If $S$ is a nonsingular symmetric matrix over a number field and $D_k$ is its principal minor of order $k$, is $\frac{D_k}{D_{k-1}} > 0$ always true?

Question

In Chapter II, Paragraph 4, Section 1 of

F. R. Gantmacher, The Theory of Matrices. Vol. 1 English translation by K. A. Hirsch. Chelsea Publishing Company. 1959. SBN 8284-0131-4,

the following notation is introduced:

"Let $A=\lVert a_{ik}\rVert_1^n$ be a given matrix of rank $r$. We introduce the following notation for the successive principal minors of the matrix $D_k = A\bigl({1\ 2 \ \dotsc\ k \atop 1\ 2 \ \dotsc\ k }\bigr)\qquad(k=1,2,\dotsc,r)$."

Then, [op. cit., p. 37] gives the following:

Corollary $3$: If $S= ||s_{ik}||_1^n$ is a symmetric matrix of rank $r$ and $$D_k= S\bigl( \begin{smallmatrix} 1 & 2 & \dots & k \\ 1 & 2 & \dots & k \end{smallmatrix} \bigr)\ne 0 \ \ \ \ \ \ \ (k= 1,2, \dots,r),$$

then $$S=BB'$$ where $B=||b_{ik}||_1^n$ is a lower triangular matrix in which $$b_{gk}= \Biggl\{\begin{matrix} \frac{1}{\sqrt{D_kD_{k-1}}} S\bigl( \begin{smallmatrix} 1 & 2 & \dots & k-1 & g \\ 1 & 2 & \dots & k-1 & k \end{smallmatrix} \bigr) & (g=k, k+1, \dots, n; k= 1,2, \dots, r), \\ 0 & (g=k, k+1, \dots, n; k= r+1, \dots, n). \\ \end{matrix}$$

Question. Why (if at all) is $D_kD_{k-1}$ always greater than $0$ in this situation?

Remark. The theorem to which the above Corollary is added in op. cit. is the following (cf. op. cit, p. 35):

Theorem 1: Every matrix $A=\lVert a_{ik}\rVert_1^n$ of rank $r$ in which the first $r$ successive principal minors are different from zero

$D_k = A\bigl({1\ 2 \ \dotsc\ k \atop 1\ 2 \ \dotsc\ k }\bigr)\neq 0\quad$ for $k=1,2,\dotsc,r$ $\hspace{135pt}$ (34)

can be represented in the form of a product of a lower triangular matrix $B$ and an upper triangular matrix $C$

$A = BC = \left\lVert\begin{matrix} b_{11} & 0 & \dotsc & 0 \\ b_{21} & b_{22} & \dotsc & 0 \\ & & {{}}\ .\ .\ .\ .\ .\ .\\ b_{n1} & b_{n2} & \dotsc & b_{nn} \end{matrix}\right\rVert\quad\left\lVert\begin{matrix} c_{11} & c_{12} & \dotsc & c_{1n} \\ 0 & c_{22} & \dotsc & c_{2n} \\ & & {{}}\ .\ .\ .\ .\ .\ .\\ 0 & 0 & \dotsc & c_{nn} \end{matrix}\right\rVert$. (35)

Here

$b_{11}c_{11}=D_1$, $b_{22}c_{22}=\frac{D_2}{D_1}$, $\dotsc$, $b_{rr}c_{rr}=\frac{D_r}{D_{r-1}}.$ $\hspace{120pt}$ (36)

The values of the first $r$ diagonal elements of $B$ and $C$ can be chosen arbitrarily subject to the conditions (36).

When the first $r$ diagonal elements of $B$ and $C$ are given, then the elements of the first $r$ columns of $B$ and of the first $r$ rows of $C$ are uniquely determined, and are given by the following formulas:

$b_{gk}=b_{kk}\frac{ A\bigl({1\ 2 \ \dotsc\ k-1 \ g\atop 1\ 2 \ \dotsc\ k-1 \ k }\bigr) }{ A\bigl({1\ 2 \ \dotsc\ k \atop 1\ 2 \ \dotsc \ k }\bigr) }$, $\quad$ $c_{kg}=c_{kk}\frac{ A\bigl({1\ 2 \ \dotsc\ k-1 \ k\atop 1\ 2 \ \dotsc\ k-1 \ g }\bigr) }{ A\bigl({1\ 2 \ \dotsc\ k \atop 1\ 2 \ \dotsc \ k }\bigr)} \hspace{120pt}$ (37) $(g=k,k+1,\dotsc,n; k=1,2,\dotsc,r)$.

If $r<n$ $(\lvert A \rvert = 0)$, then all the elements in the last $n-r$ columns of $B$ can be put equal to zero and all the elements of the last $n-r$ rows of $C$ can be chosen arbitrarily; or, conversely, the last $n-r$ rows of $C$ can be filled with zeros and the last $n-r$ columns of $B$ can be chosen arbitrarily.

P.S.: In the text, there is an inadvertent interchange between columns and rows of $B$ & $C$ in the last two paragraphs of theorem $1$, but I stated them correctly. The correction can be seen from the subscripts $g$ & $k$; since, in the formulas $37$, the "columns" of $B$ can go till $r$ while the "rows" of $C$ can go till $r$.

Answer 1

That seems to be a mistake. My guess is that the $\neq 0$ should have been $>0$ instead; then the result should hold.

(I'm posting this as an answer to prevent this question to be bumped to the home page).

Answer 2

An interpretation of the corollary might be:

If $S$ is over a number field $F$, $B$ can be over a number field $E$, and $F$ need not be the same as $E$.

Specifically, If $S$ is real valued, $B$ can be complex valued. Hence $\sqrt{D_kD_{k-1}}$ will be always defined; i.e. $D_kD_{k-1}$ need not be positive.