7.0 导引 一类具有特殊正性的Hermite矩阵常常出现在许多应用中。具有这种正性的Hermite矩阵（特别是实对称矩阵）可以看作正数概念到矩阵的推广。这样考虑常常可以深入理解正定矩阵的一些性质和应用。下面给出这方面的一些例子，其中就要出现这些特殊的Hermite矩阵。
Hessian 矩阵，极小化和凸性 设 f ( x ) f(x) f ( x ) D ⊂ R n D \subset \mathbb{R}^n D ⊂ R n y = [ y i ] y = [y_i] y = [ y i ] D D D
f ( x ) = f ( y ) + ∑ i = 1 n ( x i − y i ) ∂ f ∂ x i ∣ , + ∑ i , j = 1 n ( x i − y j ) ( x j − y j ) ∂ 2 f ∂ x i ∂ x j ∣ y + … \begin{array}{l} f (x) = f (y) + \sum_ {i = 1} ^ {n} \left(x _ {i} - y _ {i}\right) \frac {\partial f}{\partial x _ {i}} \Bigg |, \\ + \sum_ {i, j = 1} ^ {n} \left(x _ {i} - y _ {j}\right) \left(x _ {j} - y _ {j}\right) \frac {\partial^ {2} f}{\partial x _ {i} \partial x _ {j}} \Bigg | _ {y} + \dots \\ \end{array} f ( x ) = f ( y ) + ∑ i = 1 n ( x i − y i ) ∂ x i ∂ f , + ∑ i , j = 1 n ( x i − y j ) ( x j − y j ) ∂ x i ∂ x j ∂ 2 f y + … 对 y y y x ∈ D x \in D x ∈ D y y y f f f y y y f f f y y y
f ( x ) − f ( y ) = ∑ i , j = 1 n ( x i − y i ) ( x j − y j ) ∂ 2 f ∂ x i ∂ x j ∣ y + … = ( x − y ) T H ( f ; y ) ( x − y ) + … , \begin{array}{l} f (x) - f (y) = \sum_ {i, j = 1} ^ {n} \left(x _ {i} - y _ {i}\right) \left(x _ {j} - y _ {j}\right) \frac {\partial^ {2} f}{\partial x _ {i} \partial x _ {j}} \Bigg | _ {y} + \dots \\ = (x - y) ^ {T} H (f; y) (x - y) + \dots , \\ \end{array} f ( x ) − f ( y ) = ∑ i , j = 1 n ( x i − y i ) ( x j − y j ) ∂ x i ∂ x j ∂ 2 f y + … = ( x − y ) T H ( f ; y ) ( x − y ) + … , m × n m \times n m × n
H ( f ; y ) ≡ [ ∂ 2 f ∂ x i ∂ x j ∣ y H (f; y) \equiv \left[ \frac {\partial^ {2} f}{\partial x _ {i} \partial x _ {j}} \right| _ {y} H ( f ; y ) ≡ [ ∂ x i ∂ x j ∂ 2 f y 称为 f f f y \mathbf{y} y f f f
z T H ( f ; y ) z , z ≠ 0 , z ∈ R n (7.0.1) z ^ {T} H (f; y) z, \quad z \neq 0, z \in \mathbf {R} ^ {n} \tag {7.0.1} z T H ( f ; y ) z , z = 0 , z ∈ R n ( 7.0.1 ) 总是正的，则 y y y f f f y y y f f f z ∈ R n z \in \mathbb{R}^n z ∈ R n y y y n = 1 n = 1 n = 1 n = 1 n = 1 n = 1 n > 1 n > 1 n > 1
如果二次型(7.0.1)在 D D D f f f f f f D D D n = 1 n = 1 n = 1
方差-协方差矩阵 设 X 1 , X 2 , ⋯ , X n X_{1}, X_{2}, \cdots, X_{n} X 1 , X 2 , ⋯ , X n E E E μ i = E ( X i ) \mu_{i} = E(X_{i}) μ i = E ( X i ) X = ( X 1 , ⋯ , X n ) T X = (X_{1}, \cdots, X_{n})^{T} X = ( X 1 , ⋯ , X n ) T A = [ a i j ] A = [a_{ij}] A = [ a ij ]
a i j = E [ ( X ˙ i − μ i ) ( X j − μ j ) ] , i , j = 1 , … , n . a _ {i j} = E \left[ \left(\dot {X} _ {i} - \mu_ {i}\right) \left(X _ {j} - \mu_ {j}\right) \right], \quad i, j = 1, \dots , n. a ij = E [ ( X ˙ i − μ i ) ( X j − μ j ) ] , i , j = 1 , … , n . 显然， A A A z = [ z i ] ∈ C n z = [z_i] \in \mathbf{C}^n z = [ z i ] ∈ C n
z ′ A z − E [ ∑ i , j = 1 n z i ( X i − μ ˉ i ) z j ( X j − μ j ) ] = E ∣ ∑ i = 1 n z i ( X i − μ i ) ∣ 2 ⩾ 0. z ^ {\prime} A z - E \left[ \sum_ {i, j = 1} ^ {n} z _ {i} \left(X _ {i} - \bar {\mu} _ {i}\right) z _ {j} \left(X _ {j} - \mu_ {j}\right) \right] = E \left| \sum_ {i = 1} ^ {n} z _ {i} \left(X _ {i} - \mu_ {i}\right) \right| ^ {2} \geqslant 0. z ′ A z − E [ i , j = 1 ∑ n z i ( X i − μ ˉ i ) z j ( X j − μ j ) ] = E i = 1 ∑ n z i ( X i − μ i ) 2 ⩾ 0. 在这个结论中所涉及到的期望函数的仅有性质是它的线性、齐性和非负性三个性质；即只要 Y Y Y E [ Y ] ⩾ 0 E[Y] \geqslant 0 E [ Y ] ⩾ 0
392
不借助概率术语也可以作出同样的结论。如果在直线上有一族复值函数 f 1 , f 2 , ⋯ , f n f_{1}, f_{2}, \cdots, f_{n} f 1 , f 2 , ⋯ , f n g g g
a i j = ∫ − ∞ ∞ f ˉ i ( x ) f j ( x ) g ( x ) d x , i , j = 1 , … , n a _ {i j} = \int_ {- \infty} ^ {\infty} \bar {f} _ {i} (x) f _ {j} (x) g (x) d x, i, j = 1, \dots , n a ij = ∫ − ∞ ∞ f ˉ i ( x ) f j ( x ) g ( x ) d x , i , j = 1 , … , n 都有定义且收敛，则矩阵 A = [ a η ] A = [a_{\eta}] A = [ a η ]
z ′ Λ z = ∑ i , j = 1 n ∫ i z ˉ i f ˉ i ( x ) z j f j ( x ) g ( x ) d x − ∫ i ∞ ∣ ∑ i = 1 n z i f i ( x ) ∣ 2 g ( x ) d x . z ^ {\prime} \Lambda z = \sum_ {i, j = 1} ^ {n} \int_ {i} \quad \bar {z} _ {i} \bar {f} _ {i} (x) z _ {j} f _ {j} (x) g (x) d x - \int_ {i} ^ {\infty} \left| \sum_ {i = 1} ^ {n} z _ {i} f _ {i} (x) \right| ^ {2} g (x) d x. z ′ Λ z = i , j = 1 ∑ n ∫ i z ˉ i f ˉ i ( x ) z j f j ( x ) g ( x ) d x − ∫ i ∞ i = 1 ∑ n z i f i ( x ) 2 g ( x ) d x . 因此，如果 g ( x ) g(x) g ( x )
非负函数的代数矩 设 f ( x ) f(x) f ( x ) [ 0 , 1 ] [0, 1] [ 0 , 1 ]
a k ≡ ∫ 0 1 x k f ( x ) d r . (7.0.2) a _ {k} \equiv \int_ {0} ^ {1} x ^ {k} f (x) d r. \tag {7.0.2} a k ≡ ∫ 0 1 x k f ( x ) d r . ( 7.0.2 ) 序列 a 0 , a 1 , a 2 , ⋯ a_0, a_1, a_2, \cdots a 0 , a 1 , a 2 , ⋯
∑ j , k = 0 n a j + k z j z k = ∑ j , k = 0 n ∫ 0 1 x j − k z j z k f ( x ) d x = ∫ 0 1 ( ∑ k = 0 n z k x k ) 2 f ( x ) d x (7.0.3) \sum_ {j, k = 0} ^ {n} a _ {j + k} z _ {j} z _ {k} = \sum_ {j, k = 0} ^ {n} \int_ {0} ^ {1} x ^ {j - k} z _ {j} z _ {k} f (x) d x = \int_ {0} ^ {1} \left(\sum_ {k = 0} ^ {n} z _ {k} x ^ {k}\right) ^ {2} f (x) d x \tag {7.0.3} j , k = 0 ∑ n a j + k z j z k = j , k = 0 ∑ n ∫ 0 1 x j − k z j z k f ( x ) d x = ∫ 0 1 ( k = 0 ∑ n z k x k ) 2 f ( x ) d x ( 7.0.3 ) 有联系。如果令 A ≡ [ a i − j ] A \equiv [a_{i - j}] A ≡ [ a i − j ] A A A x ∈ [ 0 , 1 ] x \in [0,1] x ∈ [ 0 , 1 ] f ( x ) ⩾ 0 f(x) \geqslant 0 f ( x ) ⩾ 0 Z ∈ R n + 1 Z \in \mathbb{R}^{n + 1} Z ∈ R n + 1 z T A z ⩾ 0 z^T A z \geqslant 0 z T A z ⩾ 0 n = 1 , 2 , … n = 1,2,\dots n = 1 , 2 , … A A A a n a_{n} a n i + j i + j i + j
非负函数的三角矩 设 ( θ ) (\theta) ( θ ) [ 0 , 2 π ] [0, 2\pi] [ 0 , 2 π ]
a k ≡ ∫ 0 2 π e i k θ f ( θ ) d θ . k = ± 1 , ± 2 , … . (7.0.4) a _ {k} \equiv \int_ {0} ^ {2 \pi} e ^ {i k \theta} f (\theta) d \theta . \quad k = \pm 1, \pm 2, \dots . \tag {7.0.4} a k ≡ ∫ 0 2 π e ik θ f ( θ ) d θ . k = ± 1 , ± 2 , … . ( 7.0.4 ) [393] 序列 a 0 , a 1 , a − 1 , a 2 , a − 2 , ⋯ a_{0}, a_{1}, a_{-1}, a_{2}, a_{-2}, \cdots a 0 , a 1 , a − 1 , a 2 , a − 2 , ⋯
∑ j , k = 1 n a j − k z j z k = ∑ j , k n ∫ 0 2 π e ( j − k ) θ z j z k f ( θ ) d θ = ∫ 0 2 π ∣ ∑ k = 0 n z k e λ k θ ∣ 2 f ( θ ) d θ (7.0.5) \sum_ {j, k = 1} ^ {n} a _ {j - k} z _ {j} z _ {k} = \sum_ {j, k} ^ {n} \int_ {0} ^ {2 \pi} e ^ {(j - k) \theta} z _ {j} z _ {k} f (\theta) d \theta = \int_ {0} ^ {2 \pi} \left| \sum_ {k = 0} ^ {n} z _ {k} e ^ {\lambda k \theta} \right| ^ {2} f (\theta) d \theta \tag {7.0.5} j , k = 1 ∑ n a j − k z j z k = j , k ∑ n ∫ 0 2 π e ( j − k ) θ z j z k f ( θ ) d θ = ∫ 0 2 π k = 0 ∑ n z k e λk θ 2 f ( θ ) d θ ( 7.0.5 ) 有联系，如果令 A = [ a i − j ] A = [a_{i - j}] A = [ a i − j ] A A A θ ∈ [ 0 , 2 π ] \theta \in [0,2\pi ] θ ∈ [ 0 , 2 π ] f ( θ ) ⩾ 0 f(\theta)\geqslant 0 f ( θ ) ⩾ 0 z ∈ C n + 1 z\in \mathbf{C}^{n + 1} z ∈ C n + 1 z ′ A z ⩾ 0 z^{\prime}Az\geqslant 0 z ′ A z ⩾ 0 n = 1 n = 1 n = 1 A A A a i j a_{ij} a ij i − j i - j i − j d μ d_{\mu} d μ f ( θ ) d θ ) f(\theta)\mathrm{d}\theta) f ( θ ) d θ ) a k a_{k} a k
关于微分方程的数值解的离散化和差分法 假定我们有形如
− y ′ ′ ( x ) + σ ( x ) y ( x ) = f ( r ) , 0 ⩽ r ⩽ 1 , - y ^ {\prime \prime} (x) + \sigma (x) y (x) = f (r), \quad 0 \leqslant r \leqslant 1, − y ′′ ( x ) + σ ( x ) y ( x ) = f ( r ) , 0 ⩽ r ⩽ 1 , y ( 1 ) = β y (1) = \beta y ( 1 ) = β 的两点边值问题，其中， α \alpha α β \beta β f ( x ) f(x) f ( x ) σ ( x ) \sigma(x) σ ( x ) y ( k h ) = y k y(kh) = y_k y ( kh ) = y k k = 0 , 1 , … , n + 1 k = 0, 1, \dots, n + 1 k = 0 , 1 , … , n + 1
y ′ ′ ( x ) ≅ y ( ( k + 1 ) h ) − 2 y ( k h ) + y ( ( k − 1 ) h ) h 3 = y k + 1 − 2 y k + y k − 1 h 3 , y ^ {\prime \prime} (x) \cong \frac {y ((k + 1) h) - 2 y (k h) + y ((k - 1) h)}{h ^ {3}} = \frac {y _ {k + 1} - 2 y _ {k} + y _ {k - 1}}{h ^ {3}}, y ′′ ( x ) ≅ h 3 y (( k + 1 ) h ) − 2 y ( kh ) + y (( k − 1 ) h ) = h 3 y k + 1 − 2 y k + y k − 1 , 那么，就得到线性方程组
− y k − 1 + 2 y k − y k − 1 h 2 + σ k y k − f i , k = 1 , 2 , … , n , y 0 − α , y n − 1 = β . \begin{array}{l} \frac {- y _ {k - 1} + 2 y _ {k} - y _ {k - 1}}{h ^ {2}} + \sigma_ {k} y _ {k} - f _ {i}, \quad k = 1, 2, \dots , n, \\ y _ {0} - \alpha , \\ y _ {n - 1} = \beta . \\ \end{array} h 2 − y k − 1 + 2 y k − y k − 1 + σ k y k − f i , k = 1 , 2 , … , n , y 0 − α , y n − 1 = β . 这里，对于正整数 n n n h = 1 / ( n + 1 ) h = 1 / (n + 1) h = 1/ ( n + 1 ) y k = y ( k h ) y_{k} = y(kh) y k = y ( kh ) σ k = σ ( k h ) \sigma_{k} = \sigma (kh) σ k = σ ( kh ) f k − f ( k h ) f_{k} - f(kh) f k − f ( kh ) ( k = 1 ) (k = 1) ( k = 1 ) ( k = n ) (k = n) ( k = n )
( 2 + h 2 σ 1 ) y 1 − y 2 = h 2 f 1 + α , − y k − 1 + ( 2 + h 2 σ l ) y k − y l − 1 = h 2 f k , k = 2 , 3 , … , n − 1 , − y n − 1 + ( 2 + h 2 σ n ) y n = h 2 f n + β , \begin{array}{l} \left(2 + h ^ {2} \sigma_ {1}\right) y _ {1} - y _ {2} = h ^ {2} f _ {1} + \alpha , \\ - y _ {k - 1} + \left(2 + h ^ {2} \sigma_ {l}\right) y _ {k} - y _ {l - 1} = h ^ {2} f _ {k}, \quad k = 2, 3, \dots , n - 1, \\ - y _ {n - 1} + \left(2 + h ^ {2} \sigma_ {n}\right) y _ {n} = h ^ {2} f _ {n} + \beta , \\ \end{array} ( 2 + h 2 σ 1 ) y 1 − y 2 = h 2 f 1 + α , − y k − 1 + ( 2 + h 2 σ l ) y k − y l − 1 = h 2 f k , k = 2 , 3 , … , n − 1 , − y n − 1 + ( 2 + h 2 σ n ) y n = h 2 f n + β , 还可以更紧凑地把它写成 A y = w Ay = w A y = w y = [ y k ] ∈ R n y = [y_k] \in \mathbb{R}^n y = [ y k ] ∈ R n w = [ h 2 f 1 + α , h 2 f 2 , … , h 2 f n ] w = [h^2 f_1 + \alpha, h^2 f_2, \dots, h^2 f_n] w = [ h 2 f 1 + α , h 2 f 2 , … , h 2 f n ] h 2 f n + β ] τ ∈ R n h^2 f_n + \beta]^\tau \in \mathbb{R}^n h 2 f n + β ] τ ∈ R n Λ ∈ M n \Lambda \in M_n Λ ∈ M n
A = [ 2 + h 2 σ 1 − 1 − 1 2 + h 2 σ 2 − 1 0 ⋱ ⋱ ⋱ ⋱ 0 1 2 + h 2 σ n − 1 − 1 − 1 2 + h 2 σ n ] . (7.0.6) A = \left[ \begin{array}{c c c c c c} 2 + h ^ {2} \sigma_ {1} & - 1 & & & & \\ - 1 & 2 + h ^ {2} \sigma_ {2} & - 1 & & 0 \\ & \ddots & \ddots & \ddots & \ddots & \\ 0 & 1 & 2 + h ^ {2} \sigma_ {n - 1} & & - 1 \\ & & - 1 & & 2 + h ^ {2} \sigma_ {n} \end{array} \right]. \tag {7.0.6} A = 2 + h 2 σ 1 − 1 0 − 1 2 + h 2 σ 2 ⋱ 1 − 1 ⋱ 2 + h 2 σ n − 1 − 1 ⋱ 0 ⋱ − 1 2 + h 2 σ n . ( 7.0.6 ) 应该指出，不论 σ ( x ) \sigma(x) σ ( x ) A A A A y = w Ay = w A y = w σ ( x ) \sigma(x) σ ( x ) A A A
容易算出相应于 Λ \Lambda Λ
x T A x − [ x 1 2 + ∑ i = 1 n − 1 ( x i − x i + 1 ) 2 + x n 2 ] + h 2 ∑ i = 1 n σ i x i 2 . x ^ {T} A x - \left[ x _ {1} ^ {2} + \sum_ {i = 1} ^ {n - 1} \left(x _ {i} - x _ {i + 1}\right) ^ {2} + x _ {n} ^ {2} \right] + h ^ {2} \sum_ {i = 1} ^ {n} \sigma_ {i} x _ {i} ^ {2}. x T A x − [ x 1 2 + i = 1 ∑ n − 1 ( x i − x i + 1 ) 2 + x n 2 ] + h 2 i = 1 ∑ n σ i x i 2 . 等式右边中的括号内各项之和是非负的，且只有当 x x x σ ( x ) ≥ 0 \sigma(x) \geq 0 σ ( x ) ≥ 0
x ′ A x ⩾ [ x 1 2 + ∑ t = 1 n − 1 ( x t − x t − 1 ) 2 + x n 2 ] ⩾ 0. (7.0.7) x ^ {\prime} A x \geqslant \left[ x _ {1} ^ {2} + \sum_ {t = 1} ^ {n - 1} \left(x _ {t} - x _ {t - 1}\right) ^ {2} + x _ {n} ^ {2} \right] \geqslant 0. \tag {7.0.7} x ′ A x ⩾ [ x 1 2 + t = 1 ∑ n − 1 ( x t − x t − 1 ) 2 + x n 2 ] ⩾ 0. ( 7.0.7 ) 如果 A A A x ^ ∈ R n \hat{x} \in \mathbb{R}^n x ^ ∈ R n A x ^ = 0 A\hat{x} = 0 A x ^ = 0 x ^ T A x ^ = 0 \hat{x}^T A\hat{x} = 0 x ^ T A x ^ = 0 x ^ − 0 \hat{x} - 0 x ^ − 0 σ ( x ) ⩾ 0 \sigma(x) \geqslant 0 σ ( x ) ⩾ 0 A A A α \alpha α β \beta β
这是在研究常微分方程或偏微分方程的数值解时的典型情形。为了计算上的稳定性，最好
是设计一种方法能把微分方程问题离散化，使得在所得到的线性方程组 A y = w Ay = w A y = w A A A
在上述这些例子中所列举的矩阵具有特殊的正性，这正是本章要研究的对象。这些矩阵出现在许多应用中：调和分析中，复分析中，力学体系的振动理论中，以及矩阵理论的其他领域（例如奇异值分解和线性最小二乘方问题的解）中。
习题 如果序列 a k a_{k} a k f f f
∑ i , j = 1 n a i + 1 + 1 z i z j 和 ∑ i , j = 1 n { a i + j − a i + j − 1 } z i z j , z = [ z 1 ] ∈ R n \sum_ {i, j = 1} ^ {n} a _ {i + 1 + 1} z _ {i} z _ {j} \quad \text {和} \quad \sum_ {i, j = 1} ^ {n} \{a _ {i + j} - a _ {i + j - 1} \} z _ {i} z _ {j}, \quad z = [ z _ {1} ] \in \mathbf {R} ^ {n} i , j = 1 ∑ n a i + 1 + 1 z i z j 和 i , j = 1 ∑ n { a i + j − a i + j − 1 } z i z j , z = [ z 1 ] ∈ R n 都是非负的.
用示意图说明 Hankel 矩阵中哪些对角线取常值。对 Toeplitz 矩阵作同样的说明。
证明，(7.0.6)中的矩阵 A A A σ ( x ) ⩾ 0 \sigma(x) \geqslant 0 σ ( x ) ⩾ 0 A A A A A A
进一步阅读 关于实正定矩阵的一个简短的综述可参看 C. R. Johnson, “Positive Definite Matrices,” Amer. Math. Monthly 77(1970), 259-264. 关于一般正定矩阵的其他综述以及有关这方面的大批参考资料可参看 O. Taussky, “Positive Definite Matrices,” pp. 309-319 of Inequalities, ed. O. Shisha, Academic Press, New York, 1967; 以及 O. Taussky, “Positive Definite Matrices and Their Role in the Study of the Characteristic Roots of General Matrices,” Advan. Math. 2(1968), 175-186.