対称行列の小行列式

(i)$\implies$(ii)

$A\geq 0$より$A((i_{1},\ldots,i_{k})) \geq 0$を得る．したがって，対称行列$A((i_{1},\ldots,i_{k}))$の固有値は総て非負なので，
$$ |A((i_{1},\ldots,i_{k}))| \geq 0$$
が成り立つ．

(ii)$\implies$(i)

(cf. MathSE )

$t>0$とし，
$$ A_{t} \coloneqq tE_{n}+A$$
とおく．このとき，
$$ A^{(k)} \coloneqq A((1,\ldots,k))$$
の固有多項式を
$$ f_{A^{(k)}}(x) = x^{k} + c_{1}x^{k-1} + c_{2}x^{k-2} +\cdots+ c_{k}$$
とおくと，
$$ A_{t}^{(k)} \coloneqq (A_{t})^{(k)} = (A^{(k)})_{t} = tE_{k}+A^{(k)}$$
の行列式は
$$ |A_{t}^{(k)}| = (-1)^{k}f_{A^{(k)}}(-t) = t^{k} + (-1)c_{1}t^{k-1} + (-1)^{2}c_{2}t^{k-2} +\cdots+ (-1)^{k}c_{k}$$
で与えられる．いま，仮定より
$$ (-1)^{\ell}c_{\ell} \geq 0$$
であるから（cf. coeff命題2），
$$ |A_{t}^{(k)}| \geq t^{k} > 0$$
が成り立ち，したがって，satakep.163定理6より，$A_{t} > 0$となる：
$$ \forall x \in \mathbb{R}^{n}\smallsetminus\{0\},\ A_{t}[x] = \sum_{i=1}^{n} (a_{ii}+t)x_{i}^{2} + 2\sum_{i< j} a_{ij}x_{i}x_{j} >0.$$
よって，$t \downarrow 0$として，
$$ \forall x \in \mathbb{R}^{n},\ A[x] \geq 0$$
を得る．

$\Ker{A}$の基底$\beta_{r+1},\ldots,\beta_{n} \in \Ker A$を取る：
$$ A\beta_{j} = 0 \quad\leadsto\quad 0 = \beta_{j}^{\transpose}A^{\transpose} = \beta_{j}^{\transpose}A.$$
このとき，$1 \leq i_{1} <\cdots< i_{r} \leq n$であって
$$ \beta_{r+1},\ldots,\beta_{n},\epsilon_{i_{1}},\ldots,\epsilon_{i_{r}} \in \mathbb{R}^{n}$$
が基底となるようなものが存在する（cf. satakep.96）．そこで，
$$ P \coloneqq \begin{bmatrix} \epsilon_{i_{1}} & \cdots & \epsilon_{i_{r}} & \beta_{r+1} & \cdots & \beta_{n} \end{bmatrix} \in \mathrm{GL}_{n}(\mathbb{R})$$
とおくと，
$$ P^{\transpose}AP = \begin{bmatrix} \epsilon_{i_{1}}^{\transpose} \\ \vdots \\ \epsilon_{i_{r}}^{\transpose} \\ \beta_{r+1}^{\transpose} \\ \vdots \\ \beta_{n}^{\transpose} \end{bmatrix}\begin{bmatrix} A\epsilon_{i_{1}} & \cdots & A\epsilon_{i_{r}} & A\beta_{r+1} & \cdots & A\beta_{n} \end{bmatrix} = \begin{bmatrix} A((i_{1},\ldots,i_{r})) & O_{r,n-r} \\ O_{n-r,r} & O_{n-r,n-r} \end{bmatrix}$$
となるので，
$$ r = \rank{A} = \rank{P^{\transpose}AP} = \rank{A((i_{1},\ldots,i_{r}))} \quad\leadsto\quad |A((i_{1},\ldots,i_{r}))| \neq 0$$
が成り立つ（cf. satakep.106定理8）．

$A$の任意の$(i,j)$余因子$\Delta_{ij}=\Delta_{ji},\,i< j,\,$が$0$になることを示せばよい．そこで，
$$ A' \coloneqq \begin{bmatrix} & && a_{1i} & a_{1j} \\ &A((1,\ldots,\hat{i},\ldots,\hat{j},\ldots,m)) && \vdots & \vdots \\ && & a_{mi} & a_{mj} \\ a_{i1} & \stackrel{\hat{i}\quad\hat{j}}{\cdots\cdots} & a_{im} & a_{ii} & a_{ij} \\ a_{j1} & \cdots\cdots & a_{jm} & a_{ji} & a_{jj} \end{bmatrix}$$
に対してJacobiの公式（cf. satakep.67）を適用すると，
$$ |A((1,\ldots,\hat{i},\ldots,\hat{j},\ldots,m))|\cdot|A'| = |A((1,\ldots,\hat{i},\ldots,m))|\cdot|A((1,\ldots,\hat{j},\ldots,m))| - \Delta_{ij}^{2}$$
を得る．いま，仮定より左辺および右辺第1項は$0$なので，$\Delta_{ij}=0$となる．

置換行列$P_{\sigma} \in \mathrm{O}(n)$に対して
$$ (P_{\sigma}^{\transpose}AP_{\sigma})((1,\cdots,r)) = A((\sigma(1),\ldots,\sigma(r)))$$
が成り立つので（cf. perm例2，例3），nonzero-prin-minorより，
$$ |A^{(r)}| \neq 0$$
としてよい．

Case 1.

まづ，
$$ |A^{(1)}| \cdots |A^{(r-1)}| \neq 0$$
なる場合を考える．

1. 自明．
2. 自明．
3. 各$k \in \{0,\ldots,r\}$に対して，$n$次対称行列$\tilde{A}_{k} \in \mathrm{M}_{n}(\mathbb{R})$を次で定める：
   $$ \tilde{A}_{k}(i,j) \coloneqq \begin{dcases} A(i,j) = a_{ij} & k=0;\\[3pt] \begin{vmatrix} a_{11} & \cdots & a_{1k} & a_{1j} \\ \vdots && \vdots & \vdots \\ a_{k1} & \cdots & a_{kk} & a_{kj} \\ a_{i1} & \cdots & a_{ik} & a_{ij} \end{vmatrix} & 1 \leq k \leq r. \end{dcases}$$
   このとき，Jacobiの公式より
   $$ |A^{(k-1)}|\cdot\tilde{A}_{k}(i,j) = |A^{(k)}|\cdot\tilde{A}_{k-1}(i,j) - \tilde{A}_{k-1}(i,k)\tilde{A}_{k-1}(k,j)$$
   となるので，
   \begin{align} |A^{(k)}|\cdot\tilde{A}_{k-1}[x] - |A^{(k-1)}|\cdot\tilde{A}_{k}[x] &= \sum_{i,j} (|A^{(k)}|\cdot\tilde{A}_{k-1}(i,j) - |A^{k-1}|\cdot\tilde{A}_{k}(i,j))x_{i}x_{j} \\ &= \sum_{i,j} \tilde{A}_{k-1}(k,i)\tilde{A}_{k-1}(k,j)x_{i}x_{j} \\ &= \left(\sum_{i=1}^{n} \tilde{A}_{k-1}(k,i)x_{i}\right)^{2} \end{align}
   が成り立つ．よって，$\tilde{A}_{r}=O$と合わせて
   $$ A[x] = \sum_{k=1}^{r} \left(\frac{\tilde{A}_{k-1}[x]}{|A^{(k-1)}|} - \frac{\tilde{A}_{k}[x]}{|A^{(k)}|} \right) = \sum_{k=1}^{r} \frac{\left(\sum_{i=1}^{n} \tilde{A}_{k-1}(k,i)x_{i}\right)^{2}}{|A^{(k-1)}|\cdot|A^{(k)}|}$$
   を得る．ところで，
   $$ P \coloneqq \begin{bmatrix} \tilde{A}_{0}(1,1) & \cdots & \tilde{A}_{0}(1,r) & \tilde{A}_{0}(1,r+1) & \cdots & \tilde{A}_{0}(1,n) \\ \vdots && \vdots & \vdots && \vdots \\ \tilde{A}_{r-1}(r,1) & \cdots & \tilde{A}_{r-1}(r,r) & \tilde{A}_{r-1}(r,r+1) & \cdots & \tilde{A}_{r-1}(r,r) \\ 0 & \cdots & 0 & 1 && \\ \vdots && \vdots & & \ddots & \\ 0 & \cdots & 0 & &&1 \end{bmatrix} = \begin{bmatrix} |A^{(1)}| & \cdots & \tilde{A}_{0}(1,r) & \tilde{A}_{0}(1,r+1) & \cdots & \tilde{A}_{0}(1,n) \\ & \ddots & & \vdots && \vdots \\ && |A^{(r)}| & \tilde{A}_{r-1}(r,r+1) & \cdots & \tilde{A}_{r-1}(r,r) \\ 0 & \cdots & 0 & 1 && \\ \vdots && \vdots & & \ddots & \\ 0 & \cdots & 0 & &&1 \end{bmatrix} \in \mathrm{GL}_{n}(\mathbb{R})$$
   であるから，$x' \coloneqq Px$とおくと
   $$ A[x] = \sum_{k=1}^{r} \frac{{x'_{k}}^{2}}{|A^{(k-1)}|\cdot|A^{(k)}|}$$
   が成り立つ．したがって，Sylvesterの慣性法則より
   $$ \#\{k \in \{1,\ldots,r\} \mid |A^{(k-1)}|\cdot|A^{(k)}| < 0\} = \#\{\text{$A$の負固有値}\}$$
   を得る．

Case 2.

つぎに
$$ |A^{(1)}| \cdots |A^{(r-1)}| = 0$$
なる場合を考える．

1. 1. $1\leq i_{1}<\cdots< i_{r-1} \leq r$であって$|A^{(r)}((i_{1},\ldots,i_{r-1}))| \neq 0$なるものが存在するとき，適当な置換行列$P \in \mathrm{O}(r) < \mathrm{O}(n)$によって
      \begin{align} |(P^{-1}AP)^{(r-1)}| &= |A^{(r)}((i_{1},\ldots,i_{r-1}))| \neq 0;\\ |(P^{-1}AP)^{(r)}| &= |P^{-1}A^{(r)}P| \neq 0;\\ \end{align}
      となる．
   2. $|A^{(r)}|$の$r-1$次主小行列式が総て$0$であるとき，det-zeroより，$1 \leq i_{1} <\cdots< i_{r-2} \leq r$であって
      $$ |A^{(r)}((i_{1},\ldots,i_{r-2}))| \neq 0$$
      なるものが存在する．そこで，対応する置換行列を$P \in \mathrm{O}(r) < \mathrm{O}(n)$とおけば，
      \begin{align} |(P^{-1}AP)^{(r-2)}| &= |A^{(r)}((i_{1},\ldots,i_{r-2}))| \neq 0;\\ |(P^{-1}AP)^{(r-1)}| &= |A^{(r)}((i_{1},\ldots,i_{r-2},i_{r-1}))| = 0;\\ |(P^{-1}AP)^{(r)}| &= |P^{-1}A^{(r)}P| \neq 0;\\ \end{align}
      が成り立つ．
   3. [1]において
      $$ |(P^{-1}AP)^{(1)}|\cdots|(P^{-1}AP)^{(r-2)}| \neq 0$$
      となった場合，および[2]において
      $$ |(P^{-1}AP)^{(1)}|\cdots|(P^{-1}AP)^{(r-3)}| \neq 0$$
      となった場合は，これ以上することはない．それ以外の場合は，$A^{(r)}$を$(P^{-1}AP)^{(r-1)}$または$(P^{-1}AP)^{(r-2)}$に置き換えて，[1],[2]の手順を繰り返せばよい．
2. 前段より
   $$ |A^{(1)}| \cdots |A^{(r-1)}| = 0,\ (|A^{(k)}|,|A^{(k+1)}|) \neq (0,0)$$
   としてよい．このとき，$|A^{(k)}|=0$とすると，Case 1.(3)で見たように，
   \begin{align} 0 \neq |A^{(k-1)}|\cdot|A^{(k+1)}| &= |A^{(k-1)}|\cdot\tilde{A}_{k}(k+1,k+1) \\ &= |A^{(k)}|\cdot\tilde{A}_{k-1}(k+1,k+1) - \tilde{A}_{k-1}(k+1,k)\tilde{A}_{k-1}(k,k+1) \\ &= - \tilde{A}_{k-1}(k,k+1)^{2} \end{align}
   となるので，
   $$ |A^{(k-1)}|\cdot|A^{(k+1)}| < 0$$
   が成り立つ．
3. 1. $|A^{(k-1)}|\cdot|A^{(k)}| \neq 0$のとき，Case 1.で見たように，
      $$ \frac{\tilde{A}_{k-1}[x]}{|A^{(k-1)}|} - \frac{\tilde{A}_{k}[x]}{|A^{(k)}|} = \frac{\left(\sum_{i=1}^{n} \tilde{A}_{k-1}(k,i)x_{i}\right)^{2}}{|A^{(k-1)}|\cdot|A^{(k)}|}$$
      が成り立つ．
   2. $|A^{(k)}|=0$のとき，$k+2$次正方行列
      $$ \begin{bmatrix} a_{11} & \cdots & a_{1,k-1} & a_{1k} & a_{1,k+1} & a_{1j} \\ \vdots && \vdots & \vdots & \vdots & \vdots \\ a_{k-1,1} & \cdots & a_{k-1,k-1} & a_{k-1,k} & a_{k-1,k+1} & a_{k-1,j} \\ a_{k1} & \cdots & a_{k,k-1} & a_{kk} & a_{k,k+1} & a_{kj} \\ a_{k+1,1} & \cdots &a_{k+1,k-1} & a_{k+1,k} & a_{k+1,k+1} & a_{k+1,j} \\ a_{i1} & \cdots & a_{i,k-1} & a_{ik} & a_{i,k+1} & a_{ij} \end{bmatrix}$$
      に対してSylvesterの定理（cf. fujip.397）を適用すると，$\tilde{A}_{k-1}(k,k)=|A^{(k)}|=0$に注意して，
      \begin{align} |A^{(k-1)}|^{2}\tilde{A}_{k+1}(i,j) &= \begin{vmatrix} \tilde{A}_{k-1}(k,k) & \tilde{A}_{k-1}(k,k+1) & \tilde{A}_{k-1}(k,j) \\ \tilde{A}_{k-1}(k+1,k) & \tilde{A}_{k-1}(k+1,k+1) & \tilde{A}_{k-1}(k+1,j) \\ \tilde{A}_{k-1}(i,k) & \tilde{A}_{k-1}(i,k+1) & \tilde{A}_{k-1}(i,j) \end{vmatrix} \\ &= \tilde{A}_{k-1}(k+1,k)\tilde{A}_{k-1}(i,k+1)\tilde{A}_{k-1}(k,j) \\ &\phantom{=}\quad+\tilde{A}_{k-1}(i,k)\tilde{A}_{k-1}(k,k+1)\tilde{A}_{k-1}(k+1,j) \\ &\phantom{=}\quad- \tilde{A}_{k-1}(k,j)\tilde{A}_{k-1}(k+1,k+1)\tilde{A}_{k-1}(i,k) \\ &\phantom{=}\quad- \tilde{A}_{k-1}(i,j)\tilde{A}_{k-1}(k,k+1)\tilde{A}_{k-1}(k+1,k) \\ &= -\tilde{A}_{k-1}(k,k+1)^{2}\tilde{A}_{k-1}(i,j)-\tilde{A}_{k-1}(k+1,k+1)\tilde{A}_{k-1}(k,i)\tilde{A}_{k-1}(k,j) \\ &\phantom{=}\quad+ \tilde{A}_{k-1}(k,k+1)(\tilde{A}_{k-1}(k,i)\tilde{A}_{k-1}(k+1,j) + \tilde{A}_{k-1}(k,j)\tilde{A}_{k-1}(k+1,i)) \\ &= |A^{(k-1)}|\cdot|A^{(k+1)}|\cdot\tilde{A}_{k-1}(i,j)-\tilde{A}_{k-1}(k+1,k+1)\tilde{A}_{k-1}(k,i)\tilde{A}_{k-1}(k,j) \\ &\phantom{=}\quad+ \tilde{A}_{k-1}(k,k+1)(\tilde{A}_{k-1}(k,i)\tilde{A}_{k-1}(k+1,j) + \tilde{A}_{k-1}(k,j)\tilde{A}_{k-1}(k+1,i)) \\ \end{align}
      を得る．よって
      \begin{align} |A^{(k-1)}|\cdot|A^{(k+1)}|\cdot\tilde{A}_{k-1}[x] - |A^{(k-1)}|^{2}\cdot\tilde{A}_{k+1}[x] &= \sum_{i,j} \tilde{A}_{k-1}(k+1,k+1)\tilde{A}_{k-1}(k,i)\tilde{A}_{k-1}(k,j)x_{i}x_{j} - \sum_{i,j}\tilde{A}_{k-1}(k,k+1)(\tilde{A}_{k-1}(k,i)\tilde{A}_{k-1}(k+1,j) + \tilde{A}_{k-1}(k,j)\tilde{A}_{k-1}(k+1,i))x_{i}x_{j} \\ &= \sum_{i,j} \tilde{A}_{k-1}(k+1,k+1)\tilde{A}_{k-1}(k,i)\tilde{A}_{k-1}(k,j)x_{i}x_{j} - 2\sum_{i,j}\tilde{A}_{k-1}(k,k+1)\tilde{A}_{k-1}(k,i)\tilde{A}_{k-1}(k+1,j)x_{i}x_{j} \\ &= \tilde{A}_{k-1}(k+1,k+1)\left(\sum_{i=1}^{n} \tilde{A}_{k-1}(k,i)x_{i}\right)^{2} - 2\tilde{A}_{k-1}(k,k+1)\left(\sum_{i=1}^{n}\tilde{A}_{k-1}(k,i)x_{i}\right)\cdot\left(\sum_{j=1}^{n}\tilde{A}_{k-1}(k+1,j)x_{j}\right) \end{align}
      が成り立つ．ここで，
      $$ y \coloneqq \sum_{i=1}^{n} \tilde{A}_{k-1}(k,i)x_{i},\ z \coloneqq \sum_{j=1}^{n}\tilde{A}_{k-1}(k+1,j)x_{j}$$
      とおくと，
      1. $\tilde{A}_{k-1}(k+1,k+1) \neq 0$のとき，
         \begin{align} \frac{\tilde{A}_{k-1}[x]}{|A^{(k-1)}|} - \frac{\tilde{A}_{k+1}[x]}{|A^{(k+1)}|} &= \frac{\tilde{A}_{k-1}(k+1,k+1)y^{2}-2\tilde{A}_{k-1}(k,k+1)yz}{|A^{(k-1)}|^{2}\cdot|A^{(k+1)}|} \\ &= \frac{(\tilde{A}_{k-1}(k+1,k+1)y - \tilde{A}_{k-1}(k,k+1)z)^{2}}{\tilde{A}_{k-1}(k+1,k+1)\cdot|A^{(k-1)}|^{2}\cdot|A^{(k+1)}|} + \frac{z^{2}}{\tilde{A}_{k-1}(k+1,k+1)\cdot|A^{(k-1)}|} \end{align}
         となる．
      2. $\tilde{A}_{k-1}(k+1,k+1)=0$のとき，
         $$ \frac{\tilde{A}_{k-1}[x]}{|A^{(k-1)}|} - \frac{\tilde{A}_{k+1}[x]}{|A^{(k+1)}|} = \frac{-\tilde{A}_{k-1}(k,k+1)((y+z)^{2} - (y-z)^{2})}{2|A^{(k-1)}|^{2}\cdot|A^{(k+1)}|}$$
         となる．
   3. いま，
      \begin{align} \sum_{i=1}^{n} \tilde{A}_{k-1}(k,i)x_{i} &= \sum_{i=k}^{n} \tilde{A}_{k-1}(k,i)x_{i} = |A^{(k)}|\cdot x_{k} + \tilde{A}_{k-1}(k,k+1)x_{k+1} +\cdots+ \tilde{A}_{k-1}(k,n)x_{n}; \\ \sum_{j=1}^{n}\tilde{A}_{k-1}(k+1,j)x_{j} &= \sum_{j=k}^{n} \tilde{A}_{k-1}(k+1,j)x_{j} = |A^{(k-1)}|\cdot|A^{(k+1)}|\cdot x_{k} + \tilde{A}_{k-1}(k+1,k+1)x_{k+1} +\cdots+ \tilde{A}_{k-1}(k+1,n)x_{n} \end{align}
      より
      $$ \begin{dcases} \sum_{i=1}^{n} \tilde{A}_{k-1}(k,i)x_{i} = |A^{(k)}|\cdot x_{k} + \tilde{A}_{k-1}(k,k+1)x_{k+1} +\cdots+ \tilde{A}_{k-1}(k,n)x_{n} & |A^{(k-1)}|\cdot|A^{(k)}| \neq 0 \\ &\\ \begin{drcases} \tilde{A}_{k-1}(k+1,k+1)y - \tilde{A}_{k-1}(k,k+1)z = \tilde{A}_{k-1}(k,k+1)^{3}x_{k} + 0x_{k+1} +\cdots {} \\ \phantom{\tilde{A}_{k-1}(k+1,k+1)y - \tilde{A}_{k-1}(k,k+1)}z = |A^{(k-1)}|\cdot|A^{(k+1)}|\cdot x_{k} + \tilde{A}_{k-1}(k+1,k+1)x_{k+1} +\cdots {} \end{drcases} & |A^{(k)}|=0,\ \tilde{A}_{k-1}(k+1,k+1) \neq 0 \\ &\\ \begin{drcases} y+z = |A^{(k-1)}|\cdot|A^{(k+1)}|\cdot x_{k} + \tilde{A}_{k-1}(k,k+1)x_{k+1} +\cdots {}\\ y-z = |A^{(k-1)}|\cdot|A^{(k+1)}|\cdot x_{k} - \tilde{A}_{k-1}(k,k+1)x_{k+1} +\cdots {} \end{drcases} & |A^{(k)}|=0,\ \tilde{A}_{k-1}(k+1,k+1)=0 \end{dcases}$$
      となるので，適当な可逆行列$P \in \mathrm{GL}_{n}(\mathbb{R})$を用いて$x' \coloneqq Px$とおくと，
      \begin{align} A[x] &= \sum_{|A^{(k-1)}|\cdot|A^{(k)}|\neq 0} \left(\frac{\tilde{A}_{k-1}[x]}{|A^{(k-1)}|} - \frac{\tilde{A}_{k}[x]}{|A^{(k)}|}\right) + \sum_{\substack{|A^{(k)}|=0 \\ \tilde{A}_{k-1}(k+1,k+1) \neq 0}} \left(\frac{\tilde{A}_{k-1}[x]}{|A^{(k-1)}|} - \frac{\tilde{A}_{k+1}[x]}{|A^{(k+1)}|}\right) + \sum_{\substack{|A^{(k)}|=0 \\ \tilde{A}_{k-1}(k+1,k+1)=0}} \left(\frac{\tilde{A}_{k-1}[x]}{|A^{(k-1)}|} - \frac{\tilde{A}_{k+1}[x]}{|A^{(k+1)}|}\right) \\ &= \sum_{k=1}^{r} \alpha_{k}{x'_{k}}^{2} \end{align}
      と書ける．ただし，
      $$ \alpha_{k} \coloneqq \begin{dcases} \frac{1}{|A^{(k-1)}|\cdot|A^{(k)}|} & |A^{(k-1)}|\cdot|A^{(k)}|\neq 0 \\[3pt] \frac{1}{\tilde{A}_{k-1}(k+1,k+1)\cdot|A^{(k-1)}|^{2}\cdot|A^{(k+1)}|} & |A^{(k)}|=0,\ \tilde{A}_{k-1}(k+1,k+1) \neq 0 \\[3pt] \frac{1}{\tilde{A}_{k-2}(k,k)\cdot|A^{(k-2)}|} & |A^{(k-1)}|=0,\ \tilde{A}_{k-2}(k,k) \neq 0 \\[3pt] -\frac{\tilde{A}_{k-1}(k,k+1)}{2|A^{(k-1)}|^{2}\cdot|A^{(k+1)}|} & |A^{(k)}|=0,\ \tilde{A}_{k-1}(k+1,k+1) = 0 \\[3pt] \frac{\tilde{A}_{k-2}(k-1,k)}{2|A^{(k-2)}|^{2}\cdot|A^{(k)}|} & |A^{(k-1)}|=0,\ \tilde{A}_{k-2}(k,k) = 0 \\ \end{dcases}$$
      である．
   4. $|A^{(k)}|=0$のときの$x'_{k},x'_{k+1}$の係数について，その積
      $$ \alpha_{k}\alpha_{k+1} = \begin{dcases} \frac{1}{\tilde{A}_{k-1}(k+1,k+1)^{2}\cdot|A^{(k-1)}|^{2}} \cdot \frac{1}{|A^{(k-1)}|\cdot|A^{(k+1)}|} & \tilde{A}_{k-1}(k+1,k+1) \neq 0 \\[3pt] -\frac{\tilde{A}_{k-1}(k,k+1)^{2}}{4|A^{(k-1)}|^{4}\cdot|A^{(k+1)}|^{2}} & \tilde{A}_{k-1}(k+1,k+1) = 0 \end{dcases}$$
      はいづれの場合も負なので，Sylvesterの慣性法則より，
      $$ \#\{k\in\{1,\ldots,r\} \mid |A^{(k)}|=0 \lor |A^{(k-1)}|\cdot|A^{(k)}|<0\} = \#\{\text{$A$の負固有値}\}$$
      を得る．