ˉ ) − θ ^ k θ ˉ k σ k ′ . 因此， μ k − θ ^ k θ ^ k σ k = θ k σ k \mu_k - \sqrt{\hat{\theta}_k \hat{\theta}_k} \sigma_k = \theta_k \sigma_k μ k − θ ^ k θ ^ k σ k = θ k σ k θ k = θ ^ k θ ˉ k \theta_k = \sqrt{\hat{\theta}_k \bar{\theta}_k} θ k = θ ^ k θ ˉ k
我们从(1.3.19)得知，两个可对角化的矩阵可经同一个相似变换同时对角化，当且仅当它们可交换。关于通过相合同时对角化的相应结果是什么呢？
或许最早是由于研究关于稳定平衡的“最小振动”力学问题。才促使人们去考虑关于通过相合同时对角化的有关结果。如果动力系统的组态由广义(Lagrange)坐标 q 1 , q 2 , ⋯ , q n q_{1}, q_{2}, \cdots, q_{n} q 1 , q 2 , ⋯ , q n V V V q i q_{i} q i
V = ∑ i , j = 1 n a i j q i q j V = \sum_ {i, j = 1} ^ {n} a _ {i j} q _ {i} q _ {j} V = i , j = 1 ∑ n a ij q i q j 来逼近．动能 T T T q i q_{i} q i
T = ∑ i , j = 1 n b i , q ˙ i , q ˙ j T = \sum_ {i, j = 1} ^ {n} b _ {i}, \dot {q} _ {i}, \dot {q} _ {j} T = i , j = 1 ∑ n b i , q ˙ i , q ˙ j 来逼近．系统的变化过程由Lagrange方程组
d d t ( ∂ T ∂ q ˙ i ) − ∂ T ∂ q i + ∂ V ∂ q i = 0 \frac {\mathrm {d}}{\mathrm {d} t} \left(\frac {\partial T}{\partial \dot {q} _ {i}}\right) - \frac {\partial T}{\partial q _ {i}} + \frac {\partial V}{\partial q _ {i}} = 0 d t d ( ∂ q ˙ i ∂ T ) − ∂ q i ∂ T + ∂ q i ∂ V = 0 所决定，它是常系数二阶线性常微分方程组，如果两个二次型 T T T V V V A = [ a i j ] A = [a_{ij}] A = [ a ij ] B = [ b i j ] B = [b_{ij}] B = [ b ij ]
如果可以求得非奇异变换 S = [ s n ] ∈ M n S = [s_n] \in M_n S = [ s n ] ∈ M n S A S T SAS^T S A S T S B S T SBS^T SB S T
q i = ∑ j = 1 n s i j p j (4.5.14) q _ {i} = \sum_ {j = 1} ^ {n} s _ {i j} p _ {j} \tag {4.5.14} q i = j = 1 ∑ n s ij p j ( 4.5.14 ) 的新广义坐标 p i p_i p i T \pmb{T} T V \pmb{V} V n n n
难解出这些方程，而原问题的解可利用(4.5.14)求得
因此，如果可以通过相合同时对角化两个实对称矩阵，则一类重要的力学问题的实质性简化是可以实现的。根据物理知识，动能二次型是正定的，结果证明，这是可通过相合同时对角化的充分（而不是必要的）条件。
我们可能要考虑的同时对角化结果有多种形式。可能有两个Hermite矩阵 A A A B B B U U U U A U ′ UAU^{\prime} U A U ′ U B U ′ UBU^{\prime} U B U ′ S S S S A S ′ SAS^{\prime} S A S ′ S B S ′ SBS^{\prime} SB S ′ A A A B B B U A U ′ UAU^{\prime} U A U ′ U B U ′ UBU^{\prime} U B U ′ S A S ′ SAS^{\prime} S A S ′ S B S ′ SBS^{\prime} SB S ′ A A A B B B U A U ′ UAU^{\prime} U A U ′ U B U ′ UBU^{\prime} U B U ′ S A S ′ SAS^{\prime} S A S ′ S B S ′ SBS^{\prime} SB S ′
4.5.15 定理 设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n U U U S S S U , S ∈ M n U, S \in M_{n} U , S ∈ M n
表4.5.15f
证明：在六组条件的每一组中，所述大部分条件的等价性只是属于定义的推论。[a]组中(3)和(4)的等价性可以从以下论断推出：如果 A A A B B B A B AB A B
矩阵当且仅当 A A A B B B A A A A − 1 A^{-1} A − 1 B B B B − 1 B^{-1} B − 1 A A A A T = A ‾ A^T = \overline{A} A T = A
在六组条件的每一组中，条件(1)的必要性可直接从相应的相合具有对角形式的假设推出。例如，在情形Ⅱ（b），如果 S A S T = Λ SAS^T = \Lambda S A S T = Λ S B S T = M SBS^T = M SB S T = M
A − 1 B = ( S T Λ − 1 S ) [ S − 1 M ( S T ) − 1 ] = S T ( Λ − 1 M ) ( S T ) − 1 , A ^ {- 1} B = \left(S ^ {T} \Lambda^ {- 1} S\right) \left[ S ^ {- 1} M \left(S ^ {T}\right) ^ {- 1} \right] = S ^ {T} \left(\Lambda^ {- 1} M\right) \left(S ^ {T}\right) ^ {- 1}, A − 1 B = ( S T Λ − 1 S ) [ S − 1 M ( S T ) − 1 ] = S T ( Λ − 1 M ) ( S T ) − 1 , 因而 R = S T R = S^T R = S T C = A − 1 B C = A^{-1}B C = A − 1 B R = S ∗ R = S^{*} R = S ∗ S S S R R R
考虑情形I，其中 A A A B B B A A A R = [ r 1 r 2 … r n ] ∈ M n R = [r_1r_2\dots r_n]\in M_n R = [ r 1 r 2 … r n ] ∈ M n r i ∈ C n r_i\in \mathbf{C}^n r i ∈ C n Λ = d i a g ( λ 1 , λ 2 , … , λ n ) \Lambda = \mathrm{diag}(\lambda_1,\lambda_2,\dots ,\lambda_n) Λ = diag ( λ 1 , λ 2 , … , λ n ) λ i \lambda_{i} λ i R − 1 A − 1 B R = Λ R^{-1}A^{-1}BR = \Lambda R − 1 A − 1 BR = Λ B R = A R Λ BR = AR\Lambda BR = A R Λ R ∗ B R = R ∗ A R Λ R^{*}BR = R^{*}AR\Lambda R ∗ BR = R ∗ A R Λ λ i \lambda_{i} λ i Λ \pmb{\Lambda} Λ
Λ = [ Λ 1 0 Λ 2 0 ⋱ Λ k ] , (4.5.16) \Lambda = \left[ \begin{array}{c c c c} \Lambda_ {1} & & & 0 \\ & \Lambda_ {2} & & \\ 0 & & \ddots & \\ & & & \Lambda_ {k} \end{array} \right], \tag {4.5.16} Λ = Λ 1 0 Λ 2 ⋱ 0 Λ k , ( 4.5.16 ) Λ i ∈ M n ; 1 ⩽ n i ⩽ n ; Λ i = μ i I , i = 1 , 2 , … , k , \Lambda_ {i} \in M _ {n}; \quad 1 \leqslant n _ {i} \leqslant n; \quad \Lambda_ {i} = \mu_ {i} I, \quad i = 1, 2, \dots , k, Λ i ∈ M n ; 1 ⩽ n i ⩽ n ; Λ i = μ i I , i = 1 , 2 , … , k , 其中，所有 μ i \mu_{i} μ i i ≠ j i\neq j i = j μ i ≠ μ j \mu_{i}\neq \mu_{j} μ i = μ j λ i \lambda_{i} λ i 1 ⩽ i 1\leqslant i 1 ⩽ i j ⩽ n j\leqslant n j ⩽ n i i i j j j λ i ≠ λ j \lambda_{i}\neq \lambda_{j} λ i = λ j R ∗ B R = R ∗ A R Λ R^{*}BR = R^{*}AR\Lambda R ∗ BR = R ∗ A R Λ i i i j j j
r i ∗ A r j λ j = r i ∗ B r j = r j ∗ B r i ‾ = r j ∗ A r i λ i ‾ = r i ∗ A r j λ i , r _ {i} ^ {*} A r _ {j} \lambda_ {j} = r _ {i} ^ {*} B r _ {j} = \overline {{r _ {j} ^ {*} B r _ {i}}} = \overline {{r _ {j} ^ {*} A r _ {i} \lambda_ {i}}} = r _ {i} ^ {*} A r _ {j} \lambda_ {i}, r i ∗ A r j λ j = r i ∗ B r j = r j ∗ B r i = r j ∗ A r i λ i = r i ∗ A r j λ i , 这里，用到了 A A A B B B x , y ∈ C n x, y \in \mathbb{C}^n x , y ∈ C n x ∗ A y = y ∗ A x ‾ x^{*}Ay = \overline{y^{*}Ax} x ∗ A y = y ∗ A x λ i \lambda_{i} λ i λ j \lambda_{j} λ j λ i ≠ λ j \lambda_{i} \neq \lambda_{j} λ i = λ j r i ∗ A r j = 0 r_{i}^{*}Ar_{j} = 0 r i ∗ A r j = 0 r j ∗ A r i = r i ∗ B r j = r j ∗ B r i = 0 r_{j}^{*}Ar_{i} = r_{i}^{*}Br_{j} = r_{j}^{*}Br_{i} = 0 r j ∗ A r i = r i ∗ B r j = r j ∗ B r i = 0 R ∗ B R R^{*}BR R ∗ BR R ∗ A R R^{*}AR R ∗ A R
R ′ B R = [ B 1 0 B 2 0 ⋱ B k ] = R ′ A R A = [ μ 1 A 1 0 μ 2 A 2 0 ⋱ μ k A k ] , \begin{array}{l} R ^ {\prime} B R = \left[ \begin{array}{c c c c} B _ {1} & & & 0 \\ & B _ {2} & & \\ 0 & & \ddots & \\ & & & B _ {k} \end{array} \right] = R ^ {\prime} A R A \\ = \left[ \begin{array}{c c c c} \mu_ {1} A _ {1} & & & 0 \\ & \mu_ {2} A _ {2} & & \\ 0 & & \ddots & \\ & & & \mu_ {k} A _ {k} \end{array} \right], \\ \end{array} R ′ BR = B 1 0 B 2 ⋱ 0 B k = R ′ A R A = μ 1 A 1 0 μ 2 A 2 ⋱ 0 μ k A k , 其中，对 i = 1 , 2 , … , k i = 1,2,\dots ,k i = 1 , 2 , … , k B i B_{i} B i A i ∈ M n i A_{i}\in M_{n_{i}} A i ∈ M n i k = n k = n k = n λ i \lambda_{i} λ i k < n k < n k < n n i > 1 n_i > 1 n i > 1 B i = μ i A i B_{i} = \mu_{i}A_{i} B i = μ i A i A i A_{i} A i B i B_{i} B i A i = U i D i U i ∗ A_{i} = U_{i}D_{i}U_{i}^{*} A i = U i D i U i ∗ U i U_{i} U i D i ∈ M n i D_{i}\in M_{n_{i}} D i ∈ M n i U i U_{i} U i D i D_{i} D i B i = μ i A i = U i ( μ i D i ) U i ∗ B_{i} = \mu_{i}A_{i} = U_{i}(\mu_{i}D_{i})U_{i}^{*} B i = μ i A i = U i ( μ i D i ) U i ∗
U = [ U 1 0 U 2 0 ⋱ U k ] , D = [ D 1 0 D 2 0 ⋱ D k ] , U = \left[ \begin{array}{c c c c} U _ {1} & & & 0 \\ & U _ {2} & & \\ 0 & & \ddots & \\ & & & U _ {k} \end{array} \right], D = \left[ \begin{array}{c c c c} D _ {1} & & & 0 \\ & D _ {2} & & \\ 0 & & \ddots & \\ & & & D _ {k} \end{array} \right], U = U 1 0 U 2 ⋱ 0 U k , D = D 1 0 D 2 ⋱ 0 D k , 且当 n r = 1 n_r = 1 n r = 1 U r = [ 1 ] U_r = [1] U r = [ 1 ] U U U D D D
R ′ B R = U ( D Λ ) U ′ , R ′ A R = U D U ′ . R ^ {\prime} B R = U (D \Lambda) U ^ {\prime}, R ^ {\prime} A R = U D U ^ {\prime}. R ′ BR = U ( D Λ ) U ′ , R ′ A R = U D U ′ . 最后，欲求的表示式是
A = [ ( R − 1 ) ∗ U ] D [ ( R − 1 ) ∗ U ] ∗ 和 B = [ ( R − 1 ) ∗ U ] ( D A ) [ ( R − 1 ) ∗ U ] ∗ . A = \left[ \left(R ^ {- 1}\right) ^ {*} U \right] D \left[ \left(R ^ {- 1}\right) ^ {*} U \right] ^ {*} \text {和} B = \left[ \left(R ^ {- 1}\right) ^ {*} U \right] (D A) \left[ \left(R ^ {- 1}\right) ^ {*} U \right] ^ {*}. A = [ ( R − 1 ) ∗ U ] D [ ( R − 1 ) ∗ U ] ∗ 和 B = [ ( R − 1 ) ∗ U ] ( D A ) [ ( R − 1 ) ∗ U ] ∗ . 注意，如果假定I(a)(1)成立，证法是相同的，只是我们还知道 R R R ( R − 1 ) ∗ U = R U (R^{-1})^{*}U = RU ( R − 1 ) ∗ U = R U
余下的四种情形中所要作的证明是类似的。利用相应的假设得到相合矩阵，它们是分块对角矩阵，然后利用关于Hermite矩阵的谱定理或关于对称矩阵的Takagi分解(4.4.4)便完成了到对角形式的化简。
考虑情形Ⅱ，其中， A A A B B B A A A I ( b ) ( l ) \mathbf{I}(\mathbf{b})(\mathbf{l}) I ( b ) ( l ) R − [ r 1 r 2 … r n ] ∈ M n R - [r_1r_2\dots r_n]\in M_n R − [ r 1 r 2 … r n ] ∈ M n r i ∈ C n r_i\in \mathbb{C}^n r i ∈ C n Λ = d i a g ( λ 1 , \Lambda = \mathrm{diag}(\lambda_1, Λ = diag ( λ 1 , λ 2 , … , λ n ) \lambda_{2},\dots ,\lambda_{n}) λ 2 , … , λ n ) R − 1 A − 1 B R = A R^{-1}A^{-1}BR = A R − 1 A − 1 BR = A B R = A R Λ BR = AR\Lambda BR = A R Λ R T B R = R T A R Λ R^{T}BR = R^{T}AR\Lambda R T BR = R T A R Λ λ i \lambda_{i} λ i Λ \pmb{\Lambda} Λ μ i \mu_{i} μ i λ i \lambda_{i} λ i 1 ⩽ i 1\leqslant i 1 ⩽ i j ⩽ n j\leqslant n j ⩽ n i i i j j j λ i ≠ λ j \lambda_{i}\neq \lambda_{j} λ i = λ j R T B R = R T A R Λ R^{T}BR = R^{T}AR\Lambda R T BR = R T A R Λ i , j i,j i , j
r i T A r j λ j = r i T B r j = r j T B r i = r j T A r i λ i = r i T A r j λ j r _ {i} ^ {T} A r _ {j} \lambda_ {j} = r _ {i} ^ {T} B r _ {j} = r _ {j} ^ {T} B r _ {i} = r _ {j} ^ {T} A r _ {i} \lambda_ {i} = r _ {i} ^ {T} A r _ {j} \lambda_ {j} r i T A r j λ j = r i T B r j = r j T B r i = r j T A r i λ i = r i T A r j λ j 这里，用到了 A A A B B B x , y ∈ C n x, y \in \mathbb{C}^n x , y ∈ C n x T A y = y T A x x^T Ay = y^T Ax x T A y = y T A x λ i ≠ λ j \lambda_i \neq \lambda_j λ i = λ j r i T A r j = 0 r_i^T Ar_j = 0 r i T A r j = 0 r j T A r i = r i T B r j − r j T B r i = 0 r_j^T A r_i = r_i^T Br_j - r_j^T Br_i = 0 r j T A r i = r i T B r j − r j T B r i = 0 R T B R R^T BR R T BR R T A R R^T AR R T A R
R T B R = [ B 1 0 B 2 0 ⋱ B k ] = R T A R Λ = [ μ 1 A 1 0 μ 2 A 2 0 ⋱ μ k A k ] , \begin{array}{l} R ^ {T} B R = \left[ \begin{array}{c c c c} B _ {1} & & & 0 \\ & B _ {2} & & \\ 0 & & \ddots & \\ & & & B _ {k} \end{array} \right] = R ^ {T} A R \Lambda \\ = \left[ \begin{array}{c c c c} \mu_ {1} A _ {1} & & & 0 \\ & \mu_ {2} A _ {2} & & \\ 0 & & \ddots & \\ & & & \mu_ {k} A _ {k} \end{array} \right], \\ \end{array} R T BR = B 1 0 B 2 ⋱ 0 B k = R T A R Λ = μ 1 A 1 0 μ 2 A 2 ⋱ 0 μ k A k , 其中 B i , A i ∈ M n i B_{i}, A_{i} \in M_{n_{i}} B i , A i ∈ M n i k = n k = n k = n k < n k < n k < n n i > 1 n_{i} > 1 n i > 1 B i = μ i A i B_{i} = \mu_{i}A_{i} B i = μ i A i A i A_{i} A i B i B_{i} B i A i = U i Σ i U i T A_{i} = U_{i}\Sigma_{i}U_{i}^{T} A i = U i Σ i U i T U i , Σ i ∈ M n i , U i U_{i}, \Sigma_{i} \in M_{n_{i}}, U_{i} U i , Σ i ∈ M n i , U i Σ i \Sigma_{i} Σ i B i = μ i A i = U i ( μ i Σ i ) B_{i} = \mu_{i}A_{i} = U_{i}(\mu_{i}\Sigma_{i}) B i = μ i A i = U i ( μ i Σ i ) U i T U_{i}^{T} U i T
230
U = [ U 1 0 U 2 0 ⋱ U k ] , Σ = [ Σ 1 0 Σ 2 0 ⋱ Σ k ] , U = \left[ \begin{array}{c c c c} U _ {1} & & & 0 \\ & U _ {2} & & \\ 0 & & \ddots & \\ & & & U _ {k} \end{array} \right], \quad \Sigma = \left[ \begin{array}{c c c c} \Sigma_ {1} & & & 0 \\ & \Sigma_ {2} & & \\ 0 & & \ddots & \\ & & & \Sigma_ {k} \end{array} \right], U = U 1 0 U 2 ⋱ 0 U k , Σ = Σ 1 0 Σ 2 ⋱ 0 Σ k , 且当 n r = 1 n_r = 1 n r = 1 U r = [ 1 ] U_r = [1] U r = [ 1 ] U U U Σ \Sigma Σ
R ⊺ B R = U ( Σ Λ ) U ⊺ 和 R ⊺ A R = U Σ U ⊺ . R ^ {\intercal} B R = U (\Sigma \Lambda) U ^ {\intercal} \text {和} R ^ {\intercal} A R = U \Sigma U ^ {\intercal}. R ⊺ BR = U ( ΣΛ ) U ⊺ 和 R ⊺ A R = U Σ U ⊺ . 最后，欲求的表示式是
A = [ ( R − 1 ) T U ] Σ [ ( R − 1 ) T U ] T 和 B = [ ( R − 1 ) T U ] Σ A [ ( R − 1 ) T U ] T . A = \left[ (R ^ {- 1}) ^ {T} U \right] \Sigma \left[ (R ^ {- 1}) ^ {T} U \right] ^ {T} \text {和} B = \left[ (R ^ {- 1}) ^ {T} U \right] \Sigma A \left[ (R ^ {- 1}) ^ {T} U \right] ^ {T}. A = [ ( R − 1 ) T U ] Σ [ ( R − 1 ) T U ] T 和 B = [ ( R − 1 ) T U ] Σ A [ ( R − 1 ) T U ] T . 如果假定 I ( a ) ( 1 ) \mathbb{I}(\mathbf{a})(1) I ( a ) ( 1 ) R R R ( R − 1 ) T U = R U (R^{-1})^T U = R U ( R − 1 ) T U = R U I ( a ) ( 1 ) \mathbb{I}(\mathbf{a})(1) I ( a ) ( 1 )
在情形Ⅲ，证明需作一点修改。设 A A A B ∈ M n B \in M_n B ∈ M n A A A B B B R = [ r 1 r 2 ⋯ r n ] ∈ M n R = [r_1 r_2 \cdots r_n] \in M_n R = [ r 1 r 2 ⋯ r n ] ∈ M n Λ = d i a g ( λ 1 , λ 2 , ⋯ , λ n ) \Lambda = \mathrm{diag}(\lambda_1, \lambda_2, \cdots, \lambda_n) Λ = diag ( λ 1 , λ 2 , ⋯ , λ n ) R − 1 A − 1 B R = Λ R^{-1} A^{-1} B R = \Lambda R − 1 A − 1 BR = Λ B R = A R Λ B R = A R \Lambda BR = A R Λ R ∗ B R = R ˉ T B R ˉ = R ∗ A R Λ R^* B R = \bar{R}^T B \bar{R} = R^* A R \Lambda R ∗ BR = R ˉ T B R ˉ = R ∗ A R Λ λ 1 \lambda_1 λ 1 Λ \Lambda Λ
Λ = [ Λ 1 0 Λ 2 0 ⋱ Λ k ] , \Lambda = \left[ \begin{array}{c c c c} \Lambda_ {1} & & & 0 \\ & \Lambda_ {2} & & \\ 0 & & \ddots & \\ & & & \Lambda_ {k} \end{array} \right], Λ = Λ 1 0 Λ 2 ⋱ 0 Λ k , 其 中 , Λ i = [ μ i ( 1 ) 0 μ i ( 2 ) 0 ⋱ μ i ( n i ) ] , i = 1 , … , k , \text {其 中}, \Lambda_ {i} = \left[ \begin{array}{c c c c} {\mu_ {i} ^ {(1)}} & & & {0} \\ & {\mu_ {i} ^ {(2)}} & & \\ {0} & & {\ddots} & \\ & & & {\mu_ {i} ^ {(n _ {i})}} \end{array} \right], \quad i = 1, \dots , k, 其 中 , Λ i = μ i ( 1 ) 0 μ i ( 2 ) ⋱ 0 μ i ( n i ) , i = 1 , … , k , 并且对 j , k = 1 , 2 , … , n j, k = 1, 2, \dots, n j , k = 1 , 2 , … , n ∣ μ i ( j ) ∣ = ∣ μ i ( k ) ∣ |\mu_i^{(j)}| = |\mu_i^{(k)}| ∣ μ i ( j ) ∣ = ∣ μ i ( k ) ∣ i ≠ j i \neq j i = j ∣ μ i ( p ) ∣ ≠ ∣ μ j ( q ) ∣ |\mu_i^{(p)}| \neq |\mu_j^{(q)}| ∣ μ i ( p ) ∣ = ∣ μ j ( q ) ∣ λ i \lambda_i λ i 1 ≤ i , j ≤ n 1 \leq i, j \leq n 1 ≤ i , j ≤ n ∣ λ i ∣ ≠ ∣ λ j ∣ |\lambda_i| \neq |\lambda_j| ∣ λ i ∣ = ∣ λ j ∣ i , j i, j i , j R T B B ‾ = R ∗ A R A R^T B \overline{B} = R^* A R A R T B B = R ∗ A R A i , j i, j i , j
r 1 ∗ A r 1 λ 1 = r ˉ 1 T B r ˉ 1 − : r ˉ 1 2 B r ˉ 1 = r 1 ∗ A r 1 λ 1 = r ˉ 1 ∗ A ‾ r ˉ 1 , r _ {1} ^ {*} A r _ {1} \lambda_ {1} = \bar {r} _ {1} ^ {T} B \bar {r} _ {1} -: \bar {r} _ {1} ^ {2} B \bar {r} _ {1} = r _ {1} ^ {*} A r _ {1} \lambda_ {1} = \bar {r} _ {1} ^ {*} \overline {{A}} \bar {r} _ {1}, r 1 ∗ A r 1 λ 1 = r ˉ 1 T B r ˉ 1 − : r ˉ 1 2 B r ˉ 1 = r 1 ∗ A r 1 λ 1 = r ˉ 1 ∗ A r ˉ 1 , 这里，用到了 A A A B B B ∣ r i ∗ A r j ∣ ∣ λ i ∣ = ∣ r i ∗ A r j ∣ ∣ λ i ∣ \left|r_{i}^{*} A r_{j}\right| \left|\lambda_{i}\right| = \left|r_{i}^{*} A r_{j}\right| \left|\lambda_{i}\right| ∣ r i ∗ A r j ∣ ∣ λ i ∣ = ∣ r i ∗ A r j ∣ ∣ λ i ∣ ∣ λ i ∣ ≠ ∣ λ j ∣ \left|\lambda_{i}\right| \neq \left|\lambda_{j}\right| ∣ λ i ∣ = ∣ λ j ∣ r i ∗ A r j = 0 r_{i}^{*} A r_{j} = 0 r i ∗ A r j = 0 r j ∗ A r i = r i T ‾ B r j ‾ = r ˙ j T B r i = 0 r_{j}^{*} A r_{i} = \overline{r_{i}^{T}} B \overline{r_{j}} = \dot{r}_{j}^{T} B r_{i} = 0 r j ∗ A r i = r i T B r j = r ˙ j T B r i = 0 R ˉ T B R ˉ \bar{R}^{T} B \bar{R} R ˉ T B R ˉ R ∗ A R R^{*} A R R ∗ A R
R ˉ T B R ˉ = [ B 1 0 B 2 0 ⋱ B k ] = R ∗ A R Λ , \bar {R} ^ {T} B \bar {R} = \left[ \begin{array}{c c c c} B _ {1} & & & 0 \\ & B _ {2} & & \\ 0 & & \ddots & \\ & & & B _ {k} \end{array} \right] = R ^ {*} A R \Lambda , R ˉ T B R ˉ = B 1 0 B 2 ⋱ 0 B k = R ∗ A R Λ , [ A 1 Λ 1 0 Λ 2 Λ 2 0 ⋱ A k Λ k ] , \left[ \begin{array}{c c c c} A _ {1} \Lambda_ {1} & & & 0 \\ & \Lambda_ {2} \Lambda_ {2} & & \\ 0 & & \ddots & \\ & & & A _ {k} \Lambda_ {k} \end{array} \right], A 1 Λ 1 0 Λ 2 Λ 2 ⋱ 0 A k Λ k , 其中，所有 B t , A t , Λ t ∈ M n B_{t}, A_{t}, \Lambda_{t} \in M_{n} B t , A t , Λ t ∈ M n Λ t = σ t D t 2 , σ t ⩾ 0 \Lambda_{t} = \sigma_{t}D_{t}^{2}, \sigma_{t} \geqslant 0 Λ t = σ t D t 2 , σ t ⩾ 0
D j − diag ( e θ 1 , e i θ 2 , … , e i θ n ) , D _ {j} - \operatorname {d i a g} \left(e ^ {\theta_ {1}}, e ^ {i \theta_ {2}}, \dots , e ^ {i \theta_ {n}}\right), D j − diag ( e θ 1 , e i θ 2 , … , e i θ n ) , 所有 θ i j ∈ R \theta_{ij} \in \mathbb{R} θ ij ∈ R k = n k = n k = n k < n k < n k < n n i > 1 n_i > 1 n i > 1 B i = A i A i = σ i A i D i 2 B_i = A_i A_i = \sigma_i A_i D_i^2 B i = A i A i = σ i A i D i 2 D i D_i D i D i ′ = D ˉ i = D ˉ i T = D i D_i^{\prime} = \bar{D}_i = \bar{D}_i^T = D_i D i ′ = D ˉ i = D ˉ i T = D i
D ˉ t T B t D ˉ t = σ t D t ∗ A t D t . (4.5.17) \bar {D} _ {t} ^ {T} B _ {t} \bar {D} _ {t} = \sigma_ {t} D _ {t} ^ {*} A _ {t} D _ {t}. \tag {4.5.17} D ˉ t T B t D ˉ t = σ t D t ∗ A t D t . ( 4.5.17 ) 这个恒等式左边是对称矩阵 D ˉ i T B i D i \bar{D}_i^T B_i D_i D ˉ i T B i D i σ i D i ∗ A i D i \sigma_i D_i^* A_i D_i σ i D i ∗ A i D i σ i \sigma_i σ i σ i ≠ 0 \sigma_i \neq 0 σ i = 0 D i ∗ A i D i D_i^* A_i D_i D i ∗ A i D i σ i ≠ 0 \sigma_i \neq 0 σ i = 0 D i ∗ A i D i D_i^* A_i D_i D i ∗ A i D i σ i = 0 \sigma_i = 0 σ i = 0 i i i D i ∗ A i D i D_i^* A_i D_i D i ∗ A i D i i − 1 , ⋯ , k i - 1, \cdots, k i − 1 , ⋯ , k U i ∈ M n i U_i \in M_{n_i} U i ∈ M n i M i M_i M i D i ∗ A i D i = U i M i U i ∗ D_i^* A_i D_i = U_i M_i U_i^* D i ∗ A i D i = U i M i U i ∗ σ i ≠ 0 \sigma_i \neq 0 σ i = 0 U i U_i U i U i T = U i ∗ U_i^T = U_i^* U i T = U i ∗
D ˉ i 1 B i D ˉ i = σ i D i ′ A i D i = U i ( σ i M i ) U i T . \bar {D} _ {i} ^ {1} B _ {i} \bar {D} _ {i} = \sigma_ {i} D _ {i} ^ {\prime} A _ {i} D _ {i} = U _ {i} (\sigma_ {i} M _ {i}) U _ {i} ^ {T}. D ˉ i 1 B i D ˉ i = σ i D i ′ A i D i = U i ( σ i M i ) U i T . 如果 σ i = 0 \sigma_{i} = 0 σ i = 0 U i + − U i T U_{i}^{+} - U_{i}^{T} U i + − U i T i − 1 , 2 , … , k i - 1,2,\dots ,k i − 1 , 2 , … , k
A t − ( D t U t ) M t ( D t U t ) ∗ 和 B t = ( D t U t ) ( σ t M t ) ( D t U t ) 7 . A _ {t} - (D _ {t} U _ {t}) M _ {t} (D _ {t} U _ {t}) ^ {*} \text {和} B _ {t} = (D _ {t} U _ {t}) (\sigma_ {t} M _ {t}) (D _ {t} U _ {t}) ^ {7}. A t − ( D t U t ) M t ( D t U t ) ∗ 和 B t = ( D t U t ) ( σ t M t ) ( D t U t ) 7 . 如果令
U = [ D 1 U 1 0 D 2 U 2 0 ⋱ D k U k ] , U = \left[ \begin{array}{c c c c} D _ {1} U _ {1} & & & 0 \\ & D _ {2} U _ {2} & & \\ 0 & & \ddots & \\ & & & D _ {k} U _ {k} \end{array} \right], U = D 1 U 1 0 D 2 U 2 ⋱ 0 D k U k , M = [ M 1 0 M 2 0 ⋱ M k ] , Σ = [ σ 1 I 0 σ 2 I 0 ⋱ σ k I ] . M = \left[ \begin{array}{c c c c} M _ {1} & & & 0 \\ & M _ {2} & & \\ 0 & & \ddots & \\ & & & M _ {k} \end{array} \right], \quad \Sigma = \left[ \begin{array}{c c c c} \sigma_ {1} I & & & 0 \\ & \sigma_ {2} I & & \\ 0 & & \ddots & \\ & & & \sigma_ {k} I \end{array} \right]. M = M 1 0 M 2 ⋱ 0 M k , Σ = σ 1 I 0 σ 2 I ⋱ 0 σ k I . 则 B = [ ( R ˉ − 1 ) T U ] Σ M [ U T R ˉ − 1 ] B = \left[(\bar{R}^{-1})^{T}U\right]\Sigma M\left[U^{T}\bar{R}^{-1}\right] B = [ ( R ˉ − 1 ) T U ] Σ M [ U T R ˉ − 1 ] A = [ ( R − 1 ) ∗ U ] M [ U ∗ R ] A = \left[(R^{-1})^{*}U\right]M\left[U^{*}R\right] A = [ ( R − 1 ) ∗ U ] M [ U ∗ R ] R R R ( R − 1 ) ∗ U = R U (R^{-1})^{*}U = RU ( R − 1 ) ∗ U = R U ( R ˉ − 1 ) T U = R U (\bar{R}^{-1})^{T}U = RU ( R ˉ − 1 ) T U = R U
当 A A A B B B R ∈ M n R \in M_{n} R ∈ M n R − 1 B − 1 A R ˉ = A R^{-1} B^{-1} A \bar{R} = A R − 1 B − 1 A R ˉ = A A R = B R A A R = B R A A R = BR A R ˉ ⋅ A R ˉ = R ′ B R A \bar{R} \cdot A \bar{R} = R^{\prime} B R A R ˉ ⋅ A R ˉ = R ′ BR A A A A A A A B B B
位，且用 Takagi 分解(4.4.4)对角化 D i T B i D D_{i}^{T}B_{i}D D i T B i D
在定理(4.5.15)(表1.5.15T)的情形I和Ⅱ，有一个关于 A − 1 B A^{-1}B A − 1 B A A A B B B A − 1 B A^{-1}B A − 1 B A − 1 B A^{-1}B A − 1 B R A R − 1 \mathsf{RAR}^{-1} RAR − 1 Λ \pmb{\Lambda} Λ Λ \pmb{\Lambda} Λ A − 1 B A^{-1}B A − 1 B A − 1 B A^{-1}B A − 1 B R A R − 1 \mathsf{RAR}^{-1} RAR − 1 Λ \pmb{\Lambda} Λ A − 1 B A^{-1}B A − 1 B C C \mathbb{C}\mathbb{C} CC C C ˉ \mathbb{C}\bar{\mathbb{C}} C C ˉ rank C = rank C C ˉ \operatorname{rank} C = \operatorname{rank} \mathbb{C}\bar{\mathbb{C}} rank C = rank C C ˉ
在定理(4.5.15)中作非奇异性假定是方便的，但是在酉相合的情形I(a)，II(a)和Ⅲ(a)中，这个假定可以取消，在情形I(a)，这种计算方法给出了关于可交换的Hermite矩阵可同时酉对角化的经典结果(4.4.6)的又一个证明。
4.5.18 推论 设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n (a) 如果 A A A B B B U ∈ M n U \in M_{n} U ∈ M n U A U ∗ U A U^{*} U A U ∗ U B U ∗ U B U^{*} U B U ∗ A B AB A B A B = B A AB = BA A B = B A A A A B B B U ∈ M n U \in M_{n} U ∈ M n U A U T U A U^{T} U A U T U B U T U B U^{T} U B U T A B ‾ A\overline{B} A B A B ‾ B A = B A ‾ A B ‾ A\overline{B}BA = B\overline{A}A\overline{B} A B B A = B A A B A A A B B B U ∈ M n U \in M_{n} U ∈ M n U A U ∗ UAU^{*} U A U ∗ U B U r UBU^{r} U B U r A B AB A B A B = B A AB = BA A B = B A
证明：（a）如果 U A U ∗ − A U A U^{*} - A U A U ∗ − A U B U ∗ = M U B U^{*} = M U B U ∗ = M A = U ∗ A U , B = U ∗ M U , A = U^{*} A U, B = U^{*} M U, A = U ∗ A U , B = U ∗ M U , A B = U ∗ A U U ∗ M U − U ∗ A M U − U ∗ M A U = U ∗ M U U ∗ A U = B A . AB = U^{*} A U U^{*} M U - U^{*} A M U - U^{*} M A U = U^{*} M U U^{*} A U = B A. A B = U ∗ A U U ∗ M U − U ∗ A M U − U ∗ M A U = U ∗ M U U ∗ A U = B A . A B = B A AB = BA A B = B A ε > 0 \varepsilon > 0 ε > 0 A ε = A + ε I A_{\varepsilon} = A + \varepsilon I A ε = A + ε I A ε B = ( A + ε I ) B = A B + ε B = B A + ε B = B ( A + ε I ) = B A ε A_{\varepsilon} B = (A + \varepsilon I) B = AB + \varepsilon B = BA + \varepsilon B = B(A + \varepsilon I) = BA_{\varepsilon} A ε B = ( A + ε I ) B = A B + εB = B A + εB = B ( A + ε I ) = B A ε B B B A ε A_{\varepsilon} A ε A ε A_{\varepsilon} A ε A ε B A_{\varepsilon} B A ε B U ε U_{\varepsilon} U ε U ε A ε U ε ∗ − U ε A U ε ∗ + ε I = A ε U_{\varepsilon} A_{\varepsilon} U_{\varepsilon}^{*} - U_{\varepsilon} A U_{\varepsilon}^{*} + \varepsilon I = A_{\varepsilon} U ε A ε U ε ∗ − U ε A U ε ∗ + ε I = A ε U ε B U ε ∗ − M ε U_{\varepsilon} B U_{\varepsilon}^{*} - M_{\varepsilon} U ε B U ε ∗ − M ε U ε A U ε ∗ − A ε − ε I U_{\varepsilon} A U_{\varepsilon}^{*} - A_{\varepsilon} - \varepsilon I U ε A U ε ∗ − A ε − ε I U ε B U ε ∗ = M ε U_{\varepsilon} B U_{\varepsilon}^{*} = M_{\varepsilon} U ε B U ε ∗ = M ε
（b）如果 U A U ′ = Λ U A U^{\prime} = \Lambda U A U ′ = Λ U B U ′ = M UBU^{\prime} = M U B U ′ = M A = U ∗ Λ U ‾ A = U^{*}\Lambda \overline{U} A = U ∗ Λ U B = U ∗ M U ‾ B = U^{*}M\overline{U} B = U ∗ M U A B = AB = A B = U ∗ A U ˉ U ′ M ‾ U = U ′ ( Λ M ‾ ) U U^{*}A\bar{U} U^{\prime}\overline{M} U = U^{\prime}(\Lambda \overline{M})U U ∗ A U ˉ U ′ M U = U ′ ( Λ M ) U A B \pmb{AB} A B A A A A B = ( A − 1 ) − 1 B ‾ AB = (A^{-1})^{-1}\overline{B} A B = ( A − 1 ) − 1 B A − 1 A^{-1} A − 1 B B B U ∈ M n U\in M_{n} U ∈ M n A \pmb{A} A M ∈ M n M\in M_{n} M ∈ M n A − 1 = U A U ′ A^{-1} = UAU^{\prime} A − 1 = U A U ′ B ⃗ = U M U ′ \vec{B} = UMU^{\prime} B = U M U ′ A = U ‾ Λ − 1 U ′ A = \overline{U}\Lambda^{-1}U^{\prime} A = U Λ − 1 U ′ B = U M U ′ ‾ B = \overline{UMU^{\prime}} B = U M U ′ A A A B B B A A A U ∈ M n U\in M_{n} U ∈ M n U U U
U A U r = [ Σ 0 − 0 0 ] . Σ ∈ M k , 1 ⩽ k < n . U A U ^ {r} = \left[ \begin{array}{l l} \Sigma & 0 \\ - 0 & 0 \end{array} \right]. \quad \Sigma \in M _ {k}, \quad 1 \leqslant k < n. U A U r = [ Σ − 0 0 0 ] . Σ ∈ M k , 1 ⩽ k < n . 且 Σ \Sigma Σ U B U ⊺ UBU^{\intercal} U B U ⊺
U B U T = [ B 12 B 12 B 12 T B 22 ] , B 11 ∈ M k , B 22 ∈ M n − k , U B U ^ {T} = \left[ \begin{array}{l l} B _ {1 2} & B _ {1 2} \\ B _ {1 2} ^ {T} & B _ {2 2} \end{array} \right], \quad B _ {1 1} \in M _ {k}, \quad B _ {2 2} \in M _ {n - k}, U B U T = [ B 12 B 12 T B 12 B 22 ] , B 11 ∈ M k , B 22 ∈ M n − k , 则子块 B 11 B_{11} B 11 B 22 B_{22} B 22
( U A U T ) ( U ˉ B ˉ U ˉ l ) = U A B U l = [ Σ 0 0 0 ] [ B 11 B 12 B 12 B ˉ 22 ] = [ Σ B 11 Σ B 12 0 0 ] . (U A U ^ {T}) (\bar {U} \bar {B} \bar {U} ^ {l}) = U A B U ^ {l} = \left[ \begin{array}{l l} \Sigma & 0 \\ 0 & 0 \end{array} \right] \left[ \begin{array}{l l} B _ {1 1} & B _ {1 2} \\ B _ {1 2} & \bar {B} _ {2 2} \end{array} \right] = \left[ \begin{array}{l l} \Sigma B _ {1 1} & \Sigma B _ {1 2} \\ 0 & 0 \end{array} \right]. ( U A U T ) ( U ˉ B ˉ U ˉ l ) = U A B U l = [ Σ 0 0 0 ] [ B 11 B 12 B 12 B ˉ 22 ] = [ Σ B 11 0 Σ B 12 0 ] . 但是 U A B ‾ U ∗ U A \overline{B} U^{*} U A B U ∗ Σ B ‾ 12 = 0 \Sigma \overline{B}_{12} = 0 Σ B 12 = 0 Σ \Sigma Σ B 12 = 0 B_{12} = 0 B 12 = 0
U A U 7 = [ Σ 0 0 0 ] , U B U 7 = [ B 11 0 0 B 22 ] , U A U ^ {7} = \left[ \begin{array}{l l} \Sigma & 0 \\ 0 & 0 \end{array} \right], \quad U B U ^ {7} = \left[ \begin{array}{l l} B _ {1 1} & 0 \\ 0 & B _ {2 2} \end{array} \right], U A U 7 = [ Σ 0 0 0 ] , U B U 7 = [ B 11 0 0 B 22 ] , 14
( U A U T ) ( U B U T ‾ ) = [ Σ B 11 0 0 0 ] . (U A U ^ {T}) (\overline {{U B U ^ {T}}}) = \left[ \begin{array}{c c} \Sigma B _ {1 1} & 0 \\ 0 & 0 \end{array} \right]. ( U A U T ) ( U B U T ) = [ Σ B 11 0 0 0 ] . 根据上述关于非奇异情形的证明，得知，存在酉矩阵 V 1 ∈ M k V_{1} \in M_{k} V 1 ∈ M k Λ 1 , Λ 2 ∈ M k \Lambda_{1}, \Lambda_{2} \in M_{k} Λ 1 , Λ 2 ∈ M k Σ = V 1 Λ 1 V 1 T \Sigma = V_{1} \Lambda_{1} V_{1}^{T} Σ = V 1 Λ 1 V 1 T B 11 = V 1 Λ 2 V 1 T B_{11} = V_{1} \Lambda_{2} V_{1}^{T} B 11 = V 1 Λ 2 V 1 T B 22 B_{22} B 22 V 2 ∈ M n V_{2} \in M_{n} V 2 ∈ M n Λ 3 ∈ M n \Lambda_{3} \in M_{n} Λ 3 ∈ M n B 22 = V 2 Λ 3 V T B_{22} = V_{2} \Lambda_{3} V^{T} B 22 = V 2 Λ 3 V T Λ = Λ 1 ⊕ 0 ∈ M n \Lambda = \Lambda_{1} \oplus 0 \in M_{n} Λ = Λ 1 ⊕ 0 ∈ M n M = Λ 2 ⊕ Λ 3 M = \Lambda_{2} \oplus \Lambda_{3} M = Λ 2 ⊕ Λ 3 V = V 1 ⊕ V 2 V = V_{1} \oplus V_{2} V = V 1 ⊕ V 2 U A U T = V A V T U A U^{T} = V A V^{T} U A U T = V A V T U B U T = V M V T U B U^{T} = V M V^{T} U B U T = V M V T A = ( U ∗ V ) A ( U ∗ V ) T A = (U^{*} V) A (U^{*} V)^{T} A = ( U ∗ V ) A ( U ∗ V ) T B = ( U ∗ V ) M ( U ∗ V ) T B = (U^{*} V) M (U^{*} V)^{T} B = ( U ∗ V ) M ( U ∗ V ) T
(c) 如果 U A U ∗ = Λ UAU^* = \Lambda U A U ∗ = Λ U B U T = M UBU^T = M U B U T = M Λ \Lambda Λ A = U ∗ Λ U A = U^*\Lambda U A = U ∗ Λ U B = U ∗ M U B = U^*MU B = U ∗ M U
A B = U ∗ Λ U U ∗ M U − U ∗ Λ M U − U ∗ M A U ˉ = U ∗ M U U t Λ U ˉ = ( U ∗ M U ˉ ) ( U ∗ ‾ A U ‾ ) = B A ˉ . \begin{array}{l} A B = U ^ {*} \Lambda U U ^ {*} M U - U ^ {*} \Lambda M U - U ^ {*} M A \bar {U} \\ = U ^ {*} M U U ^ {t} \Lambda \bar {U} = (U ^ {*} M \bar {U}) (\overline {{U ^ {*}}} \overline {{A U}}) = B \bar {A}. \\ \end{array} A B = U ∗ Λ U U ∗ M U − U ∗ Λ M U − U ∗ M A U ˉ = U ∗ M U U t Λ U ˉ = ( U ∗ M U ˉ ) ( U ∗ A U ) = B A ˉ . 反之，如果 A B = B A ‾ AB = B\overline{A} A B = B A ε > 0 \varepsilon > 0 ε > 0 A ε = A + ε I A_{\varepsilon} = A + \varepsilon I A ε = A + ε I A ε B = A B + ε B = B A ‾ + ε B = B A ε A_{\varepsilon}B = AB + \varepsilon B = B\overline{A} + \varepsilon B = BA_{\varepsilon} A ε B = A B + εB = B A + εB = B A ε U ε ∈ M n U_{\varepsilon} \in M_{n} U ε ∈ M n U ε A ε U ε ∗ = U ε A U ε ∗ + ε I = Λ ε U_{\varepsilon}A_{\varepsilon}U_{\varepsilon}^{*} = U_{\varepsilon}A U_{\varepsilon}^{*} + \varepsilon I = \Lambda_{\varepsilon} U ε A ε U ε ∗ = U ε A U ε ∗ + ε I = Λ ε U ε B U ε T = M ε U_{\varepsilon}B U_{\varepsilon}^{T} = M_{\varepsilon} U ε B U ε T = M ε U ε A U ε ∗ = A ε − ε I U_{\varepsilon}A U_{\varepsilon}^{*} = A_{\varepsilon} - \varepsilon I U ε A U ε ∗ = A ε − ε I U ε B U ε T − M ε U_{\varepsilon}B U_{\varepsilon}^{T} - M_{\varepsilon} U ε B U ε T − M ε
两个奇异Hermite矩阵经（不一定是酉的）相合同时对角化的问题在习题8中讨论.
我们已经看到，在相合下，一个Hermite矩阵总可以取非常简单的形式（在对角线上有±1或0的对角矩阵），并且，在一定的条件下，一对Hermite矩阵经'相合可以同时变成对角矩阵。于是，自然要提出的问题是：一般的Hermite矩阵偶 A A A B ∈ M n B\in M_{n} B ∈ M n C C C
C ∙ A C 和 C ∙ B C C ^ {\bullet} A C \text {和} C ^ {\bullet} B C C ∙ A C 和 C ∙ BC 可以取什么样的标准形？虽然这个问题是针对(可能都是奇异的)一般Hermite矩阵偶来讨论的，但是不论是提出还是证明其一般结果都是相当复杂的。这里，对其中至少有一个矩阵是非奇异的Hermite矩阵偶，不加证明地叙述标准形偶定理。我们已经讨论了可经相合同时对角化的特殊情形。
4.5.19 定理 假定 A , B ∈ M n A, B \in M_{n} A , B ∈ M n A A A k k k C ∈ M n C \in M_{n} C ∈ M n
C ⋅ A C = [ A 1 0 A 2 0 ⋱ A k ] , C ⋅ B C = [ B 1 0 B 2 0 ⋱ B k ] , C ^ {\cdot} A C = \left[ \begin{array}{c c c c} A _ {1} & & & 0 \\ & A _ {2} & & \\ 0 & & \ddots & \\ & & & A _ {k} \end{array} \right], \quad C ^ {\cdot} B C = \left[ \begin{array}{c c c c} B _ {1} & & & 0 \\ & B _ {2} & & \\ 0 & & \ddots & \\ & & & B _ {k} \end{array} \right], C ⋅ A C = A 1 0 A 2 ⋱ 0 A k , C ⋅ BC = B 1 0 B 2 ⋱ 0 B k , 其中，每对 A i , B i ∈ M n , i = 1 , 2 , … , k A_{i}, B_{i} \in M_{n}, i = 1, 2, \dots, k A i , B i ∈ M n , i = 1 , 2 , … , k
B i = ϵ [ 0 α ⋱ 1 ⋱ ⋱ a 1 0 ] , A i − ϵ [ 0 1 ⋱ 1 0 ] , (4.5.20) B _ {i} = \epsilon \left[ \begin{array}{c c c c} 0 & & & \alpha \\ & & \ddots & 1 \\ & \ddots & \ddots & \\ a & 1 & & 0 \end{array} \right], A _ {i} - \epsilon \left[ \begin{array}{c c c} 0 & & 1 \\ & \ddots & \\ 1 & & 0 \end{array} \right], \tag {4.5.20} B i = ϵ 0 a ⋱ 1 ⋱ ⋱ α 1 0 , A i − ϵ 0 1 ⋱ 1 0 , ( 4.5.20 ) 且 α \alpha α
B i = [ 0 α 0 ⋱ 1 ⋱ ⋱ α 1 0 … … … … … 0 α ˉ ⋱ 1 0 ⋱ ⋱ α ˉ 1 0 ] , A i = [ 0 1 ⋱ 1 0 ] , (4.5.21) B _ {i} = \left[ \begin{array}{c c c c c c c c} & & & & 0 & & & \alpha \\ & & 0 & & & & \ddots & 1 \\ & & & & & \ddots & \ddots \\ & & & & \alpha & 1 & & 0 \\ \dots & & & \dots & \dots & \dots & \dots & \\ 0 & & & \bar {\alpha} & & & \\ & & \ddots & 1 & & 0 \\ & \ddots & \ddots & \\ \bar {\alpha} & 1 & & 0 & & \end{array} \right], A _ {i} = \left[ \begin{array}{c c c c} 0 & & 1 \\ & \ddots & \\ 1 & & 0 \\ \end{array} \right], \tag {4.5.21} B i = … 0 α ˉ ⋱ 1 0 ⋱ ⋱ … α ˉ 1 0 0 α … ⋱ 1 … 0 ⋱ ⋱ … α 1 0 , A i = 0 1 ⋱ 1 0 , ( 4.5.21 ) 且 α \alpha α ε \varepsilon ε + 1 +1 + 1 − 1 -1 − 1 n t n_{t} n t M ( 1 , 2 ) n t M_{(1,2)n_{t}} M ( 1 , 2 ) n t
说明：
在 α \alpha α n i = 1 n_i = 1 n i = 1 α \alpha α ± 1 \pm 1 ± 1 α \alpha α ε = 1 \varepsilon = 1 ε = 1 1 × 1 1 \times 1 1 × 1 C ∗ B C C^* BC C ∗ BC α I \alpha I α I C ∗ A C C^* AC C ∗ A C I I I
在 α \alpha α n 1 = 2 n_{1} = 2 n 1 = 2
B i = [ 0 α α ˉ 0 ] , A i = [ 0 1 1 0 ] . B _ {i} = \left[ \begin{array}{l l} 0 & \alpha \\ \bar {\alpha} & 0 \end{array} \right], \quad A _ {i} = \left[ \begin{array}{l l} 0 & 1 \\ 1 & 0 \end{array} \right]. B i = [ 0 α ˉ α 0 ] , A i = [ 0 1 1 0 ] . 在定理中，同时产生的子块结构恰好对应于 A A A A ⋅ B A \cdot B A ⋅ B A i ⋅ B i A_{i} \cdot B_{i} A i ⋅ B i ( C ⋅ A C ) − 1 ( C ⋅ B C ) = C − 1 ( A ⋅ B ) C (C \cdot AC)^{-1}(C \cdot BC) = C^{-1}(A \cdot B)C ( C ⋅ A C ) − 1 ( C ⋅ BC ) = C − 1 ( A ⋅ B ) C C C C A − 1 B A^{-1}B A − 1 B A − 1 B A^{-1}B A − 1 B ε \varepsilon ε
4.5.22 注释 就像两个 Hermite 矩阵 A , B A, B A , B A − 1 B A^{-1}B A − 1 B A , B A, B A , B A − 1 B A^{-1}B A − 1 B B B B
替，而其他可能形式的子块取原来的形式。
习题 设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n B B B C ∈ M n C \in M_{n} C ∈ M n A = B C A = BC A = BC S ∈ M n S \in M_{n} S ∈ M n S A S ∗ = ( S B S ∗ ) C ∗ SAS^{*} = (SBS^{*})C^{*} S A S ∗ = ( SB S ∗ ) C ∗ C ∗ C^{*} C ∗ C C C
Sylvester 惯性定律 (4.5.8) 的证明中较难理解的部分是要证明, 若 D 1 D_{1} D 1 D 2 D_{2} D 2 n × n n \times n n × n D 2 = S ∗ D 1 S D_{2} = S^{*} D_{1} S D 2 = S ∗ D 1 S D 1 D_{1} D 1 s s s D 1 D_{1} D 1 s s s D 2 D_{2} D 2 t t t 1 ⩽ s , t < n 1 \leqslant s, t < n 1 ⩽ s , t < n s < t s < t s < t x = [ x t ] ∈ C n x = [x_{t}] \in \mathbb{C}^{n} x = [ x t ] ∈ C n x t + 1 = x t + 2 = ⋯ = x n = 0 x_{t + 1} = x_{t + 2} = \cdots = x_{n} = 0 x t + 1 = x t + 2 = ⋯ = x n = 0 ( S x ) 1 = ( S x ) 2 = ⋯ = ( S x ) s = 0 (Sx)_{1} = (Sx)_{2} = \cdots = (Sx)_{s} = 0 ( S x ) 1 = ( S x ) 2 = ⋯ = ( S x ) s = 0 x ∗ D 2 x > 0 x^{*} D_{2} x > 0 x ∗ D 2 x > 0 ( S x ) ∗ D 1 ( S x ) < 0 (Sx)^{*} D_{1}(Sx) < 0 ( S x ) ∗ D 1 ( S x ) < 0
设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n A A A B B B a , b , a A + b B a, b, aA + bB a , b , a A + b B B B B A A A B B B A + i B A + iB A + i B
试用定理(4.5.15)证明中的证明方法以及交换族定理(1.3.19)和(4.1.6)证明定理(4.5.15)之I(b)的如下推广．设 A 1 , A 2 , ⋯ , A k ∈ M n A_{1}, A_{2}, \cdots, A_{k} \in M_{n} A 1 , A 2 , ⋯ , A k ∈ M n A i A_{i} A i T ∈ M n T \in M_{n} T ∈ M n i = 1 , 2 , ⋯ , k i = 1, 2, \cdots, k i = 1 , 2 , ⋯ , k T ∗ A i T T^{*}A_{i}T T ∗ A i T i = 2 , ⋯ , k i = 2, \cdots, k i = 2 , ⋯ , k A 1 − 1 A i A_{1}^{-1}A_{i} A 1 − 1 A i { A i − 1 A i : i = 2 , ⋯ , n } \{A_{i}^{-1}A_{i}: i = 2, \cdots, n\} { A i − 1 A i : i = 2 , ⋯ , n } C i = A 1 − 1 A i C_{i} = A_{1}^{-1}A_{i} C i = A 1 − 1 A i i = 2 , ⋯ , k } i = 2, \cdots, k\} i = 2 , ⋯ , k } S C i S − 1 SC_{i}S^{-1} S C i S − 1 B i = ( S ∗ ) − 1 A i S − 1 B_{i} = (S^{*})^{-1}A_{i}S^{-1} B i = ( S ∗ ) − 1 A i S − 1 { B i } \{B_{i}\} { B i } U U U i = 2 , ⋯ , k i = 2, \cdots, k i = 2 , ⋯ , k U B i U ∗ UB_{i}U^{*} U B i U ∗ T = U S T = US T = U S
由(4.0.4)给出的具有实对称系数矩阵 A ( x ) = [ a i j ( x ) ] A(x) = [a_{ij}(x)] A ( x ) = [ a ij ( x )] L L L x ∈ D ⊂ R n x \in D \subset \mathbb{R}^n x ∈ D ⊂ R n A ( x ) A(x) A ( x ) L L L x x x A ( x ) A(x) A ( x ) n − 1 n - 1 n − 1
∇ 2 f = ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 + ∂ 2 f ∂ z 2 = 0 \nabla^ {2} f = \frac {\partial^ {2} f}{\partial x ^ {2}} + \frac {\partial^ {2} f}{\partial y ^ {2}} + \frac {\partial^ {2} f}{\partial z ^ {2}} = 0 ∇ 2 f = ∂ x 2 ∂ 2 f + ∂ y 2 ∂ 2 f + ∂ z 2 ∂ 2 f = 0 给出了椭圆微分算子的一个例子，而波动方程
□ 2 f = − ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 − ∂ 2 f ∂ t 2 = 0 \square^ {2} f = - \frac {\partial^ {2} f}{\partial x ^ {2}} + \frac {\partial^ {2} f}{\partial y ^ {2}} - \frac {\partial^ {2} f}{\partial t ^ {2}} = 0 □ 2 f = − ∂ x 2 ∂ 2 f + ∂ y 2 ∂ 2 f − ∂ t 2 ∂ 2 f = 0 是双曲型算子的一个例子。这两个方程都是在笛卡儿坐标系中给出的。在球极坐标，柱面坐标或其他坐标系下，这两个方程的差别就很大。
设 X = [ X 1 , ⋯ , X n ] T X = [X_{1}, \cdots, X_{n}]^{T} X = [ X 1 , ⋯ , X n ] T Y = [ Y 1 , ⋯ , Y n ] T Y = [Y_{1}, \cdots, Y_{n}]^{T} Y = [ Y 1 , ⋯ , Y n ] T X X X Y Y Y S ∈ M n S \in M_{n} S ∈ M n S X SX SX S Y SY S Y S S S S X SX SX S Y SY S Y
相关.
利用习题 4 给出三个或多个随机向量满足什么条件就能保证有一个非奇异线性变换使变换后的诸随机向量的各分量是不相关的。
定理(4.5.15)的情形I(b)考虑了两个Hermite矩阵在至少有一个矩阵是非奇异的情形下经相合同时对角化的问题。推论(4.5.18a)考虑了两个矩阵可能都是奇异的情形下用酉相合同时对角化的问题。若两个矩阵都是奇异的，则经（不一定是酉的）相合同时对角化它们的问题最终可化成(4.5.15)，但必须考察这两个矩阵的两个零空间之交的正交补的性质。设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n N ( A ) N(A) N ( A ) N ( B ) N(B) N ( B ) A A A B B B [ 0 0 0 1 ] \left[ \begin{array}{cc}0 & 0 \\ 0 & 1\end{array} \right] [ 0 0 0 1 ] [ 1 0 0 0 ] \left[ \begin{array}{cc}1 & 0 \\ 0 & 0\end{array} \right] [ 1 0 0 0 ] N ( A ) ∩ N ( B ) = { 0 } N(A) \cap N(B) = \{0\} N ( A ) ∩ N ( B ) = { 0 } A A A B B B a a a a A + B aA + B a A + B C ∈ M n C \in M_{n} C ∈ M n C ∗ A C = A 1 C^{*}AC = A_{1} C ∗ A C = A 1 C ∗ B C = A 2 C^{*}BC = A_{2} C ∗ BC = A 2 A 1 A_{1} A 1 A 2 A_{2} A 2 A 1 A_{1} A 1 A 2 A_{2} A 2 a a a a A 1 + A 2 aA_{1} + A_{2} a A 1 + A 2
A = [ 0 1 0 1 0 0 0 0 0 ] 和 B = [ 0 0 0 0 0 1 0 1 0 ] A = \left[ \begin{array}{l l l} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] \quad \text {和} \quad B = \left[ \begin{array}{l l l} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right] A = 0 1 0 1 0 0 0 0 0 和 B = 0 0 0 0 0 1 0 1 0 不能经“相合同时对角化. (d) 若
N ( A ) ∩ N ( B ) = { 0 } , N (A) \cap N (B) = \{0 \}, N ( A ) ∩ N ( B ) = { 0 } , 又 a ∈ R a \in \mathbb{R} a ∈ R a A + B aA + B a A + B A A A B B B ( a A + B ) − 1 B (aA + B)^{-1}B ( a A + B ) − 1 B dim N ( A ) ∩ N ( B ) = k ⩾ 1 \dim N(A) \cap N(B) = k \geqslant 1 dim N ( A ) ∩ N ( B ) = k ⩾ 1 { u 1 , u 2 , … , u n } \{u_1, u_2, \dots, u_n\} { u 1 , u 2 , … , u n } R n \mathbb{R}^n R n { u 1 , u 2 , … , u k } \{u_1, u_2, \dots, u_k\} { u 1 , u 2 , … , u k } N ( A ) ∩ N ( B ) N(A) \cap N(B) N ( A ) ∩ N ( B ) U = [ u 1 , u 2 , … , u n ] ∈ M n U = [u_1, u_2, \dots, u_n] \in M_n U = [ u 1 , u 2 , … , u n ] ∈ M n
240
U ⋆ A U = [ 0 0 0 A ′ ] 和 U ⋆ B U = [ 0 0 0 B ′ ] , U ^ {\star} A U = \left[ \begin{array}{l l} {0} & {0} \\ {0} & {A ^ {\prime}} \end{array} \right] \quad \text {和} \quad U ^ {\star} B U = \left[ \begin{array}{l l} {0} & {0} \\ {0} & {B ^ {\prime}} \end{array} \right], U ⋆ A U = [ 0 0 0 A ′ ] 和 U ⋆ B U = [ 0 0 0 B ′ ] , 其中 A ′ , B ′ ∈ M n − k A^{\prime}, B^{\prime} \in M_{n - k} A ′ , B ′ ∈ M n − k N ( A ′ ) ∩ N ( B ′ ) = { 0 } N(A^{\prime}) \cap N(B^{\prime}) = \{0\} N ( A ′ ) ∩ N ( B ′ ) = { 0 } k × k k \times k k × k A A A B B B A ′ A^{\prime} A ′ B ′ B^{\prime} B ′ A ′ A^{\prime} A ′ B ′ B^{\prime} B ′
如果 A , B ∈ M n A, B \in M_{n} A , B ∈ M n B B B A A A B B B A A A B B B
证明 [ 0 1 1 0 ] \left[ \begin{array}{ll}0 & 1\\ 1 & 0 \end{array} \right] [ 0 1 1 0 ] [ 1 0 0 − 1 ] \left[ \begin{array}{ll}1 & 0\\ 0 & -1 \end{array} \right] [ 1 0 0 − 1 ]
证明 [ 1 1 1 0 ] \left[ \begin{array}{ll}1 & 1\\ 1 & 0 \end{array} \right] [ 1 1 1 0 ] [ 0 1 1 0 ] \left[ \begin{array}{ll}0 & 1\\ 1 & 0 \end{array} \right] [ 0 1 1 0 ]
设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n A A A A A A B B B
设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n U ∈ M n U \in M_{n} U ∈ M n U A U ⊤ = A U A U^{\top} = A U A U ⊤ = A U B U ⊤ = M U B U^{\top} = M U B U ⊤ = M V V V B A ‾ − A V B ‾ B \overline{A} - A V \overline{B} B A − A V B A = diag ( λ 1 , λ 2 , … , λ n ) A = \operatorname{diag}(\lambda_{1}, \lambda_{2}, \dots, \lambda_{n}) A = diag ( λ 1 , λ 2 , … , λ n ) D D D A ‾ − D A = A D \overline{A} - D A = A D A − D A = A D
B A ˙ = U ∗ M A ˉ U − U ∗ A D 1 D 2 M ¨ U = A ( U ⊤ D 1 D 2 U ˉ ) B ˙ , B \dot {A} = U ^ {*} M \bar {A} U - U ^ {*} A D _ {1} D _ {2} \ddot {M} U = A (U ^ {\top} D _ {1} D _ {2} \bar {U}) \dot {B}, B A ˙ = U ∗ M A ˉ U − U ∗ A D 1 D 2 M ¨ U = A ( U ⊤ D 1 D 2 U ˉ ) B ˙ , 其中 D 1 D_{1} D 1 D 2 D_{2} D 2
利用习题 13 的必要条件证明，习题 8(c) 的两个对称矩阵不能经酉 了 ^{\text{了}} 了 B A ‾ B\overline{A} B A A U B AUB A U B
如果 A , B ∈ M n A, B \in M_{n} A , B ∈ M n A A A B B B T {}^T T B A ‾ = A U B B\overline{A} = AUB B A = A U B A A A B B B A − 1 B A B ‾ − 1 = U A^{-1}BA\overline{B}^{-1} = U A − 1 B A B − 1 = U I = U U − 1 I = UU^{-1} I = U U − 1 A B ‾ : B : A = B : A A ‾ B ‾ \overline{AB}: B: A = B: A\overline{A}\overline{B} A B : B : A = B : A A B A − 1 B A^{-1}B A − 1 B
设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n U ∈ M n U \in M_{n} U ∈ M n U A U T = A UAU^{T} = A U A U T = A U B U T = M UBU^{T} = M U B U T = M A A AA AA B B BB BB A A AA AA B B BB BB n n n
设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n Λ \Lambda Λ B B B U ∈ M n U \in M_{n} U ∈ M n U A U ∗ = Λ UAU^{*} = \Lambda U A U ∗ = Λ U B U r = M UBU^{r} = M U B U r = M A A A B B BB BB ( ∗ (^{*} ( ∗ T ) {}^{T}) T ) B B BB BB
设 A , B ∈ M n A, B \in M_{n} A , B ∈ M n A A A B B B A A A p 1 , B ( t ) ≡ det ( t A − B ) p_{1,B}(t) \equiv \operatorname{det}(tA - B) p 1 , B ( t ) ≡ det ( t A − B ) n n n A A A B B B A − 1 B A^{-1}B A − 1 B
对 Sylvester 惯性定律 (4.5.8) 的下述另一个证明作详细的论述。若 A ∈ M n A \in M_n A ∈ M n S ∈ M n S \in M_n S ∈ M n S − Q R S - QR S − QR Q ∈ M n Q \in M_n Q ∈ M n R ∈ M n R \in M_n R ∈ M n 0 ⩽ t ⩽ 1 0 \leqslant t \leqslant 1 0 ⩽ t ⩽ 1 S ( t ) = t Q + ( 1 − t ) Q R S(t) = tQ + (1 - t)QR S ( t ) = tQ + ( 1 − t ) QR A ( t ) = S ( t ) A S ( t ) ∗ A(t) = S(t)AS(t)^* A ( t ) = S ( t ) A S ( t ) ∗ A ( 0 ) A(0) A ( 0 ) A ( 1 ) A(1) A ( 1 ) A ( t ) A(t) A ( t ) t t t A ( t ) A(t) A ( t ) A ( 0 ) A(0) A ( 0 ) A ( 1 ) A(1) A ( 1 )
征值．对较小的 ε > 0 \varepsilon >0 ε > 0 A ± ε I A\pm \varepsilon I A ± ε I
如果 A = [ B C 0 0 ] ∈ M n A = \begin{bmatrix} B & C \\ 0 & 0 \end{bmatrix} \in M_n A = [ B 0 C 0 ] ∈ M n B ∈ M k B \in M_k B ∈ M k 1 ⩽ k ⩽ n 1 \leqslant k \leqslant n 1 ⩽ k ⩽ n A A A B B B C = 0 C = 0 C = 0 A A ∗ AA^* A A ∗ A ∗ A A^* A A ∗ A C ∗ C = 0 C^* C = 0 C ∗ C = 0 x ∈ C n − k x \in \mathbf{C}^{n-k} x ∈ C n − k ( C x ) ∗ ( C x ) = 0 (Cx)^*(Cx) = 0 ( C x ) ∗ ( C x ) = 0 x ∈ C n − k x \in \mathbf{C}^{n-k} x ∈ C n − k C x = 0 Cx = 0 C x = 0
说明(4.5.18)(b)中所采用的证法也可以用来证明(a)和(c)这两部分.
设 F = { A 1 , … , A k } ⊂ M n \mathcal{F} = \{A_1, \dots, A_k\} \subset M_n F = { A 1 , … , A k } ⊂ M n G = { A i A ‾ j : i , j = 1 , 2 , … , k } \mathcal{G} = \{A_i \overline{A}_j : i, j = 1, 2, \dots, k\} G = { A i A j : i , j = 1 , 2 , … , k } U ∈ M n U \in M_n U ∈ M n i = 1 , … , k i = 1, \dots, k i = 1 , … , k U A i U T UA_i U^T U A i U T G \mathcal{G} G k = 2 k = 2 k = 2 G \mathcal{G} G F \mathcal{F} F
设 F = { A 1 , … , A k } ⊂ M n \mathcal{F} = \{A_1, \dots, A_k\} \subset M_n F = { A 1 , … , A k } ⊂ M n H = { B 1 , … , B m } ⊂ M n \mathcal{H} = \{B_1, \dots, B_m\} \subset M_n H = { B 1 , … , B m } ⊂ M n g = { A i A ‾ i : i = 1 , … , k } g = \{A_i \overline{A}_i : i = 1, \dots, k\} g = { A i A i : i = 1 , … , k } U ∈ M n U \in M_n U ∈ M n U A i U T UA_i U^T U A i U T U B i U UB_i U U B i U g g g H \mathcal{H} H i = 1 , … , k i = 1, \dots, k i = 1 , … , k j = 1 , … , m j = 1, \dots, m j = 1 , … , m B i A i B_i A_i B i A i k = m = 1 k = m = 1 k = m = 1 F \mathcal{F} F H \mathcal{H} H
进一步阅读 定理(4.5.9)的Ostrowski的证明以及有关的结果可参看“A Quantitative Formulation of Sylvester's Law of Inertia,” Proc. Nat. Acad. Sci. 45 (1959), 740-744. 定理(4.5.25)的另一种形式在[GLR 82]中给出；包括两个矩阵是奇异矩阵的情形的一个详细证明还在R.C.Thompson未发表的手稿中。有关两个以上矩阵同时对角化的结果可参看Y.P.Hong and R.A.Horn, “On Simultaneous Reduction of Families of Matrices to Triangular or Diagonal Form by Unitary Congruence,” Linear and Multilinear Algebra 17 (1985), 271-288.