Can one find the signature of a real symmetric matrix just from the signs of some minors? Background to the question:  We know that a real symmetric matrix has all the eigenvalues $>0 ( \ge 0)$ if and only if all the diagonal minors are $>0 (\ge 0)$.  Also, one can tell the number of $0$ eigenvalues from the rank ( = largest size of a diagonal non-zero minor). 
${\bf Question:}$ can we tell how many of the eigenvalues of a real symmetric matrix will be positive and how many negative in a not too complicated way? ( I am aware that it can be done in principle  with the Sturm sequence but I am hoping for something more direct). 
Any help would be appreciated!
 A: Let me state the Cauchy Interlacing Theorem:
Let $B$ be a symmetric $n\times n$ real matrix. Let $c$ be a dim. $n$. real vector. Define $$A=\left[
\begin{array}{c|c}
B & c \\ \hline
c^T & \delta 
\end{array}\right],$$ where $\delta \in \mathbb{R}$. Then the spectrum of $B$ interlaces $A$, which is to say: $$\alpha_1 \leq \beta_1 \leq \alpha_2 \leq ... \leq \alpha_n \leq \beta_n \leq \alpha_{n+1}, $$ where $\{\alpha_1\leq \alpha_2 \leq ... \leq \alpha_{n+1} \}$ and $\{ \beta_1 \leq \beta_2 \leq ... \leq \beta_n \}$ are the spectra of $A$ and $B$ respectively.
Assume we know $\det B\neq 0$ and $\det A \neq 0$. Say the signature of $B$ is $(k,n-k,0)$ (recall that the signature of a symmetric matrix is the ordered pair of the number of its positive, negative, and zero eigenvalues). That is to say: $$0<\beta_{n-k+1}\leq \beta_{n-k+2} \leq ... \leq \beta_n, $$
and $$\beta_{1}\leq \beta_{2} \leq ... \leq \beta_{n-k}<0. $$
Then from the Cauchy Interlacing Theorem, 
$$0<\alpha_{n-k+2}\leq \alpha_{n-k+3} \leq ... \leq \alpha_{n+1}, $$
and $$\alpha_{1}\leq \alpha_{2} \leq ... \leq \alpha_{n-k}<0. $$
It follows that the signature of $A$ is either $(k+1,n-k,0)$, in which case the determinant of $A$ is clearly of the same sign as the determinant of $B$, or it is $(k,n-k+1,0)$, in which case the determinant of $A$ is of a different sign than the determinant of $B$. 
An inductive application of this result yields thus this statement about the sequence of leading principle minors of a symmetric matrix: the number of negative eigenvalues of an invertible symmetric matrix whose leading principle minors are all non-zero is equal to the number of sign changes in the sequence of leading principle minors (here without loss of generality we are letting the first leading principle minor be positive). 
So... under mild conditions, one can find the signature of a real symmetric matrix just from the signs of some minors!
