正惯性指数：正特征值个数 负惯性指数：负特征值个数

合同：如果存在可逆矩阵 P \mathbf{P} P，使得 P T A P = B \mathbf{P}^T\mathbf{A}\mathbf{P}=\mathbf{B} PTAP=B，则 A \mathbf{A} A合同于 B \mathbf{B} B，记为 A ≃ B \mathbf{A}\simeq \mathbf{B} A≃B

合同不改变矩阵的惯性指数

A \mathbf{A} A是实对称矩阵 A ≻ 0 \mathbf{A}\succ 0 A≻0的充要条件是顺序主子式大于0

证明: 记 A k = ( a 11 a 12 ⋯ a 1 k a 21 a 22 ⋯ a 2 k ⋯ ⋯ ⋯ ⋯ a k 1 a k 2 ⋯ a k k ) \mathbf{A}_k =\begin{pmatrix} a_{11}&a_{12}&\cdots & a_{1k}\\ a_{21}&a_{22}&\cdots & a_{2k}\\ \cdots & \cdots & \cdots & \cdots\\ a_{k1}&a_{k2}&\cdots & a_{kk}\\ \end{pmatrix} Ak​= ​a11​a21​⋯ak1​​a12​a22​⋯ak2​​⋯⋯⋯⋯​a1k​a2k​⋯akk​​ ​ Δ k = ∣ A k ∣ \mathbf{\Delta}_k=\left| \mathbf{A}_k\right| Δk​=∣Ak​∣

必要性： 设 x k = ( x 1 , x 2 , … , x k ) T ≠ 0 \mathbf{x}_k=(x_1,x_2,\dots,x_k)^T \neq 0 xk​=(x1​,x2​,…,xk​)T=0 x = ( x 1 , x 2 , … , x k , … , x n ) T \mathbf{x}= (x_1,x_2,\dots,x_k,\dots,x_n)^T x=(x1​,x2​,…,xk​,…,xn​)T,其中 x k + 1 = ⋯ = x n = 0 x_{k+1}=\cdots=x_{n}=0 xk+1​=⋯=xn​=0 A ≻ 0 ⇒ ∀ x ≠ 0 , x T A x = x k T A k x k > 0 ⇒ A k ≻ 0 \mathbf{A}\succ 0\Rightarrow \forall \mathbf{x}\neq \mathbf{0} , \mathbf{x}^T \mathbf{A} \mathbf{x}= \mathbf{x}_k^T \mathbf{A}_k \mathbf{x}_k>0\Rightarrow \mathbf{A}_k \succ 0 A≻0⇒∀x=0,xTAx=xkT​Ak​xk​>0⇒Ak​≻0 所以 A k ≻ 0 \mathbf{A}_k\succ 0 Ak​≻0 由行列式等于特征值的乘积，以及正定矩阵的特征值大于0，所以行列式也大于0 Δ k = ∣ A k ∣ > 0 \mathbf{\Delta}_k=\left|\mathbf{A}_k\right| >0 Δk​=∣Ak​∣>0

充分性： 方法1比较暴力，好懂， 方法2，感觉稍微花了点，但是也还好

方法1： 当 n = 1 n=1 n=1时，显然成立 假设 n = k − 1 n=k-1 n=k−1时成立 A = ( A n − 1 α α T a n n ) \mathbf{A}=\begin{pmatrix} \mathbf{A}_{n-1}& \mathbf{\alpha} \\ \mathbf{\alpha}^T & a_{nn}\\ \end{pmatrix} A=(An−1​αT​αann​​) A n − 1 \mathbf{A}_{n-1} An−1​是一个实对称矩阵，并且根据假设所有的顺序主子式大于0 所以一定合同于单位矩阵 即存在可逆 C \mathbf{C} C 使得 C T A n − 1 C = I n − 1 \mathbf{C}^T \mathbf{A}_{n-1}\mathbf{C} =\mathbf{I}_{n-1} CTAn−1​C=In−1​ 设 P = ( I n − 1 − A n − 1 − 1 α 0 T 1 ) \mathbf{P}=\begin{pmatrix} \mathbf{I}_{n-1} & -\mathbf{A}_{n-1}^{-1}\mathbf{\alpha}\\ \mathbf{0}^T& 1\\ \end{pmatrix} P=(In−1​0T​−An−1−1​α1​),显然可逆

P T A P = ( A n − 1 0 0 T a n n − α T A n − 1 − 1 α ) \mathbf{P}^T \mathbf{A}\mathbf{P}=\begin{pmatrix} \mathbf{A}_{n-1} &\mathbf{0}\\ \mathbf{0}^T& a_{nn}-\mathbf{\alpha}^T A_{n-1}^{-1} \mathbf{\alpha} \\ \end{pmatrix} PTAP=(An−1​0T​0ann​−αTAn−1−1​α​) 设 b = a n n − α T A n − 1 − 1 α b=a_{nn}-\mathbf{\alpha}^T \mathbf{A}_{n-1}^{-1} \mathbf{\alpha} b=ann​−αTAn−1−1​α ⇒ A ≃ ( A n − 1 0 0 T b ) \Rightarrow \mathbf{A} \simeq \begin{pmatrix} \mathbf{A}_{n-1} &\mathbf{0}\\ \mathbf{0}^T& b \\ \end{pmatrix} ⇒A≃(An−1​0T​0b​) 同取行列式 ∣ A ∣ = ∣ P T ∣ ∣ A ∣ ∣ P ∣ = ∣ P T A P ∣ = ∣ A n − 1 ∣ b > 0 ⇒ b > 0 \left|\mathbf{A}\right|=\left|\mathbf{P}^T\right|\left|\mathbf{A}\right|\left|\mathbf{P}\right|=\left|\mathbf{P}^T \mathbf{A}\mathbf{P}\right| =\left| \mathbf{A}_{n-1}\right| b>0 \Rightarrow b>0 ∣A∣= ​PT ​∣A∣∣P∣= ​PTAP ​=∣An−1​∣b>0⇒b>0 ( C 0 0 T 1 ) T ( A n − 1 0 0 T b ) ( C 0 0 T 1 ) = ( I n − 1 0 0 T b ) \begin{pmatrix} \mathbf{C}& \mathbf{0}\\ \mathbf{0}^T & 1 \end{pmatrix}^T \begin{pmatrix} \mathbf{A}_{n-1} &\mathbf{0}\\ \mathbf{0}^T& b \\ \end{pmatrix} \begin{pmatrix} \mathbf{C}& \mathbf{0}\\ \mathbf{0}^T & 1 \end{pmatrix}=\begin{pmatrix} \mathbf{I}_{n-1} &\mathbf{0}\\ \mathbf{0}^T& b \\ \end{pmatrix} (C0T​01​)T(An−1​0T​0b​)(C0T​01​)=(In−1​0T​0b​) ( A n − 1 0 0 T b ) ≃ ( I n − 1 0 0 T b ) \begin{pmatrix} \mathbf{A}_{n-1} &\mathbf{0}\\ \mathbf{0}^T& b \\ \end{pmatrix} \simeq \begin{pmatrix} \mathbf{I}_{n-1} &\mathbf{0}\\ \mathbf{0}^T& b \\ \end{pmatrix} (An−1​0T​0b​)≃(In−1​0T​0b​) 由合同传递性 A \mathbf{A} A的正惯性指数为 n n n， A ≻ 0 \mathbf{A}\succ 0 A≻0

方法2：

当 n = 1 n=1 n=1时，显然成立 假设 n = k − 1 n=k-1 n=k−1时成立 当 n = k n=k n=k时 f ( x 1 , x 2 , … , x k ) = ∑ i = 1 k ∑ j = 1 k a i j x i x j f(x_1,x_2,\dots,x_k)=\sum_{i=1}^{k}\sum_{j=1}^{k}a_{ij}x_{i}x_{j} f(x1​,x2​,…,xk​)=i=1∑k​j=1∑k​aij​xi​xj​ 由假设, a 11 > 0 a_{11}>0 a11​>0 f t = f ( x 1 , … , x t , 0 , 0 , … , 0 ) = ∑ i = 1 t ∑ j = 1 t a i j x i x j ( 1 < t ≤ k ) f_t=f(x_1,\dots , x_t,0,0,\dots,0)=\sum_{i=1}^{t}\sum_{j=1}^{t}a_{ij}x_{i}x_{j}(10 a11​>0 f t = 1 a 11 ( a 11 x 1 + a 12 x 2 + x 2 ⋯ + a 1 t x t ) 2 + ∑ i = 2 t ∑ j = 2 t b i j x i x j f_t=\frac{1}{a_{11}}(a_{11} x_{1} +a_{12}x_{2}+x_{2}\dots + a_{1t} x_{t})^2+\sum_{i=2}^{t}\sum_{j=2}^{t}b_{ij}x_{i}x_{j} ft​=a11​1​(a11​x1​+a12​x2​+x2​⋯+a1t​xt​)2+i=2∑t​j=2∑t​bij​xi​xj​ 其中 b i j = a i j − a 1 i a 1 j a 11 b_{ij}=a_{ij}-\frac{a_{1i} a_{1j}}{a_{11}} bij​=aij​−a11​a1i​a1j​​ a i j = a j i ⇒ b i j = b j i a_{ij}=a_{ji}\Rightarrow b_{ij}=b_{ji} aij​=aji​⇒bij​=bji​

当 t = k t=k t=k的时候 可以看出前半部分时大于等于0的 现在证明后半部分大于0 Δ t = ∣ a 11 a 12 ⋯ a 1 t a 21 a 22 ⋯ a 2 t ⋯ ⋯ ⋯ ⋯ a t 1 a t 2 ⋯ a t t ∣ = i = 2 , 3 , … , t r i − a i 1 a 11 r 1 ∣ a 11 a 12 ⋯ a 1 t 0 b 22 ⋯ b 2 t ⋯ ⋯ ⋯ ⋯ 0 b t 2 ⋯ b t t ∣ = a 11 ∣ b 22 ⋯ b 2 t ⋯ ⋯ ⋯ b t 2 ⋯ b t t ∣ > 0 \begin{aligned} \mathbf{\Delta}_t &=\left| \begin{array}{cccc} a_{11}&a_{12}&\cdots & a_{1t}\\ a_{21}&a_{22}&\cdots & a_{2t}\\ \cdots & \cdots & \cdots & \cdots\\ a_{t1}&a_{t2}&\cdots & a_{tt}\\ \end{array}\right|\\ &\xlongequal[i=2,3,\dots,t]{r_{i} -\frac{a_{i1}}{a_{11}}r_1} \left| \begin{array}{cccc} a_{11}&a_{12}&\cdots & a_{1t}\\ 0&b_{22}&\cdots & b_{2t}\\ \cdots & \cdots & \cdots & \cdots\\ 0&b{t2}&\cdots & b_{tt}\\ \end{array}\right|\\ &=a_{11}\left| \begin{array}{cccc} b_{22}&\cdots & b_{2t}\\ \cdots & \cdots & \cdots\\ b_{t2}&\cdots & b_{tt}\\ \end{array}\right|\\ &>0 \end{aligned} Δt​​= ​a11​a21​⋯at1​​a12​a22​⋯at2​​⋯⋯⋯⋯​a1t​a2t​⋯att​​ ​ri​−a11​ai1​​r1​ i=2,3,…,t​ ​a11​0⋯0​a12​b22​⋯bt2​⋯⋯⋯⋯​a1t​b2t​⋯btt​​ ​=a11​ ​b22​⋯bt2​​⋯⋯⋯​b2t​⋯btt​​ ​>0​ ⇒ ∣ b 22 ⋯ b 2 t ⋯ ⋯ ⋯ b k 2 ⋯ b t t ∣ > 0 \Rightarrow \left| \begin{array}{cccc} b_{22}&\cdots & b_{2t}\\ \cdots & \cdots & \cdots\\ b_{k2}&\cdots & b_{tt}\\ \end{array}\right|>0 ⇒ ​b22​⋯bk2​​⋯⋯⋯​b2t​⋯btt​​ ​>0 显然对于所有的 t = 2 , 3 , … , k t=2,3,\dots,k t=2,3,…,k都成立 设 B = ( b 22 ⋯ b 2 t ⋯ ⋯ ⋯ b t 2 ⋯ b t t ) \mathbf{B}=\begin{pmatrix} b_{22}&\cdots & b_{2t}\\ \cdots & \cdots & \cdots\\ b_{t2}&\cdots & b_{tt}\\ \end{pmatrix} B= ​b22​⋯bt2​​⋯⋯⋯​b2t​⋯btt​​ ​ 因为 B \mathbf{B} B是一个 k − 1 k-1 k−1阶实对称矩阵, B \mathbf{B} B所有顺序主子式大于0 由假设 B ≻ 0 \mathbf{B}\succ 0 B≻0 所以 A ≻ 0 \mathbf{A}\succ 0 A≻0 由数学归纳法，成立